Develop Reference

How to write a custom function that has the same behavior of inbuilt Array Join in Node.js

function arr(a) {
let sum = "";
for (let i = 0; i < a.length; i++) {
sum += a[i] + "-";
}
console.log(sum);
}
arr(["Hello", "World", "!"]);
It's printing "-" after "!" as well

You should probably return your joined string from the function and then log the result.
Maybe also pass in a delimeter argument so you don't just have to use "-"; you can use anything. And that will mirror the join method a lot better.
Most importantly: check that the current index is less than the length of the array (-1). If it is add the delimeter. If it the equal to length - 1 don't add it.
function join(arr, delimeter) {
let str = '';
for (let i = 0; i < arr.length; i++) {
if (i < arr.length - 1) {
str += arr[i] + delimeter;
} else {
str += arr[i];
}
}
return str;
}
console.log(join(['Hello', 'World', '!'], '-'));
console.log(join(['We', 'Have', 'Popcorn', '!'], '🍿'));

I think you only need to add one condition for the last item. For example
function arr(a){
let sum= "";
for(let i=0; i<a.length; i++){
sum+= a[i] + ((i<a.length-1) ? "-" : "");
}
return sum
}
const result = arr(["Hello","World","!"]);
console.log(result)
Or it's neater with array reduce
const join = (arr, separator) => {
return arr.reduce((prev, current)=> prev + separator + current)
}
const rs =join(["Hello", "World", "!"], "-")
console.log(rs)
Same result
"Hello-World-!"

you can do this as follow
function arr(a,glue='') {
let sum = "";
let i=0;
for (i = 0; i < a.length-1; i++) {
sum += a[i] + glue;
}
sum+=a[i];
return sum;
}
console.log(arr(["Hello", "World", "!"],"-"));

Just use ternary operator to determine rendering last dash
function arr(a){
let sum= "";
for(let i=0; i<a.length; i++){
i === a.length-1 ? sum+=a[i] : sum+= a[i]+"-";
}
console.log(sum);
}
arr(["Hello","World","!"]);

function arr(a) {
let sum= "";
for (let i=0; i<a.length; i++) {
if (i==a.length-1) {
sum = sum + a[i];
} else {
sum = sum + a[i] + "-";
}
}
console.log(sum);
}
arr(["Hello","World","!"]);

how do I insert dashes between each two even digits of a specific number

I am trying to insert dash(-) between two even numbers.
Problem is dashes don't locate between two even numbers but at the end of the number.
here is the code
function insertHyphen(str) {
var strArr = str.split('');
var numArr = strArr.map(Number);
for(var i = 0; i < numArr.length; i++) {
if(numArr[i-1]%2===0 && numArr[i]%2===0) {
numArr.push('-');
}
}
return numArr.join('');
}
insertHyphen('112233445566'); // 112233445566---

Using regex:
var p = '112233445566';
var regex = /([02468])([02468])/g
console.log(p.replace(regex, '$1-$2'));
Try it online: https://tio.run/##y0osSyxOLsosKNHNy09J/f#/LLFIoUDBVkHd0NDIyNjYxMTU1MxM3ZqLCyRRlJqeWgGU1NeINjAyMbOI1YQz9NO5uJLz84rzc1L1cvLTNQr0ilILchKTUzXAmnQU1FUMdVWM1DU1rbn#/wcA

Use splice() instead of push(). See here Splice MDN
function insertHyphen(str) {
var strArr = str.split('');
var numArr = strArr.map(Number);
for(var i = 0; i < numArr.length; i++) {
if(numArr[i-1]%2===0 && numArr[i]%2===0) {
numArr.splice(i, 0, '-');
}
}
return numArr.join('');
}
console.log(insertHyphen('112233445566')); // 112-2334-4556-6

replace
numArr.push('-');
with
numArr.splice(i, 0, '-');

function insertHyphen(str) {
var strArr = str.split('');
var numArr = strArr.map(Number);
for(var i = 0; i < numArr.length; i++) {
if(numArr[i-1]%2===0 && numArr[i]%2===0) {
numArr.splice(i, 0, '-');
}
}
return numArr.join('');
}
console.log(insertHyphen('025468 '));

Convert number to String
keep first element of array into separate array i.e. newArray
Run the loop on string length
Check current and the next element is divisible by 2 or not. If yes, push '-' else push the next item.
Finally join the array elements of newArray with ''.
let a = 224578;
let str = a.toString()
var newArray=[arr[0]]
if(arr?.length > 0){
for (let i = 0; i < arr.length; i++) {
if(arr[i] % 2 === 0 && arr[i+1] % 2 === 0){
newArray.push('-', arr[i+1])
} else{
newArray.push(arr[i+1])
}
}
}
console.log('newArray', newArray.join(''))

Push inserts element at the end of the array. You can use splice to enter element at particular position.
arr.splice(index, 0, item);
Here index is the position where you want to insert item element. And 0 represents without deleting 0 elements.
function insertHyphen(str) {
var strArr = str.split('');
var numArr = strArr.map(Number);
for(var i = 0; i < numArr.length; i++) {
if(numArr[i-1]%2===0 && numArr[i]%2===0) {
numArr.splice(i, 0, '-');
}
}
return numArr.join('');
}
console.log(insertHyphen('112233445566'));

A simple way:
function insertHyphen(str) {
var strArr = str.split('');
var numArr = strArr.map(Number);
var result ="";
for(var i = 0; i < numArr.length; i++) {
if((numArr[i+1]!==undefined)&&(numArr[i]%2===0 && numArr[i+1]%2===0)) {
//numArr.push('-');
result = result + numArr[i] + "-";
}else{
result = result + numArr[i];
}
}
return result;
}
console.log(insertHyphen('112233445566'));

You could map the new string by check the value and predecessor.
function insertHyphen(string) {
return Array
.from(string, (v, i) => !i || v % 2 || string[i - 1] % 2 ? v : '-' + v)
.join('');
}
console.log(insertHyphen('22112233445566'));

Using push() method and single loop
let input = '346845';
let result = [input[0]],
len = input.length;
for (var i = 1; i < len; i++) {
if (input[i - 1] % 2 === 0 && input[i] % 2 === 0) {
result.push('-', input[i]);
} else {
result.push(input[i]);
}
}
console.log(result.join(''));

Pair of elements from a specified array whose sum equals a specific target number

