I am trying to figure out how to make a function that takes a string. Then it needs to return a string with each letter that appears in the function along with the number of times it appears in the string. For instance "eggs" should return e1g2s1.
function charRepString(word) {
var array = [];
var strCount = '';
var countArr = [];
// Need an Array with all the characters that appear in the String
for (var i = 0; i < word.length; i++) {
if (array.indexOf(word[i]) === false) {
array.push(word[i]);
}
}
// Need to iterate through the word and compare it with each char in the Array with characters and save the count of each char.
for (var j = 0; j < word.length; i++) {
for (var k = 0; k < array.length; k++){
var count = 0;
if (word[i] === array[k]){
count++;
}
countArr.push(count);
}
// Then I need to put the arrays into a string with each character before the number of times its repeated.
return strCount;
}
console.log(charRepString("taco")); //t1a1co1
console.log(charRepString("egg")); //e1g2
let str = prompt('type a string ') || 'taco'
function getcount(str) {
str = str.split('')
let obj = {}
for (i in str) {
let char = str[i]
let keys = Object.getOwnPropertyNames(obj)
if (keys.includes(char)) {
obj[char] += 1
} else {
obj[char] = 1
}
}
let result = ''
Object.getOwnPropertyNames(obj).forEach((prop) => {
result += prop + obj[prop]
})
return result
}
console.log(getcount(str))
If the order of the alphanumeric symbols matters
const str = "10zza";
const counted = [...[...str].reduce((m, s) => (
m.set(s, (m.get(s) || 0) + 1), m
), new Map())].flat().join("");
console.log(counted); // "1101z2a1"
Or also like (as suggested by Bravo):
const str = "10zza";
const counted = [...new Set([...str])].map((s) =>
`${s}${str.split(s).length-1}`
).join("");
console.log(counted); // "1101z2a1"
A more clear and verbose solution-
Let m be max number of symbols in charset
Time complexity- O(n log(m))
Space complexity- O(m)
function countFrequencies(str) {
const freqs = new Map()
for (const char of str) {
const prevFreq = freqs.get(char) || 0
freqs.set(char, prevFreq + 1)
}
return freqs
}
function getCountStr(str) {
const freqs = countFrequencies(str)
const isListed = new Set()
const resultArray = []
for (const char of str) {
if (isListed.has(char)) continue
resultArray.push(char)
resultArray.push(freqs.get(char))
isListed.add(char)
}
return resultArray.join("")
}
console.log(getCountStr("egg"))
console.log(getCountStr("taco"))
console.log(getCountStr("10za"))
Using Set constructor, first we will get the unique data.
function myfun(str){
let createSet = new Set(str);
let newArr = [...createSet].map(function(elem){
return `${elem}${str.split(elem).length-1}`
});
let newStr = newArr.join('');
console.log(newStr);
}
myfun('array');
I am trying to solve the Leet Code challenge 14. Longest Common Prefix:
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string "".
Example 1:
Input: strs = ["flower","flow","flight"]
Output: "fl"
Example 2:
Input: strs = ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.
Constraints:
1 <= strs.length <= 200
0 <= strs[i].length <= 200
strs[i] consists of only lower-case English letters.
My solution:
let strs = ["flower", "flow", "flight"];
var longestCommonPrefix = function (strs) {
for (let i = 0; i < strs.length; i++) {
for (let j = 0; j < strs[i].length; j++) {
// console.log(strs[i+2][j]);
if (strs[i][j] == strs[i + 1][j] && strs[i + 1][j] ==strs[i + 2][j]) {
return (strs[i][j]);
} else {
return "0";
}
}
}
};
console.log(longestCommonPrefix(strs));
Output: f
How can I iterate over every character and check if it is same and then go for next and if it fails then the longest common prefix will be returned?
