Question: Create a function that takes a positive integer and returns the next bigger number that can be formed by rearranging its digits. For example:
12 ==> 21
513 ==> 531
2017 ==> 2071
//nextBigger(num: 12) // returns 21
//nextBigger(num: 513) // returns 531
//nextBigger(num: 2017) // returns 2071
I am trying to compare two Array and get correct array as answer. In do...while loop I am comparing the two array by increment second array by one.
function nextBigger(n){
let nStrg = n.toString();
let nArr = nStrg.split('');
function compareArr(Ar1,Ar2){
if(Ar2.length>Ar1.length){
return false;
}
for(let i=0; i<Ar1.length; i++){
let num = Ar1[i];
for(let j=0; j<Ar2.length; j++){
if(Ar2.lastIndexOf(num) !== -1){
Ar2.splice(Ar2.lastIndexOf(num), 1);
break;
}
else{
return false;
break;
}
}
}
return true;
}
let nextNumArr;
let m = n;
do{
let nextNum = m+1
m=nextNum
let nextNumStrg = nextNum.toString();
nextNumArr = nextNumStrg.split('')
console.log(compareArr(nArr, nextNumArr))
}
while(compareArr(nArr, nextNumArr) == false)
console.log(nextNumArr)
return parseInt(nextNumArr.join())
}
nextBigger(12);
This gives me empty array at the end;
[2,0,1,7].join() will give you '2,0,1,7', can use [2,0,1,7].join('') and get '2017'
All looks a bit complicated. How about:
const nextLarger = num => {
const numX = `${num}`.split(``).map(Number).reverse();
for (let i = 0; i < numX.length; i += 1) {
if ( numX[i] > numX[i + 1] ) {
numX.splice(i, 2, ...[numX[i+1], numX[i]]);
return +(numX.reverse().join(``));
}
}
return num;
};
const test = [...Array(100)].map(v => {
const someNr = Math.floor(10 + Math.random() * 100000);
const next = nextLarger(someNr);
return `${someNr} => ${
next === someNr ? `not possible` : next}`;
}).join('\n');
document.querySelector(`pre`).textContent = test;
<pre></pre>
See also
function nextbig(number) {
let nums = []
number.toString().split('').forEach((num) => {
nums.push(parseInt(num))
})
number = nums
n = number.length
for (var i = n - 1; i >= 0; i--) {
if (number[i] > number[i - 1])
break;
}
if (i == 1 && number[i] <= number[i - 1]) {
return 'No greater possible'
}
let x = number[i - 1];
let smallest = i;
for (let j = i + 1; j < n; j++) {
if (number[j] > x &&
number[j] < number[smallest])
smallest = j;
}
let temp = number[smallest];
number[smallest] = number[i - 1];
number[i - 1] = temp;
x = 0
for (let j = 0; j < i; j++)
x = x * 10 + number[j];
number = number.slice(i, number.length + 1);
number.sort()
for (let j = 0; j < n - i; j++)
x = x * 10 + number[j];
return x
}
console.log(nextbig(12))
console.log(nextbig(513))
console.log(nextbig(2017))
In compareArr you are deleting elements as you find them, which is correct to do, to make sure duplicates actually occur twice etc. However, that also deletes the elements from nextNumArr in the calling context, because the array is passed by reference and not by value. You need to do a manual copy of it, for example like this: compareArr(nArr, [...nextNumArr]).
I have used a different approach, first I search for all possible combinations of the given numbers with the permutator function. This function returns an array of possible numbers.
Then I sort this array of combinations and look for the index of the given number in the main function.
Once I have this index I return the position before the given number.
function nextbig(num){
function permutator(inputArr){
let result = [];
const permute = (arr, m = []) => {
if (arr.length === 0) {
result.push(m)
} else {
for (let i = 0; i < arr.length; i++) {
let curr = arr.slice();
let next = curr.splice(i, 1);
permute(curr.slice(), m.concat(next))
}
}
}
permute(inputArr)
return result;
}
let arrNums = num.toString().split('')
let combinations = permutator(arrNums).map(elem => parseInt(elem.join("")))
combinations.sort((a, b) => {
return b - a
})
let indexOfNum = combinations.findIndex(elem => elem === num)
let nextBigIndex = indexOfNum <= 0 ? 0 : indexOfNum - 1
return (combinations[nextBigIndex])
}
console.log(nextbig(12))
console.log(nextbig(517))
console.log(nextbig(2017))
I have been working on the two sum problem for the past few hours and can't seem to account for the case where there are only two numbers and their sum is the same as the first number doubled.