I am in mid of my JavaScript session. Find this code in my coding exercise. I understand the logic but I didn't get this map[nums[x]] condition.
function twoSum(nums, target_num) {
var map = [];
var indexnum = [];
for (var x = 0; x < nums.length; x++)
{
if (map[nums[x]] != null)
// what they meant by map[nums[x]]
{
index = map[nums[x]];
indexnum[0] = index+1;
indexnum[1] = x+1;
break;
}
else
{
map[target_num - nums[x]] = x;
}
}
return indexnum;
}
console.log(twoSum([10,20,10,40,50,60,70],50));
I am trying to get the Pair of elements from a specified array whose sum equals a specific target number. I have written below code.
function arraypair(array,sum){
for (i = 0;i < array.length;i++) {
var first = array[i];
for (j = i + 1;j < array.length;j++) {
var second = array[j];
if ((first + second) == sum) {
alert('First: ' + first + ' Second ' + second + ' SUM ' + sum);
console.log('First: ' + first + ' Second ' + second);
}
}
}
}
var a = [2, 4, 3, 5, 6, -2, 4, 7, 8, 9];
arraypair(a,7);
Is there any optimized way than above two solutions? Can some one explain the first solution what exactly map[nums[x]] this condition points to?

Using HashMap approach using time complexity approx O(n),below is the following code:
let twoSum = (array, sum) => {
let hashMap = {},
results = []
for (let i = 0; i < array.length; i++){
if (hashMap[array[i]]){
results.push([hashMap[array[i]], array[i]])
}else{
hashMap[sum - array[i]] = array[i];
}
}
return results;
}
console.log(twoSum([10,20,10,40,50,60,70,30],50));
result:
{[10, 40],[20, 30]}
I think the code is self explanatory ,even if you want help to understand it,let me know.I will be happy enough to for its explanation.
Hope it helps..

that map value you're seeing is a lookup table and that twoSum method has implemented what's called Dynamic Programming
In Dynamic Programming, you store values of your computations which you can re-use later on to find the solution.
Lets investigate how it works to better understand it:
twoSum([10,20,40,50,60,70], 50)
//I removed one of the duplicate 10s to make the example simpler
In iteration 0:
value is 10. Our target number is 50. When I see the number 10 in index 0, I make a note that if I ever find a 40 (50 - 10 = 40) in this list, then I can find its pair in index 0.
So in our map, 40 points to 0.
In iteration 2:
value is 40. I look at map my map to see I previously found a pair for 40.
map[nums[x]] (which is the same as map[40]) will return 0.
That means I have a pair for 40 at index 0.
0 and 2 make a pair.
Does that make any sense now?
Unlike in your solution where you have 2 nested loops, you can store previously computed values. This will save you processing time, but waste more space in the memory (because the lookup table will be needing the memory)
Also since you're writing this in javascript, your map can be an object instead of an array. It'll also make debugging a lot easier ;)

function twoSum(arr, S) {
const sum = [];
for(let i = 0; i< arr.length; i++) {
for(let j = i+1; j < arr.length; j++) {
if(S == arr[i] + arr[j]) sum.push([arr[i],arr[j]])
}
}
return sum
}
Brute Force not best way to solve but it works.

Please try the below code. It will give you all the unique pairs whose sum will be equal to the targetSum. It performs the binary search so will be better in performance. The time complexity of this solution is O(NLogN)
((arr,targetSum) => {
if ((arr && arr.length === 0) || targetSum === undefined) {
return false;
} else {
for (let x = 0; x <=arr.length -1; x++) {
let partnerInPair = targetSum - arr[x];
let start = x+1;
let end = (arr.length) - 2;
while(start <= end) {
let mid = parseInt(((start + end)/2));
if (arr[mid] === partnerInPair) {
console.log(`Pairs are ${arr[x]} and ${arr[mid]} `);
break;
} else if(partnerInPair < arr[mid]) {
end = mid - 1;
} else if(partnerInPair > arr[mid]) {
start = mid + 1;
}
}
};
};
})([0,1,2,3,4,5,6,7,8,9], 10)

function twoSum(arr, target) {
let res = [];
let indexes = [];
for (let i = 0; i < arr.length - 1; i++) {
for (let j = i + 1; j < arr.length; j++) {
if (target === arr[i] + arr[j] && !indexes.includes(i) && !indexes.includes(j)) {
res.push([arr[i], arr[j]]);
indexes.push(i);
indexes.push(j);
}
}
}
return res;
}
console.log('Result - ',
twoSum([1,2,3,4,5,6,6,6,6,6,6,6,6,6,7,8,9,10], 12)
);
Brute force.

const findTwoNum = ((arr, value) => {
let result = [];
for(let i= 0; i< arr.length-1; i++) {
if(arr[i] > value) {
continue;
}
if(arr.includes(value-arr[i])) {
result.push(arr[i]);
result.push(value-arr[i]);
break;;
}
}
return result;
});
let arr = [20,10,40,50,60,70,30];
const value = 120;
console.log(findTwoNum(arr, value));
OUTPUT : Array [50, 70]

function twoSum(arr){
let constant = 17;
for(let i=0;i<arr.length-2;i++){
for(let j=i+1;j<arr.length;j++){
if(arr[i]+arr[j] === constant){
console.log(arr[i],arr[j]);
}
}
}
}

let myArr = [2, 4, 3, 5, 7, 8, 9];
function getPair(arr, targetNum) {
for (let i = 0; i < arr.length; i++) {
let cNum = arr[i]; //my current number
for (let j = i; j < arr.length; j++) {
if (cNum !== arr[j] && cNum + arr[j] === targetNum) {
let pair = {};
pair.key1 = cNum;
pair.key2 = arr[j];
console.log(pair);
}
}
}
}
getPair(myArr, 7)

let sumArray = (arr,target) => {
let ar = []
arr.forEach((element,index) => {
console.log(index);
arr.forEach((element2, index2) => {
if( (index2 > index) && (element + element2 == target)){
ar.push({element, element2})
}
});
});
return ar
}
console.log(sumArray([8, 7, 2, 5, 3, 1],10))