As the longest common prefix must occur in every string of the array you can jus iterate over the length and check if all words have the same char at that index until you find a difference
function prefix(words){
// check border cases size 1 array and empty first word)
if (!words[0] || words.length == 1) return words[0] || "";
let i = 0;
// while all words have the same character at position i, increment i
while(words[0][i] && words.every(w => w[i] === words[0][i]))
i++;
// prefix is the substring from the beginning to the last successfully checked i
return words[0].substr(0, i);
}
console.log(1, prefix([]));
console.log(2, prefix([""]));
console.log(3, prefix(["abc"]));
console.log(4, prefix(["abcdefgh", "abcde", "abe"]));
console.log(5, prefix(["abc", "abc", "abc"]));
console.log(6, prefix(["abc", "abcde", "xyz"]));
Some of the issues:
Your inner loop will encounter a return on its first iteration. This means your loops will never repeat, and the return value will always be one character.
It is wrong to address strs[i+1] and strs[i+2] in your loop, as those indexes will go out of bounds (>= strs.length)
Instead of performing character by character comparison, you could use substring (prefix) comparison (in one operation): this may seem a waste, but as such comparison happens "below" JavaScript code, it is very fast (and as string size limit is 200 characters, this is fine).
The algorithm could start by selecting an existing string as prefix and then shorten it every time there is a string in the input that doesn't have it as prefix. At the end you will be left with the common prefix.
It is good to start with the shortest string as the initial prefix candidate, as the common prefix can certainly not be longer than that.
var longestCommonPrefix = function(strs) {
let prefix = strs.reduce((acc, str) => str.length < acc.length ? str : acc);
for (let str of strs) {
while (str.slice(0, prefix.length) != prefix) {
prefix = prefix.slice(0, -1);
}
}
return prefix;
};
let res = longestCommonPrefix(["flower","flow","flight"]);
console.log(res);
An approach based on sorting by word length, and for the shortest word, for exiting early, an entirely Array.every-based prefix-validation and -aggregation ...
function longestCommonPrefix(arr) {
const charList = [];
const [shortestWord, ...wordList] =
// sort shallow copy by item `length` first.
[...arr].sort((a, b) => a.length - b.length);
shortestWord
.split('')
.every((char, idx) => {
const isValidChar = wordList.every(word =>
word.charAt(idx) === char
);
if (isValidChar) {
charList.push(char);
}
return isValidChar;
});
return charList.join('');
}
console.log(
longestCommonPrefix(["flower","flow","flight"])
);
.as-console-wrapper { min-height: 100%!important; top: 0; }
not the best solution but this should work
function longestPrefix(strs){
if(strs.length <1){
return "";
}
const sharedPrefix=function(str1,str2){
let i=0;
for(;i<Math.min(str1.length,str2.length) /*todo optimize*/;++i){
if(str1[i] !== str2[i]){
break;
}
}
return str1.substr(0,i);
};
let curr = strs[0];
for(let i=1;i<strs.length;++i){
curr=sharedPrefix(curr,strs[i]);
if(curr.length < 1){
// no shared prefix
return "";
}
}
return curr;
}
this:
strs[i][j] == strs[i + 1][j] ==strs[i + 2][j]
makes no sense in JS... or at least, makes no sense in what you are doing... to do this you should use a && operator, like this:
strs[i][j] == strs[i + 1][j] && strs[i + 1][j] ==strs[i + 2][j]
Otherwise JS will evaluate the first condition, and then will evaluate the result of that operation (either true or false) with the third value
In addition to this, consider that you are looping with i over the array, and so i will also be str.length - 1 but in the condition you are referencing strs[i + 2][j] that will be in that case strs[str.length + 1][j] that in your case, makes no sense.
