i solving a challenge of Hacker rank . it about how anagrams. i give two string input and i have to find ...
Print a single integer denoting the number of characters you must delete to make the two strings anagrams of each other.
i have detected if it's anagrams or not and difference. but now can do the rest of it dont have any ideas.please help.
function main() {
var a = readLine();
var b = readLine();
var sum1 = 0 ;
var sum2 = 0 ;
for (var i= 0; i<= a.length-1; i++ )
{
sum1 = sum1 + a.charCodeAt(i);
}
console.log(sum1);
for (var i= 0; i<= b.length-1; i++ )
{
sum2 = sum2 + b.charCodeAt(i);
}
console.log(sum2);
if(sum1== sum2)
{
console.log("anagrams");
}
else
{
console.log("not anagram");
var diff = sum1 - sum2;
console.log(diff);
/// what to do now ?
}
}
I have solved this on hackerRank by using the object approach to count the frequency of letters if you are still looking for a reasonable solution.
function makeAnagrams(a,b){
let charA=buildcharMap(a)
let charB=buildcharMap(b)
let characters=[]
let counter=0
for(let char in charA){
if(charA[char] && charB[char]){
if(charA[char]===charB[char]){ //same frequency
continue;
}
else{
if(charA[char]>charB[char]){
counter=counter+ charA[char]-charB[char]
}
else{
counter=counter+ charB[char]-charA[char]
}
}
}
else{
counter=counter+charA[char]
}
}
for(let char in charB){
if(charB[char] && charA[char]===undefined){
counter=counter+charB[char]
}
}
return counter;
}
function buildcharMap(str){
var charMap={}
for(let char of str){
if (charMap[char]===undefined){
charMap[char]=1
}
else{
charMap[char]+=1
}
}
return charMap
}
console.log(makeAnagrams('cde','abc'))
I have solve this question on hackerearth, i took slightly different approach.
What i have done in this code is that i have check for all the characters and replace the character with "#" sign if both characters in string are same, then i count all the "#" signs in one of the strings that is modified, and then subtracted value of that count multiplied by two(because... they are similar in both strings) from total length of both strings.
Here is my code, hope you can convert it into javascript. ->
import java.util.Scanner;
public class Anagrams {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int Testcases = scanner.nextInt();
String[] str1 = new String[1000];
String[] str2 = new String[1000];
// Taking input
for (int i = 0; i < Testcases; i++) {
str1[i] = scanner.next().toLowerCase();
str2[i] = scanner.next().toLowerCase();
}
// For Total Length of both strings
int TotalLength[] = new int[Testcases];
// For counting "#" signs
int count[] = new int[Testcases];
// Loop through TestCases
for (int i = 0; i < Testcases; i++) {
TotalLength[i] = str1[i].length() + str2[i].length();
for (int j = 0; j < str1[i].length(); j++) {
for (int k = 0; k < str2[i].length(); k++) {
// If both characters are similar, then replace those characters with "#" signs
if (str1[i].charAt(j) == str2[i].charAt(k)) {
str1[i] = str1[i].replaceFirst(Character.toString(str1[i].charAt(j)),"#");
str2[i] = str2[i].replaceFirst(Character.toString(str2[i].charAt(k)),"#");
}
}
}
}
// Counting "#" signs from one string
for (int i = 0; i < Testcases; i++) {
count[i] = 0;
char[] c1 = str1[i].toCharArray();
for (char c: c1) {
if(c == '#'){
count[i]++;
}
}
}
// Output
for (int i = 0; i < Testcases; i++) {
System.out.println(TotalLength[i] - 2*count[i]);
}
scanner.close();
}
}
You are really just looking for the sum of the differences in later frequency. You can just count the frequencies into an 26 item array (the rules tell you there will only be lower case numbers). Then subtract each array from the other item by item but item and add the whole thing up. Seems like it only need about four lines of code:
function makeAnagram(a, b) {
const makeCountArray = (str) => [...str].reduce((a, c) => (a[c.charCodeAt(0) - 97]++, a), Array(26).fill(0))
let a1 = makeCountArray(a)
let a2 = makeCountArray(b)
return a1.reduce((a, c, i) => a + Math.abs(c - a2[i]), 0)
}
// test case expected: 30
let count = makeAnagram('fcrxzwscanmligyxyvym', 'jxwtrhvujlmrpdoqbisbwhmgpmeoke')
console.log(count)
Here's another approach using Map() of all the 26 alphabets in each string
function makeAnagram(a, b) {
let result = 0;
const alphabets = 'abcdefghijklmnopqrstuvwxyz'.split('');
const getCharCountMap = str => {
const strArray = str.split('');
let charMap = new Map();
alphabets.forEach(alphabet => charMap.set(alphabet, 0));
strArray.forEach(letter => charMap.set(letter, charMap.get(letter) + 1));
return charMap;
};
const aMap = getCharCountMap(a);
const bMap = getCharCountMap(b);
alphabets.forEach(alphabet => {
result = result + Math.abs(aMap.get(alphabet) - bMap.get(alphabet));
});
return result;
}
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