Use {} hash object for storing and fast lookups.
Use simple for loop so you can return as soon as you find the right combo; array methods like .forEach() have to finish iterating no matter what.
And make sure you handle edges cases like this: twoSum([1,2,3,4], 8)---that should return undefined, but if you don't check for !== i (see below), you would erroneously return [4,4]. Think about why that is...
function twoSum(nums, target) {
const lookup = {};
for (let i = 0; i < nums.length; i++) {
const n = nums[i];
if (lookup[n] === undefined) {//lookup n; seen it before?
lookup[n] = i; //no, so add to dictionary with index as value
}
//seen target - n before? if so, is it different than n?
if (lookup[target - n] !== undefined && lookup[target - n] !== i) {
return [target - n, n];//yep, so we return our answer!
}
}
return undefined;//didn't find anything, so return undefined
}

We can fix this with simple JS object as well.
const twoSum = (arr, num) => {
let obj = {};
let res = [];
arr.map(item => {
let com = num - item;
if (obj[com]) {
res.push([obj[com], item]);
} else {
obj[item] = item;
}
});
return res;
};
console.log(twoSum([2, 3, 2, 5, 4, 9, 6, 8, 8, 7], 10));
// Output: [ [ 4, 6 ], [ 2, 8 ], [ 2, 8 ], [ 3, 7 ] ]

Solution In Java
Solution 1
public static int[] twoNumberSum(int[] array, int targetSum) {
for(int i=0;i<array.length;i++){
int first=array[i];
for(int j=i+1;j<array.length;j++){
int second=array[j];
if(first+second==targetSum){
return new int[]{first,second};
}
}
}
return new int[0];
}
Solution 2
public static int[] twoNumberSum(int[] array, int targetSum) {
Set<Integer> nums=new HashSet<Integer>();
for(int num:array){
int pmatch=targetSum-num;
if(nums.contains(pmatch)){
return new int[]{pmatch,num};
}else{
nums.add(num);
}
}
return new int[0];
}
Solution 3
public static int[] twoNumberSum(int[] array, int targetSum) {
Arrays.sort(array);
int left=0;
int right=array.length-1;
while(left<right){
int currentSum=array[left]+array[right];
if(currentSum==targetSum){
return new int[]{array[left],array[right]};
}else if(currentSum<targetSum){
left++;
}else if(currentSum>targetSum){
right--;
}
}
return new int[0];
}

function findPairOfNumbers(arr, targetSum) {
var low = 0, high = arr.length - 1, sum, result = [];
while(low < high) {
sum = arr[low] + arr[high];
if(sum < targetSum)
low++;
else if(sum > targetSum)
high--;
else if(sum === targetSum) {
result.push({val1: arr[low], val2: arr[high]});
high--;
}
}
return (result || false);
}
var pairs = findPairOfNumbers([1,2,3,4,4,5], 8);
if(pairs.length) {
console.log(pairs);
} else {
console.log("No pair of numbers found that sums to " + 8);
}

Simple Solution would be in javascript is:
var arr = [7,5,10,-5,9,14,45,77,5,3];
var arrLen = arr.length;
var sum = 15;
function findSumOfArrayInGiven (arr, arrLen, sum){
var left = 0;
var right = arrLen - 1;
// Sort Array in Ascending Order
arr = arr.sort(function(a, b) {
return a - b;
})
// Iterate Over
while(left < right){
if(arr[left] + arr[right] === sum){
return {
res : true,
matchNum: arr[left] + ' + ' + arr[right]
};
}else if(arr[left] + arr[right] < sum){
left++;
}else{
right--;
}
}
return 0;
}
var resp = findSumOfArrayInGiven (arr, arrLen, sum);
// Display Desired output
if(resp.res === true){
console.log('Matching Numbers are: ' + resp.matchNum +' = '+ sum);
}else{
console.log('There are no matching numbers of given sum');
}
Runtime test JSBin: https://jsbin.com/vuhitudebi/edit?js,console
Runtime test JSFiddle: https://jsfiddle.net/arbaazshaikh919/de0amjxt/4/

function sumOfTwo(array, sumNumber) {
for (i of array) {
for (j of array) {
if (i + j === sumNumber) {
console.log([i, j])
}
}
}
}
sumOfTwo([1, 2, 3], 4)

function twoSum(args , total) {
let obj = [];
let a = args.length;
for(let i = 0 ; i < a ; i++){
for(let j = 0; j < a ; j++){
if(args[i] + args[j] == total) {
obj.push([args[i] , args[j]])
}
}
}
console.log(obj)}
twoSum([10,20,10,40,50,60,70,30],60);
/* */

return the first non repeating character in a string in javascript

So I tried looking for this in the search but the closest I could come is a similar answer in several different languages, I would like to use Javascript to do it.
The problem is I have an arbitrary string that I would like to return the first non repeating character. EX: 'aba' -> would return b
'aabcbd' -> would return c.
This is what I have so far, just a simple for loop to start.
var someString = 'aabcbd';
var firstNonRepeatedCharacter = function(string) {
for(var i = 0; i < someString.length; i++){
}
};
http://jsfiddle.net/w7F87/
Not sure where to go from here

You can use the indexOf method to find the non repeating character. If you look for the character in the string, it will be the first one found, and you won't find another after it:
function firstNonRepeatedCharacter(string) {
for (var i = 0; i < string.length; i++) {
var c = string.charAt(i);
if (string.indexOf(c) == i && string.indexOf(c, i + 1) == -1) {
return c;
}
}
return null;
}
Demo: http://jsfiddle.net/Guffa/Se4dD/

If you're looking for the first occurrence of a letter that only occurs once, I would use another data structure to keep track of how many times each letter has been seen. This would let you do it with an O(n) rather than an O(n2) solution, except that n in this case is the larger of the difference between the smallest and largest character code or the length of the string and so not directly comparable.
Note: an earlier version of this used for-in - which in practice turns out to be incredibly slow. I've updated it to use the character codes as indexes to keep the look up as fast as possible. What we really need is a hash table but given the small values of N and the small, relative speed up, it's probably not worth it for this problem. In fact, you should prefer #Guffa's solution. I'm including mine only because I ended up learning a lot from it.
function firstNonRepeatedCharacter(string) {
var counts = {};
var i, minCode = 999999, maxCode = -1;
for (i = 0; i < string.length; ++i) {
var letter = string.charAt(i);
var letterCode = string.charCodeAt(i);
if (letterCode < minCode) {
minCode = letterCode;
}
if (letterCode > maxCode) {
maxCode = letterCode;
}
var count = counts[letterCode];
if (count) {
count.count = count.count + 1;
}
else {
counts[letterCode] = { letter: letter, count: 1, index: i };
}
}
var smallestIndex = string.length;
for (i = minCode; i <= maxCode; ++i) {
var count = counts[i];
if (count && count.count === 1 && count.index < smallestIndex) {
smallestIndex = count.index;
}
}
return smallestIndex < string.length ? string.charAt(smallestIndex) : '';
}
See fiddle at http://jsfiddle.net/b2dE4/
Also a (slightly different than the comments) performance test at http://jsperf.com/24793051/2

var firstNonRepeatedCharacter = function(string) {
var chars = string.split('');
for (var i = 0; i < string.length; i++) {
if (chars.filter(function(j) {
return j == string.charAt(i);
}).length == 1) return string.charAt(i);
}
};
So we create an array of all the characters, by splitting on anything.
Then, we loop through each character, and we filter the array we created, so we'll get an array of only those characters. If the length is ever 1, we know we have a non-repeated character.
Fiddle: http://jsfiddle.net/2FpZF/