About the solution:
You should consider that the prefix is common to all the values in the array, so you can take in consideration one value, and just check if all the other are equals... the most obvious is the first one, and you will end up with something like this:
let strs = ["flower", "flow", "flight", "dix"];
function longestCommonPrefix (strs) {
// loop over the characters of the first element
for (let j = 0; j < strs[0].length; j++) {
// ignore the first elements since is obvious that is equal to itself
for (let i = 1; i < strs.length; i++) {
/* in case you have like
[
'banana',
'bana'
]
the longest prefix is the second element
*/
if(j >= strs[i].length){
return strs[i]
}
// different i-th element
if(strs[0][j] != strs[i][j]){
return strs[0].substr(0, j)
}
}
}
// all good, then the first element is common to all the other elements
return strs[0]
};
console.log(longestCommonPrefix(strs));
you can do it like this, it works fast enough ~ 110ms
function longestCommonPrefix(strs){
if (strs.length === 0) {
return ''
}
const first = strs[0];
let response = '';
let prefix = '';
for (let i = 0; i < first.length; i++) {
prefix += first[i];
let find = strs.filter(s => s.startsWith(prefix));
if (find.length === strs.length) {
response = prefix;
}
}
return response;
};
let strs = ["flower", "flow", "flight"];
var longestCommonPrefix = function (strs) {
for (let i = 0; i < strs.length; i++) {
for (let j = 0; j < strs[i].length; j++) {
console.log(strs[i+2][j]);
if (strs[i][j] == strs[i + 1][j] && strs[i][j] ==strs[i + 2][j]) {
return (strs[i][j]);
} else {
return "0";
}
}
}
};
console.log(longestCommonPrefix(strs));
This **return ** f
Increase the index while the letter is the same at that index for all words in the list. Then slice on it.
function prefix(words) {
if (words.length === 0) { return '' }
let index = 0;
while (allSameAtIndex(words, index)) {
index++;
}
return words[0].slice(0, index);
}
function allSameAtIndex(words, index) {
let last;
for (const word of words) {
if (last !== undefined && word[index] !== last[index]) {
return false;
}
last = word;
}
return true;
}
I assume you are here for Leetcode problem solution.
var longestCommonPrefix = function(strs) {
let arr = strs.concat().sort();
const a1 = arr[0];
const a2 = arr[arr.length -1];
const length = a1.length;
let i=0;
while(i<length && a1.charAt(i) == a2.charAt(i)) i++;
return a1.substring(0,i);
};
function prefixLen(s1, s2) {
let i = 0;
while (i <= s1.length && s1[i] === s2[i]) i++;
return i;
}
function commonPrefix(arr) {
let k = prefixLen(arr[0], arr[1]);
for (let i = 2; i < arr.length; i++) {
k = Math.min(k, prefixLen(arr[0], arr[i]));
}
return arr[0].slice(0, k);
}
console.log(commonPrefix(['pirate', 'pizza', 'pilates'])) // -> "pi"
var longestCommonPrefix = function(strs) {
let prefix = "";
for(let i = 0; i < strs[0].length; i++) {
for(let j = 1; j < strs.length; j++) {
if(strs[j][i] !== strs[0][i]) return prefix;
}
prefix = prefix + strs[0][i];
}
return prefix;
};
console.log(longestCommonPrefix);
It is as simple as one loop and compare each element of the strings
const longestPrefix = (strs) => {
[word1, word2, word3] = strs;
let prefix = [];
if(strs === null || strs.length <= 2 || strs.length > 3) return 'please
insert 3 elements'
for (let i=0; i < word1.length; i++){
if(word1[i] === word2[i] && word1[i] === word3[i]){
prefix.push(word1[i])
}
}
return prefix.join('')
}
I read in another answer: 'Increase the index while the letter is the same at that index for all words in the list. Then slice on it.'