The result should be [0,1], but i'm getting [0,0].
let nums = [3,3];
let targetNum = 6;
function twoSum(nums, target) {
for (let i = 0; i < nums.length; i++) {
for (let b = i+1; b < nums.length; b++) {
if ((nums[i] + nums[b]) == target) {
return [nums.indexOf(nums[i]), nums.indexOf(nums[b])];
}
}
}
}
console.log(twoSum(nums, targetNum))
Two Sum
My approach uses a javascript object and completes the algorithm in O(n) time complexity.
const twoSum = (nums, target) => {
let hash = {}
for(i=0;i<nums.length;i++) {
if (hash[nums[i]]!==undefined) {
return [hash[nums[i]], i];
}
hash[target-nums[i]] = i;
}
};
console.log(twoSum([2,7,11,15], 9)); // example
This is not the way to solve the problem. Step through the array and save the complement of the target wrt the number in the array. This will also solve your corner case.
You should consider, indexOf(i) -> start from the first element, returns the index when match found! That is why in your code, nums.indexOf(nums[i]) and nums.indexOf(nums[b]) which is basically 3 in all two cases, it will return 0, cause 3 is the first element in array.
instead of doing this, return the index itself.
let nums = [3,3];
let targetNum = 6;
function twoSum(nums, target) {
for (let i = 0; i < nums.length; i++) {
for (let b = i+1; b < nums.length; b++) {
if ((nums[i] + nums[b]) == target) {
return i + "" +b;
}
}
}
}
console.log(twoSum(nums, targetNum))
I have an array of strings array_strings and I want to get the maximum length of x consecutive values from the list and return a string. I have tried with this code but, its not working.
var myArr = ['we', 'make', 'the', 'best', 'dishes', 'in', 'cooking', 'class'];
x = 2;
function myFunc(array_strings, x) {
// your code
let val = '',
y = 0;
array_strings.sort((a, b) => {
return a.length < b.length
});
if (array_strings.length > 0) {
while (x) {
val += array_strings[y];
y++;
x--;
}
}
return val;
}
console.log(myFunc(myArr, x))
// expected output 'cookingclass'
on sorting, I cannot satisfy the consecutive order. Where am I going wrong?
To keep it simple, first get all possible values for x consecutive items.
function longestConsecutives(arr, x) {
var consecutives = [];
for (var i=0; i<arr.length-x; i++)
consecutives.push(arr.slice(i, x).join(''));
and then search the one with maximum length amongst them:
var longest = consecutives[0];
for (var i=1; i<consecutives.length; i++)
if (consecutives[i].length > longest.length)
longest = consecutives[i];
return longest;
}
There is also a more efficient approach to find the result, have a look at this for inspiration.
I don't think there's any reason to sort for this. You can just move a sliding window of x length over the array and keep track of the longest sequence you've seen:
let array_strings = ['we', 'make', 'the', 'best', 'dishes', 'in', 'cooking', 'class']
function findMax(array_strings, x){
let max_index = 0, max_length = 0
for (let i= 0; i < array_strings.length - x + 1; i++){
let length = array_strings.slice(i, i+x).reduce((l, c) => l + c.length, 0)
if (length > max_length) {
max_index = i,
max_length = length
}
}
return array_strings.slice(max_index, max_index + x)
}
console.log(findMax(array_strings,2))
console.log(findMax(array_strings,3))
console.log(findMax(array_strings,4))
Rolling sum O(n) solution:
function findLongestConsec(arr, l) {
var curSum = array_strings.slice(0, l).map(ele => ele.length).reduce((a, b) => a+b);
var curMax = curSum;
var curMaxIndex = 0;
for (var i = 1; i < arr.length - l + 1; i++) {
curSum = curMax - arr[i-1].length + arr[i+l-1].length;
if (curSum > curMax) {
curMax = curSum;
curMaxIndex = i;
}
}
return arr.slice(curMaxIndex, curMaxIndex + l).join('');
}
The previous solutions all involve string concatenation. I don't think it's worth doing it every time. We just have to keep track of the starting index of the longest subarray.
Try this:
arr = ['we', 'make', 'the', 'best', 'dishes', 'in', 'cooking', 'class']
x = 2
var sorting = function (array_strings, items) {
// your code
let val = '', y = 0;
const sort = array_strings.sort((a, b) => a.length < b.length);
console.log({sort});
if (arr.length > 0){
while (items){
val += array_strings[y];
y++;
items--;
}
}
return val;
}
var s = sorting(arr,x);
console.log({s});
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);
/* */