Two further possibilities, using ECMA5 array methods. Will return undefined if none exist.
Javascript
function firstNonRepeatedCharacter(string) {
return string.split('').filter(function (character, index, obj) {
return obj.indexOf(character) === obj.lastIndexOf(character);
}).shift();
}
console.log(firstNonRepeatedCharacter('aabcbd'));
On jsFiddle
Or if you want a bit better performance, especially on longer strings.
Javascript
function firstNonRepeatedCharacter(string) {
var first;
string.split('').some(function (character, index, obj) {
if(obj.indexOf(character) === obj.lastIndexOf(character)) {
first = character;
return true;
}
return false;
});
return first;
}
console.log(firstNonRepeatedCharacter('aabcbd'));
On jsFiddle

I came accross this while facing similar problem. Let me add my 2 lines.
What I did is a similar to the Guffa's answer. But using both indexOf method and lastIndexOf.
My mehod:
function nonRepeated(str) {
for(let i = 0; i < str.length; i++) {
let j = str.charAt(i)
if (str.indexOf(j) == str.lastIndexOf(j)) {
return j;
}
}
return null;
}
nonRepeated("aabcbd"); //c
Simply, indexOf() gets first occurrence of a character & lastIndexOf() gets the last occurrence. So when the first occurrence is also == the last occurence, it means there's just one the character.

Here's a Solution using Regex to replace all repeating characters and then returning the first character.
function firstNonRepeat(str) {
// Sorting str so that all repeating characters will come together & replacing it with empty string and taking first character using substr.
var rsl = str.split('').sort().join('').replace(/(\w)\1+/g,'').substr(0,1);
if(rsl) return rsl;
else return 'All characters are repeated in ' + str;
}
console.log(firstNonRepeat('aaabcccdeeef'));
console.log(firstNonRepeat('aaacbdcee'));
console.log(firstNonRepeat('aabcbd'));

First of all, start your loop at 1, not 0. There is no point in checking the first character to see if its repeating, obviously it can't be.
Now, within your loop, you have someString[i] and someString[i - 1]. They are the current and previous characters.
if someString[i] === someString[i - 1] then the characters are repeating, if someString[i] !== someString[i - 1] then they are not repeating, so you return someString[i]
I won't write the whole thing out for you, but hopefully the thought process behind this will help

function FirstNotRepeatedChar(str) {
var arr = str.split('');
var result = '';
var ctr = 0;
for (var x = 0; x < arr.length; x++) {
ctr = 0;
for (var y = 0; y < arr.length; y++) {
if (arr[x] === arr[y]) {
ctr+= 1;
}
}
if (ctr < 2) {
result = arr[x];
break;
}
}
return result;
}
console.log(FirstNotRepeatedChar('asif shaik'));

Here's an O(n) solution with 2 ES6 Sets, one tracking all characters that have appeared and one tracking only chars that have appeared once. This solution takes advantage of the insertion order preserved by Set.
const firstNonRepeating = str => {
const set = new Set();
const finalSet = new Set();
str.split('').forEach(char => {
if (set.has(char)) finalSet.delete(char);
else {
set.add(char);
finalSet.add(char);
}
})
const iter = finalSet.values();
return iter.next().value;
}

let arr = [10, 5, 3, 4, 3, 5, 6];
outer:for(let i=0;i<arr.length;i++){
for(let j=0;j<arr.length;j++){
if(arr[i]===arr[j+1]){
console.log(arr[i]);
break outer;
}
}
}
//or else you may try this way...
function firstDuplicate(arr) {
let findFirst = new Set()
for (element of arr)
if (findFirst.has(element ))
return element
else
findFirst.add(element )
}

function firstUniqChar(str) {
let myMap = new Map();
for(let i = 0; i < str.length; i++) {
let char = str.charAt(i);
if(!myMap.has(char)) {
myMap.set(char, 0);
}
myMap.set(char, myMap.get(char) + 1 );
}
for(let [key, value] of myMap) {
if(value === 1) {
return key;
}
}
return null;
}
let result = firstUniqChar("caabbdccee");
console.log(result);
You can use Map Object and set key and value, where in value you store the count for that particular character, After that you can iterate over map and check where is value 1 and return that key.
Map Object remembers the original insertion order of the keys.

This solution should works with array with integers and string.
function firstNoneRepeating(list, map = new Map()) {
for (let item of list) {
if (map.has(item)) {
map.set(item, map.get(item) + 1);
} else {
map.set(item, 1);
}
}
for (let [key, value] of map.entries()) {
if (value === 1) {
return key;
}
}
}
console.log(firstNoneRepeating("aabcbd"));
console.log(firstNoneRepeating([5, 2, 3, 4, 2, 6, 7, 1, 2, 3]));

let str='aabcbd'
let ans=''
for (let i=0;i<str.length;i++){
if(str.indexOf(str.charAt(i))===str.lastIndexOf(str.charAt(i))){
ans+=str.charAt(i)
break
}
}
console.log(ans)

Fill an empty array with zeros, with same length as the string array, and tally up how many times they appear through the loop. Grab the first one in the tallied array with a value of 1.
function firstNotRepeatingCharacter(s) {
const array = s.split("");
let scores = new Array(array.length).fill(0);
for (let char of array) {
scores[array.indexOf(char)]++;
}
const singleChar = array[scores.indexOf(1)];
return singleChar ? singleChar : "_"
}