that's how I came up with this:
const findPrefix = (strs) => {
let i = 0;
while (strs.every((item) => strs[0][i] === item[i])) {
i++;
}
return strs[0].slice(0, i);
};
console.log(findPrefix(["flo", "flow", "flomingo"]));
const findPrefix = (strs) => {
let broke = false;
return strs[0].split("").reduce(
(acc, curr, index) =>
broke || !strs.every((word) => word[index] === curr)
? (broke = true && acc)
: (acc += curr),
""
);
};
console.log(findPrefix(["flower", "flow", "flamingo"]));
Here is my solution, Today I had an interview and the dev asked me the same question, I think I failed because I got stuck hahaha kinda nervous when someone is watching me 😂, anyway I decided to figure it out after the interview is done and this is my answer (without google it I swear) and for those who don't feel comfortable with the common "for loop"
const arr = ["absence", "absolute", "absolutely", "absorb"]
function getPrefix(arr) {
if (arr.length === 0 || arr.some(s => !s)) return null //if it's an empty array or one of its values is null or '', return null
const first = arr[0].split("") // turns the first position of the array, into an array
const res = arr.map(w => {
// mapping the original array
const compare = w.split("") // every item of the array will be converted in another array of its characters
return first.map((l, idx) => (compare[idx] === l ? l : "")).join("") // loop through the "first" array and compare each character
})
const prefix = first.join("").startsWith(res[res.length - 1]) // compare if the first letter starts with the last item of the returned array
? res[res.length - 1] // if true, return the final response which is the prefix
: null // else, return null, which means there is no common prefix
console.log("prefix: ", prefix)
return prefix
}
getPrefix(arr)
let arr = ["flower", "flow", "flight"]
function checkPrefix(array) {
let index = []
for (let i = 0; i <= array[0].length; i++) {
if (check(array[0][i], i, array)) {
index.push(i)
} else {
break;
}
}
console.log(array[0].substring(index[0], index[index.length - 1] + 1));
}
const check = (str, index, stringArr) => {
debugger
let status = true
stringArr.map(ele => {
debugger
if (ele[index] != str) {
status = false
}
})
return status
}
checkPrefix(arr)
/**
* #param {string[]} strs
* #return {string}
*/
var longestCommonPrefix = function(strs) {
let compare = strs[0];
let str = "";
for (let i = 1; i < strs.length; i++) {
let j = 0;
while(compare[j] != undefined && strs[i][j] != undefined) {
if(strs[i][j] == compare[j]) {
str += strs[i][j];
}
else break;
j++;
}
compare = str;
str = "";
}
return compare;
};
Longest common prefix in Javascript (All test case accepted. Asked by many company interviews.)
var longestCommonPrefix = function (strs) {
let string = '';
if (strs.length > 1) {
for (let i = 0; i < strs[0].length; i++) {
let str = strs[0].charAt(i);
for (let s = 0; s < strs.length - 1; s++) {
if (!(strs[s + 1].charAt(i) && strs[s].charAt(i) && strs[s + 1].charAt(i) == strs[s].charAt(i))) {
str = '';
}
}
if (!str) {
break;
}
string += str;
}
return string;
} else {
return strs[0];
}
};
longestCommonPrefix(["flower","flow","flight"]);
Code to find longest prefix
var longestCommonPrefix = function(strs) {
let match = false;
let len = strs[0].length ;
let ans = "";
let prev_ans ="";
if(strs.length ==1){
return strs[0];
}
for (let i = 1; i < strs.length; i++){
if( strs[i-1].length > strs[i].length){
len = strs[i].length;
}
}
for (let i = 1; i < strs.length; i++){
for (let j = 0; j < len; j++){
if(strs[i-1].charAt(j) == strs[i].charAt(j)){