You can iterate through each character to find() the first letter that returns a single match(). This will result in the first non-repeated character in the given string:
const first_nonrepeated_character = string => [...string].find(e => string.match(new RegExp(e, 'g')).length === 1);
const string = 'aabcbd';
console.log(first_nonrepeated_character(string)); // c

Here is my solution which have time complexity of o(n)
function getfirstNonRepeatingCharacterInAString(params) {
let count = {};
for (let i = 0; i < params.length; i++) {
let count1 = 0;
if (!count[params.charAt(i)]) {
count[params.charAt(i)] = count1 + 1;
}
else {
count[params.charAt(i)] = count[params.charAt(i)] + 1;
}
}
for (let key in count) {
if (count[key] === 1) {
return key;
}
}
return null;
}
console.log(getfirstNonRepeatingCharacterInAString("GeeksfoGeeks"));

Here is my solution using forEach and convert the string into an array
function firstNotRepeatingCharacter(s) {
var strArr = s.split("");
var found = "_";
strArr.forEach(function(item, index) {
if (strArr.indexOf(item) == index && strArr.indexOf(item, index + 1) == -1) {
if (found === "_") {
found = item;
}
}
})
return found;
}
firstNotRepeatingCharacter("abacabad")

Here is another approach:
Everytime you find equal chars store it in an array and break out of the loop. If the char is not found in the array then you have your first nonRepeating char
function nonRepeatingChars(value) {
const memory = []
for (let i = 0; i < value.length; i++) {
for (let j = i + 1; j < value.length; j++) {
if (value[i] === value[j]) {
memory.push(value[j])
break;
}
}
if (!memory.some(x => x === value[i])) {
return value[i];
}
}
return "all chars have duplicates";
}
console.log('First non repeating char is:',nonRepeatingChars("esen"))
console.log('First non repeating char is:',nonRepeatingChars("esesn"))
console.log('First non repeating char is:',nonRepeatingChars("eseulsn"))
console.log('First non repeating char is:',nonRepeatingChars("esesnn"))

> var firstNonRepeatedCharacter = function (str){
> for(i=0;i<str.length;i++){
> if(str.indexOf(str.charAt(i)) === str.lastIndexOf(str.charAt(i))){
> console.log(str.charAt(i));
> break;
> } } }
>
> firstNonRepeatedCharacter ("areerak");
you can check below link
https://codepen.io/t3veni/pen/wvvxJzm

Easy way to solve this algorithm, very straight forward.
function firstNonRepeatChar(str){
let map = {};
for(let i=0; i<str.length; i++){
if(Object.keys(map).includes(str[i])){
map[str[i]]++
}
else{
map[str[i]] = 1;
}
}
for(let j=0; j< Object.values(map).length; j++){
if(Object.values(map)[j] == 1){
console.log(Object.keys(map)[j]);
return
}
if (j == Object.values(map).length-1 && Object.values(map)[j] != 1){
console.log('_');
return;
}
else{
continue;
}
}
}
nonRepeat("aaabbcccdeeef");

Here is one other solution just using array, using 26 unique character as length of array:
var firstUniqChar = (function(s) {
var arr = [];
var str = s.toLowerCase();
for(let c of str){
let index = c.charCodeAt(0) - "a".charCodeAt(0);
arr[index]? ++arr[index]: arr[index]=1;
}
for(let c of str){
let index = c.charCodeAt(0) - 97;
if(arr[index] == 1){
return c;
};
}
return -1;
}("aabcbd"));
console.log(firstUniqChar);

We can keep track of frequency of each character of the string in an object.
For example : for "aabcbd" we can store the frequency as
{ "a":2, "b":2, "c":1, "d":1 }
This will take O(n) time.
Then we can traverse over this object and find the first character with frequency 1, which will also take O(n) time. So, the time complexity for this approach will be O(n).
const firstNonRepeating=(str)=>{
const obj={};
str.split("").forEach(item=>{
obj[item]
? obj[item]++
: obj[item]=1;
});
const item = Object.keys(obj).find(key=> obj[key] === 1);
return item;
}
Note: I use ES6 Object.keys method which may not work in older
browsers.

//To find first non repeating letter
//It will check for both upper and lower case
//only use one String.indexOf()
var mystr="ohvhvtccggt";
var checkFirstNonRepeating=function(){
var ele=[];
for(var i=0;i<mystr.length;i++) {
var key=mystr.charAt(i);
if(!ele[key])
ele[key]=0;
ele[key]++;
//Just check for second occurance of character
//no need to use indexOf twice
if(mystr.indexOf(key,i+1)==-1 && ele[key]<2)
return mystr[i];
}
return "All repeating letters";
}
console.log(checkFirstNonRepeating());
/*
Input : "ohvhvtoccggt"
Output : All repeating letters
Input :"oohjtht"
Output :j
*/

I used object to keep track of characters count in a string then return the char that has the fa value of 1. Here is a demo:
function firstNotRepeatingCharacter(s) {
// initialize an empty object to store chars
let seen = {};
let letter = '';
// iterate over each char in a string
// if it is already there increase value by one
// else set the value to 1
for(let char of s){
if (seen[char]){
seen[char] +=1;
} else {
seen[char] = 1;
}
}
// iterate over the new constructed object
// if the value is 1 and the output variable is empty
// return the associated key to the value 1
// else return '_'
for(let v in seen){
while(seen[v] == 1 && letter === ''){
letter += v;
return letter;
}
}
return('_');
}
console.log(firstNotRepeatingCharacter("abacabad"));
console.log(firstNotRepeatingCharacter("cbc"));
console.log(firstNotRepeatingCharacter("bcccccccccccccyb"));
console.log(firstNotRepeatingCharacter("aaa"));

The most satisfactory and easy to understand answer is the following.
function firstNotRepeatingCharacter(s) {
const arr = s.split("");
for(let i = 0; i < arr.length; i++){
let chr = arr[i];
if( arr.indexOf(arr[i]) == arr.lastIndexOf(arr[i])){
return arr[i]
}
}
return "_"
}
Explanation: It loops through all the characters of a string from forward and backward and then compares the values. If the index of both forward and backward search is true then it returns that character.