ans += strs[i-1].charAt(j);
match = true;
}
else{
break;
}
}
if(prev_ans != "" && prev_ans !=ans){
if(prev_ans.length > ans.length){
return ans;
}else{
return prev_ans;
}
}
prev_ans = ans;
ans = "";
if (match == false){
return "";
}
}
return prev_ans;
};
console.log(longestCommonPrefix(["flow","fly","flu"]));
My solution:
function longestCommonPrefix(...words) {
words.sort(); // shortest string will be first and the longest last
return (
words[0].split('') // converts shortest word to an array of chars
.map((char, idx) => words[words.length - 1][idx] === char ? char : '\0') // replaces non-matching chars with NULL char
.join('') // converts back to a string
.split('\0') // splits the string by NULL characters
.at(0) // returns the first part
);
}
Usage example:
longestCommonPrefix('abca', 'abda', 'abea'); // 'ab'
let testcases = [
["flower", "flow"], //should return "flow"
["flower", "flow", "flight"], //should return "fl"
["flower", "flow", "fight"], //should return "f"
["flower", "flow", "floor"], //should return "flo"
["flower"], //should return "flower"
]
var longestCommonPrefix = function(strs) {
for(var i=0; i<strs.length; i++){
for(var j=0; j<strs[i].length; j++){
if(strs[i][j] + strs[i][j+1] === strs[i+1][j]+strs[i+1][j+1]){
return strs[i][j]+strs[i][j+1];
}
else {
return "";
}
}
}
};
for (let strs of testcases)
console.log(longestCommonPrefix(strs));
I want to write a function that replaces all the vowels (a,e,i,o,u) in a string and keeps one separator ('') between consonants. For example:
Input: 'kkkeoiekkk'
Output: ['kkk', '', 'kkk']
My try:
function split(para) {
let regex = /[a|e|i|o|u]/g;
let arr = para.replace(regex, '-').split('-');
for (let i = 0; i < str.length; i++) {
if (arr[i] === '' && arr[i - 1] === '') {}
}
return arr
}
Right now for the input split('hheeooook') I get back [ 'h', '', '', '', 'k' ] instead of ['h', '', 'k']. Thanks for everyone reading or pointing out what is wrong with my function.
Here is a slight variation on your attempt. It is mainly the regex which needed change.
function split (para) {
let regex = /[aeiou]+/g;
let arr = para.replace(regex, ' ').trim().split(' ');
return arr
}
console.log(split('kkhkeoiektr'));
console.log(split('akkeoihhkoo'));
console.log(split('kkeoihhkoo'));
Here is a solution without regex:
function split(para) {
let res = [];
let tmp = '';
for (let i = 0; i < para.length; i++) {
if ('aeiou'.includes(para[i])) {
if (tmp.length) {
res.push(tmp);
res.push('');
tmp = '';
}
} else {
tmp += para[i];
}
}
if (tmp.length) {
res.push(tmp);
} else {
if (res.length > 0)
res.pop();
}
return res;
}
console.log(split('kkkeoiekkk'));
console.log(split('kkkeoiekkka'));
console.log(split('kkkeoiekkkak'));
You can iterate to split the string.
var dict = {"a":1,"e":1,"i":1,"o":1,"u":1};
const src = 'kkkeoiekkk0qwert';
var pre = "";
var result = [];
for(var i = 0; i < src.length; i++){
var char = src.charAt(i);
console.log(char);
console.log(pre);
if(dict[char]){
if(pre){
result.push(pre);
pre = "";
}
}else{
pre += char;
}
}
if(pre){
result.push(pre);
}
An alternative using reg.exec:
function split (para) {
let res = [""], reg = /[^aeiou]+/g, match;
while(match = reg.exec(para)) res.push(match[0], "");
return res.slice(1, res.length - 1);
}
console.log(split('kkhkeoiektr'));
console.log(split('akkeoihhkoo'));
console.log(split('kkeoihhkoo'));
console.log(split('kkkk'));
console.log(split('aeiou'));
How do I capture repeating letters in a word like abababa = 2matches( a and b is repeating )
I know how to do it when the letters are adjacent like so /(\w)\1+/ .
Thanks
Try to use this String expansion:
String.prototype.getRepeating = function() {
var length = this.length;
var found = '';
var repeating = '';
var index;
var letter;
for (index = 0; index < length; index++) {
letter = this.charAt(index);
if (-1 == found.indexOf(letter)) {
found = found.concat(letter);
} else {
if (-1 == repeating.indexOf(letter)) {
repeating = repeating.concat(letter);
}
}
}
return repeating;
}
The tests:
var tests = ['ab', 'aa', 'bb', 'abab', 'abb', 'aab', 'bab'];
for (var index in tests) {
console.log(tests[index], '=>', tests[index].getRepeating());
}
ab => (an empty string)
aa => a
bb => b
abab => ab
abb => b
aab => a
bab => b
If I understand correctly, you want to extract letters which appear more than once in a given word. If so, you simply need to iterate over the letters of the word, accumulate their occurrence, then filter out letters which only appear once.
var testString = "abababa";
var letters = countGroupByLetter(testString);
var result = filterMap(letters, function(v) {
return v > 1;
});
console.log(result);
function countGroupByLetter(testString) {
var result = {};
for (var ii = 0; ii < testString.length; ii++) {
var letter = testString.charAt(ii);
if (result[letter]) {
result[letter] ++;
} else {
result[letter] = 1;
}
}
return result;
}
function filterMap(map, filterFunction) {
var result = {};
for (var p in map) {
if (filterFunction(map[p])) {
result[p] = map[p];
}
}
return result;
}
Since you already know about back references, I suppose that you know you can find out if there is a letter repetition in a string, using /(\w).*\1/. Capturing all repetitions in one pass would not be possible though, you'd still need to execute a pattern repeatedly and accumulate the matched characters (for instance using /(\w)(?=.*\1)/g). That, however, would not be optimal.
var repeatingLetters = /(\w)(?=.*\1)/g;
var testString = "abababa";
var captures = null;
var result = {};
while ((captures = repeatingLetters.exec(testString)) != null) {
result[captures[1]] = true;
}
console.log(result);
I have an array containing the individual letters of a word and i want to search the array to return the index values of certain letters. However, if the word contains more a letter more than once (such as 'tree') the programme only returns one index value.
This is a sample of the code:
var chosenWord = "tree";
var individualLetters = chosenWord.split('');
var isLetterThere = individualLetters.indexOf(e);
console.log(isLetterThere);
this code will return the number '2', as that is the first instance of the letter 'e'. How would i get it to return 2 and 3 in the integer format, so that i could use them to replace items in another array using the .splice function.
indexOf takes a second parameter, as the position where it should start searching from.
So my approach would be:
function findLetterPositions(text, letter) {
var positions = new Array(),
pos = -1;
while ((pos = text.indexOf(letter, pos + 1)) != -1) {
positions.push(pos);
}
return positions;
}
console.log(findLetterPositions("Some eerie eels in every ensemble.", "e"));
http://jsfiddle.net/h2s7hk1r/
You could write a function like this:
function indexesOf(myWord, myLetter)
{
var indexes = new Array();
for(var i = 0; i < myWord.length; i++)
{
if(myWord.charAt(i) == myLetter)
{
indexes.push(i);
}
}
return indexes;
}
console.log(indexesOf("tree", "e"));
Loop through it as here:
var chosenWord = "tree";
var specifiedLetter = "e";
var individualLetters = chosenWord.split('');
var matches = [];
for(i = 0;i<individualLetters.length;i++){
if(individualLetters[i] == specifiedLetter)
matches[matches.length] = i;
}
console.log(matches);
An alternative using string methods.
var str = "thisisasimpleinput";
var cpy = str;
var indexes = [];
var n = -1;
for (var i = cpy.indexOf('i'); i > -1; i = cpy.indexOf('i')) {
n += i;
n++;
indexes.push(n);
cpy = cpy.slice(++i);
}
alert(indexes.toString());
var getLetterIndexes = function(word, letter) {
var indexes = [];
word.split("").forEach(function(el, i) {
el === letter && indexes.push(i);
});
return indexes;
};
getLetterIndexes("tree", "e"); // [2, 3]