let str = 'aabbcdd';
let val = str.split('').reduce((a, e)=>{ if(str.indexOf(e) == str.lastIndexOf(e)) {a = e }; return a})
console.log(val); // c

the implementation below has a good time complexity and it accounts for letters with different cases:
steps
must touch every character in the string to know if it's repeated or not
function firstNonRepeatingLetter(wordd) {
const chars = {}
let word = wordd.toLowerCase()
// go through chars
// store chars in hash with values of array storing index of char and true if only 1 encountered so far
for (let i = 0; i < word.length; i += 1) {
let char = word[i]
if (chars[char]) {
chars[char][0] = false
} else {
chars[char] = [true, i]
}
}
let output = ''
let index;
for (let key in chars) {
// return char with true and lowest index
if (chars[key][0]) {
index = index === undefined ? chars[key][1] : index
if (index >= chars[key][1]) {
output = key
}
}
}
return index === undefined ? '' : wordd[index]
}
console.log(firstNonRepeatingLetter('sTreSS')) //T```

The bellow solution is a kind of frequency counter pattern and it will run only one loop, so O(n) will be the time complexity.
function firstNotRepeatingCharacter(str) {
const obj = {};
for (let i = 0, L = str.length; i < L; i++) {
const char = str[i];
obj[char] = obj[char] ? obj[char] + 1 : 1;
}
for (let key of Object.keys(obj)) {
if (obj[key] == 1) {
return key;
}
}
return -1;
}

Here is another solution
function firstNotRepeatingCharacter(s) {
const obj = {};
for (let i of s) {
if(!obj[i]) {
obj[i] = 1;
} else if (obj[i]) {
obj[i] = +obj[i] + 1;
}
}
for (let [key, value] of Object.entries(obj)) {
if(value == 1) return key;
}
return "_"
}

Using below method can achieve first non repeated character
function main(str) {
str = String(str).toLowerCase();
let non_repeated_char = 'N/A';
for (let i = 0; i < str.length; i++) {
let currentChar = str[i];
let repeated_times = String(str).split('').filter(e => e == currentChar).length;
if (repeated_times === 1) {
non_repeated_char = currentChar;
break;
}
}
return non_repeated_char;
};
let Result = main("basketball");
console.log("The Non Repeated char is-->", Result);

Finding the most frequent character in a string javascript

Assuming I have the following string "355385". I need a simple JavaScript that can tell me that the most mentioned character is 5. Thank you in advance.
I tried with this one but no results.
var exp = '355385' ;
var exps =exp.split("");
var expCounts = { };
for (var i=0;i<exp.length;i++)
{expCounts["_" + exps[i]] = (expCounts["_" + exps[i]] || 0) + 1 ;
if (expCounts==3) exps=exps[i]; }; exps;

This will loop over every character in the string and keep track of each character's count and the character with the maximum count:
var exp = '3553853335' ;
var expCounts = {};
var maxKey = '';
for(var i = 0; i < exp.length; i++)
{
var key = exp[i];
if(!expCounts[key]){
expCounts[key] = 0;
}
expCounts[key]++;
if(maxKey == '' || expCounts[key] > expCounts[maxKey]){
maxKey = key;
}
}
console.debug(maxKey + ":" + expCounts[maxKey]);
Update:
Here is an ES6 version that will handle strings where multiple character have the same max count
function maxCount(input) {
const {max, ...counts} = (input || "").split("").reduce(
(a, c) => {
a[c] = a[c] ? a[c] + 1 : 1;
a.max = a.max < a[c] ? a[c] : a.max;
return a;
},
{ max: 0 }
);
return Object.entries(counts).filter(([k, v]) => v === max);
}
Example (please excuse the crude output):
maxCount('--aaaa1111--').join(' | ').replace(/,/g, ':');
outputs 1:4 | -:4 | a:4

var getMax = function (str) {
var max = 0,
maxChar = '';
str.split('').forEach(function(char){
if(str.split(char).length > max) {
max = str.split(char).length;
maxChar = char;
}
});
return maxChar;
};
logs
getMax('355385') //5;
getMax('35538533') //3;
in equal case it will return first number
getMax('3553') //3;

var string = "355385",
counter = {};
for (var i = 0, len = string.length; i < len; i += 1) {
counter[string[i]] = (counter[string[i]] || 0) + 1;
}
var biggest = -1, number;
for (var key in counter) {
if (counter[key] > biggest) {
biggest = counter[key];
number = key;
}
}
console.log(number);
# 5

var exp = '355385';
var findMostFrequent = function (string) {
var chars = {}, first = string.charAt(0);
chars[first] = 1;
var maxChar = first, maxCount = 1;
for (var i = 1; i < string.length; i++) {
var char = string.charAt(i);
if (chars[char]) {
chars[char]++;
} else {
chars[char] = 1;
}
if (chars[char] > maxCount) {
maxChar = char;
}
}
return maxChar;
};

Another Solution
function maxChar(str) {
const charMap = {};
let max = 0;
let maxChar = '';
for(let char of str){
if(charMap[char]){
charMap[char]++;
}else{
charMap[char] = 1;
}
}
for(let char in charMap){
if(charMap[char] > max){
max = charMap[char];
maxChar = char;
}
}
return maxChar;
}
Result:
maxChar('355385')
"5"

Another way to get the most frequent character in a string - sort frequency map into an array and then return the first (greatest) value from that array:
function highest (string) {
let array = Array.from(string);
let frequencyMap = {};
array.forEach((value, index) => {
if (!frequencyMap[value]) {
frequencyMap[value] = 0;
}
frequencyMap[value] += 1;
})
let frequencyArray = Object.entries(frequencyMap);
frequencyArray.sort((a, b) => {
if (a[1] < b[1]) {
return 1;
}
if (a[1] > b[1]) {
return -1;
}
return 0;
});
return(frequencyArray[0][0]);
}
console.log(highest("hello World"));
returns "l"

None of the answers above take into consideration that JavaScript internally uses UTF-16
const s = "😄😅😄😄😅😅😄😄😱😱😄";
function getMostFrequentChar(s) {
const len = s.length;
const freq = {};
let maxFreq = 0;
let maxChar;
for (let i = 0; i < len; ++i) {
const isPair = (s.charCodeAt(i) & 0xF800) == 0xD800;
const c = isPair ? s.substr(i++, 2) : s[i];
const f = (freq[c] || 0) + 1;
freq[c] = f;
if (f > maxFreq) {
maxFreq = f;
maxChar = c;
}
}
return {maxFreq, maxChar, freq}
}
console.log(getMostFrequentChar(s));
Note: the code above assumes the string is valid UTF-16. It's possible to construct a string that is not valid UTF-16 in which case maybe you could change isPair to
const isPair = len - i > 1 &&
s.charCodeAt(i ) & 0xF800) == 0xD800 &&
s.charCodeAt(i + 1) & 0xF800) == 0xD800;
But it's not clear what a character with an invalid UTF-16 value means.
It also won't handle more funky unicode
s = "👦🏿👦👦🏿👦👦🏻👦🏽👦🏾👦🏿"
There are many graphmemes that take multiple unicode code points
Also, splitting the string using split is SSSSSSLLLLLOOOOWWWW and a huge memory hog if the string is long.

Here is yet another answer to this question:
For this I have considered that the character can be of whatevert kind except a space
function findHighestFreqInString(str) {
if (!str) return null
let cleanedStr = str.replace(/\s/g, '') //assumes no spaces needed
if (cleanedStr.length === 0) return null
let strObj = {}
let topChar = ''
for (let val of cleanedStr) {
strObj[val] = (strObj[val] || 0) + 1
if (topChar === '' || strObj[val] >= strObj[topChar]) topChar = val
}
return topChar
}
Here is how you would use it:
findHighestFreqInString('my name is Someone') // returns: e
findHighestFreqInString('') // returns: Null
findHighestFreqInString(' ') // returns: Null

Here is:
let str = '355385';
function mostFrequentCharacter(str) {
let charactersArr = str.split(''),
bins = {};
charactersArr.map(el => bins[el] = (bins[el] || 0) + 1);
return Object.keys(bins).map(key => ({
name: key,
count: bins[key]
})).sort((a, b) => b.count - a.count)[0];
}

You can use the following solution to find the most frequent character in a string:
function getMostRepeatedCharacter(string) {
return string.split('').reduce((acc,char)=>{
let len = string.split(char).length - 1;
return len > acc[1] ? [char,len] : acc
},['',0])[0]
}
getMostRepeatedCharacter('wediuaududddd') // d

Want to share this ES6 functional approach. Please provide your input.
function maxChar(myStr) {
let charObj = {};
return [...myStr].reduce((_, char) => {
if (char in charObj) charObj[char]++;
else if (char !== " ") charObj[char] = 1;
return Object.keys(charObj).reduce((a, b) => {
return charObj[a] > charObj[b] ? a : b;
});
});
}

The simplest approach will be like this:
function maxChar(str) {
const charMap = {};
let max = 0;
let maxChar = '';
start by making an object of words and how many they repeated, to do that we have to loop through the string using for of and implementing the conditions:
for (let char of str) {
if (charMap[char]) {
charMap[char]++;
} else {
charMap[char] = 1;
}
}
and now loop through the object using for in
for (let char in charMap) {
if (charMap[char] > max) {
max = charMap[char];
maxChar = char;
}
}
return maxChar;
}

this is another (bizarre) way
It substitute the current character with blank for check how many times is present in the string making the difference of length with original pattern
var str = "355385";
var mostLength = 0;
var characterMostLength;
for(t = 0; t< 10; t++)
{
var res = str.length - str.replace(new RegExp(t, "g"), "").length;
if (res > mostLength){
characterMostLength = t;
mostLength = res;
}
}

function solution(N) {
var textToArr = N.split('');
var newObj = {};
var newArr = [];
textToArr.map((letter) => {
if(letter in newObj){
newObj[letter] = newObj[letter]+1;
} else {
if(letter !== ' '){
newObj = Object.assign(newObj, {[letter]: 1})
}
}
});
for(let i in newObj){
newArr.push({name: i, value: newObj[i]})
}
var res = newArr.sort((a,b) => b.value-a.value)[0];
return res.name+':'+res.value
}
solution("hello world");

this is a simple Idea that only includes one pass-through with a hashmap. The only thing this does not do is handle several max numbers. I really hope you enjoy my solution :) .
function maxChar(str) {
//Create the output and the hashmap
let m = {}, ans
//Loop through the str for each character
//Use reduce array helper because of the accumulator
str.split('').reduce((a, c) => {
//Increments Map at location c(character) unless it does not already exist
m[c] = m[c] + 1|| 1
//This checks to see if the current passthrough of m[c] is greater than or equal to the accumulator, if it is, set the answer equal to the current character. If it's not keep the ans the same.
ans = m[c] >= a ? c : ans
//Only increment the accumulator if Map at location c(character) is greater than the accumulator. Make sure to return it otherwise it won't increment.
return a = m[c] > a ? a + 1 : a
}, 1)
//Lastly return the answer
return ans
}

Simplest way to find maximum number of occurring character in string
var arr = "5255522322";
var freq:any = {};
var num;
for(let i=0;i<arr.length;i++) {
num = arr[i];
freq[num] = freq[num] >= 1 ? freq[num] + 1 : 1;
}
var sortable:any = [];
for(let i in freq)
{
sortable.push(i);
}
var max = freq[sortable[0]];
var data:any = "";
var value = sortable[0];
for(let i=0;i<sortable.length;i++) {
if(max > freq[sortable[i]]){
data = "key" + value + " " + "value" + max;
}else{
value = sortable[i]
max = freq[sortable[i]];
}
}
console.log(data);

function maxChara(string) {
charMap = {};
maxNum = 0;
maxChar = "";
string.toString().split("").forEach(item => {
if (charMap[item]) {
charMap[item]++;
} else {
charMap[item] = 1;
}
});
for (let char in charMap) {
if (charMap[char] > maxNum) {
maxNum = charMap[char];
maxChar = char;
}
}
return maxChar;
}
let result = maxChara(355385);
console.log(result);

Here str will the string that needs to be verified.
function maxCharacter(str){
let str1 = str; let reptCharsCount=0; let ele='';let maxCount=0;
let charArr = str1.split('');
for(let i=0; i< str1.length; i++){
reptCharsCount=0;
for(let j=0; j< str1.length; j++){
if(str1[i] === str1[j]) {
reptCharsCount++;
}
}
if(reptCharsCount > maxCount) {
ele = str1[i];
maxCount = reptCharsCount;
}
}
return ele;
}
input
maxCharacter('asdefdfdsdfseddssdfsdnknmwlqweeeeeeeesssssssssssseeee');
output
"s"

function freq(str) {
var freqObj = {};
str.forEach((item) => {
if (freqObj[item]) {
freqObj[item]++;
}
else {
freqObj[item] = 1;
}
});
return freqObj;
}
function findmaxstr(str) {
let max = 0,res,freqObj;
freqObj = freq(str.split(""));
for(let keys in freqObj){
if (freqObj[keys] > max) {
max = freqObj[keys];
res = keys;
}
}
console.log(res);
return res;
}
findmaxstr("javasdasdsssssscript");

const maxChar = (str) => {
let obj = {};
for (let char of str) {
(!obj[char]) ? obj[char] = 1: obj[char]++;
}
maxCharcount = Math.max(...Object.values(obj));
const key = Object.keys(obj).filter(key => obj[key] === maxCharcount);
console.log(`Most repeated character/characters in the given string "${str}" is/are given below which repeated ${maxCharcount} times`);
console.log(...key);
}
maxChar("355385");

Here is the code, where it also checks for lower and upperCase characters with the same max count and returns a Lower ASCII character as a Max.
function mostFrequent(text) {
let charObj={}
for(let char of text){
if(char!==' '){
if(charObj.hasOwnProperty(char)) charObj[char]=charObj[char]+1;
else charObj[char]= 1
}
}
let maxOccurance= Object.keys(charObj)[0], i=0;
for(let property in charObj){
if(i>0){
if(charObj[property]> charObj[maxOccurance])
maxOccurance= property
else if(charObj[property]=== charObj[maxOccurance])
{
if(property<maxOccurance)
maxOccurance=property
}
}
i++
}
return [maxOccurance, charObj[maxOccurance]]
}

let str = '355385';
let max = 0;
let char = '';
str.split('').forEach((item) => {
let current = str.split(item).length;
if (current > max) {
max = current;
char = item;
}
});
console.log(char + ' occurred ' + (max - 1) + ' times');

var exp = '35585' ;
var expCounts = { };
let maxChar = ''
let count = 0
for(let i = 0; i < exp.length; i++){
let char = exp[i]
expCounts[char] = expCounts[char] + 1 || 1
if(expCounts[char] > count){
maxChar = char
count = expCounts[char]
}
console.log(maxChar)
}

function checkNoofOccurenance(string) {
const arr = [...new Set(string.split(''))].sort();
const finalObj = {};
arr.forEach((item) => {
finalObj[item] = string.split(item).length - 1;
});
const item=Object.keys(finalObj).reduce((occ, toBeComapir)=>finalObj[occ]>finalObj[toBeComapir]?occ:toBeComapir)
return item;
}

Using Hasmaps we can find the most frequent char and occurrence all in O(N) time complexity. Below is the code. I have used one hasmap to save all the values and while i am doing it, i am also calculating the max occurrence and the max char.
var mostFreq = function(s) {
let myMap = new Map();
let temp;
let counter = 0;
let mostFrequentChar;
for(let i =0;i <s.length;i++){
if(myMap.has(s.charAt(i))){
temp = myMap.get(s.charAt(i));
temp = temp + 1;
myMap.delete(s.charAt(i));
myMap.set(s.charAt(i) , temp)
if(temp > counter){
counter = temp;
mostFrequentChar = s.charAt(i);
}
}else{
myMap.set(s.charAt(i), 1)
}
}
//if you want number of occerance of most frequent char = counter
//if you want list of each individual char and its occurrence = myMap
//if you just want the char that is most frequence = mostFrequentChar;
return mostFrequentChar;
};

If you want the count of the letter as well, You can do this
const { letter, count } = input.split("").reduce(
(acc, letter) => {
const count = input.split(letter).length - 1;
return count > acc.count
? { letter, count }
: { letter: acc.letter, count: acc.count };
},
{ letter: "", count: 0 }
);
Here We are splitting the string, applying a reduce to the result. The Reduce Counts how many instances of a character are there in a string, using input.split(letter).length - 1; And if the count is greater than the previous count, updates the accumulated value to be the current value

let string = "355385";
function printFirstRepeat(str){
let output= {};
for (let char of str) {
char = char.toLowerCase();
output[char] = ++output[char] || 1;
if(output[char] > 1) return char;
}
return "Not Found"
}
console.log(printFirstRepeat(string));

Algorithm: Find maximum occurring character in a string (time complex: O(N))
I'll provide my solution to this algo-problem by utilizing the most recent concepts of javascript
const getMaxCharacter = (str) => {
let max = 0;
let maxChar = '';
str.split('').forEach((char) => {
if (str.split(char).length > max) {
max = str.split(char).length - 1;
maxChar = char;
}
});
return `The max letter is : ${maxChar} and the max number of times it is seen is: ${max} times`;
};
Let's express an easy way of testing the function logic I wrote it:
const letter = 'Hello Student';
getMaxCharacter(letter);
In the function developed, I've used the concepts below:
Arrow Function
Anonymous Funciton
Declare property by using let/const
Template Literals
forEach(); (array helper) & split()

This is simple and optimized solution and it returns the first occurring char if there are chars equals in counts
function maxOccurance(str) {
let maxOccurringChar = "";
const charMap = {};
for (let index = 0; index < str.length; index++) {
const ele = str.charAt(index);
if (!charMap[ele]) {
charMap[ele] = {
startIndex: index,
value: 1
};
} else {
charMap[ele].value = charMap[ele].value + 1;
}
if (
!maxOccurringChar ||
charMap[maxOccurringChar].value < charMap[ele].value
) {
maxOccurringChar = ele;
} else if (
charMap[maxOccurringChar].value === charMap[ele].value &&
charMap[ele].startIndex < charMap[maxOccurringChar].startIndex
) {
maxOccurringChar = ele;
}
}
return maxOccurringChar;
}
console.log( maxOccurance("bacdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz")
);

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<p id = "myString">Hello World! I am Julio!</p>
<p id = "mRCharacter"></p>
<script>
var string = document.getElementById("myString").innerHTML;
var mRCharater = mostRepetedCharacter(string);
document.getElementById("mRCharacter").innerHTML = mRCharater;
console.log(mRCharater);
function mostRepetedCharacter(string){
var mRCharater = "";
var strLength = string.length;
var i = 0;
var count = 0;
var max = 0;
var rest = strLength - 1;
while (i < strLength){
var j = i + 1;
while (j <= rest){
if (string[i] === string[j]){
count++;
}
if (count > max){
max = count;
mRCharater = string[i];
}
j++;
}
i++;
count = 0;
}
return mRCharater;
}
</script>
</body>
</html>
enter code here