I sorted the elements and comparing the first and last string to check the common prefixes. It works for most of the cases, but not for the input ["dog","racecar","car"]. The expected output is "", but what I'm getting is "c" (The "r" in "car" and "r" in "racecar"). I can tell the code to remove the last char, but this will break the other cases such as ["car", "car", "car"]. Not sure what am I missing. Any insights would help me improve.
Thanks
var longestCommonPrefix = function(strs) {
let count=0
const sortedString = strs.sort()
const firstString = sortedString[0]
const lastString = sortedString[sortedString.length-1]
for(let i=0; i< firstString.length; i++) {
if(firstString.charAt(i) === lastString.charAt(i)) {
count++
}
}
console.log(firstString.substring(0, count))
};
longestCommonPrefix(
["dog","racecar","car"])
You need to break out of the loop as soon as a match is not found. Otherwise, for example, ra and ca match on the second index, the a - which is undesirable.
var longestCommonPrefix = function(strs) {
let count = 0
const sortedString = strs.sort()
const firstString = sortedString[0]
const lastString = sortedString[sortedString.length - 1]
for (let i = 0; i < firstString.length; i++) {
if (firstString.charAt(i) === lastString.charAt(i)) {
count++
} else {
break;
}
}
console.log(firstString.substring(0, count))
};
longestCommonPrefix(
["dog", "racecar", "car"])
or, refactored a bit
const longestCommonPrefix = (strs) => {
strs.sort();
const firstString = strs[0];
const lastString = strs[strs.length - 1];
let prefixSoFar = '';
for (let i = 0; i < firstString.length; i++) {
if (firstString[i] === lastString[i]) {
prefixSoFar += firstString[i];
} else {
return prefixSoFar;
}
}
return prefixSoFar;
};
console.log(longestCommonPrefix(["dog", "racecar", "car"]));
I'm trying to solve this using the .every method but it's not returning true and therefore it's not adding onto my string and I'm not sure why.
var longestCommonPrefix = function(arr) {
if (arr.length === 0) {
return undefined;
}
let result = '';
for (let i = 0; i < arr.length; i++) {
if (arr.every(x => arr[i].charAt(i) === x)) {
result += arr[i].charAt(i);
}
}
return result
}
console.log(longestCommonPrefix(["flower", "flow", "flight"])); //fl
You need to iterate over one string, not over the whole array: check if the first character of the string is present everywhere, then the second character, etc:
var longestCommonPrefix = function(arr) {
if (arr.length === 0) {
return undefined;
}
let result = '';
for (let i = 0; i < arr[0].length; i++) {
if (arr.every(x => x.charAt(i) === arr[0][i])) {
result += arr[i].charAt(i);
} else break;
}
return result;
}
console.log(longestCommonPrefix(["flower", "flow", "flight"])); //fl
Your use of Array.every is along the right lines. You want to check that every string in the array has the same character at position i. I think you got confused when you named the parameter x, when it is in fact a string :)
var longestCommonPrefix = function(words) {
if (words.length === 0) {
return "";
}
const letters = [];
const lengthOfShortestWord = Math.min(...words.map(word => word.length));
for (let i = 0; i < lengthOfShortestWord; i++) {
const char = words[0][i];
if (words.every(word => word[i] === char)) {
letters.push(char);
} else {
break;
}
}
return letters.join("");
}
console.log(longestCommonPrefix(["flower", "flow", "flight"])); //fl
Unless I am mistaken the longest prefix is never going to be greater than the smallest string in the array.
In this case "fl" is both the smallest string and the longest common prefix:
["flower", "fl", "flight"]
So start with finding the smallest string in arr:
let [sm] = [...arr].sort((a, b) => a.length - b.length);
Then check that all strings in arr start with sm:
arr.every(str => str.startsWith(sm));
If that isn't the case then shorten sm by one character:
sm = sm.slice(0, -1);
And keep going until you eventually find the longest prefix or sm becomes an empty string:
const prefix = arr => {
let [sm] = [...arr].sort((a, b) => a.length - b.length);
while (sm && !arr.every(str => str.startsWith(sm))) sm = sm.slice(0, -1);
return sm;
};
I have 2 arrays.
1: [a, ab, abc, abcde]
2: [a, ab, abc, abcde, abcdefe, axde]
in the first array, I used this code to get the longest line.
function longestChain(words) {
// Write your code here
var xintTOstring = "";
var result = 0;
for (var x = 0; x < words.length; x++){
xintTOstring = words[x].toString();
if (xintTOstring.length > result) {
result = xintTOstring.length;
}
}
return result;
}
but then in the second array, the longest is "axde". because the abcde in that array cannot be the longest because it has an equal like value.
I try this code but did not get the expected result. and also the longest line is the abcdefer.
question: how can I get the longest line and check if it is valued like equal in the string. I tried this code but did not get the right output.
function longestChain(words) {
// Write your code here
var xintTOstring = "";
var result = 0;
for (var x = 0; x < words.length; x++){
xintTOstring = words[x].toString();
if (!words[x].toString().inclcudes(xintTOstring)) {
if (xintTOstring.length > result) {
result = xintTOstring.length;
}
}
}
return result;
}
regards
function equalLike(word) {
// should the equality be checked within the array or in global stream?
}
function longestChain(words) {
return words.reduce((longest,word) => longest = longest.length > equalLike(word).length ?
longest : word,'');
}
the longest word acts as the accumulator.
If I understand correctly, each call to longest word should return the longest word not yet found. Go through each list, keep object of longest words, check against that object, and check substrings against keys
const longestWords = {};
const longestChain = function(words) {
let longestInList = "";
words.forEach(function(word) {
if (validLongestWord(word) && word.length > longestInList.length) {
longestInList = word;
}
});
longestWords[longestInList] = longestInList.length; //maybe handy for sorting later
return longestInList;
}
const validLongestWord = function(word) {
if(longestWords[word]) return false;
return !Object.keys(longestWords).some(key=>key.indexOf(word) >=0);
}
console.log(longestChain(["a", "ab", "abc", "abcde", "abcdefe", "axde"])); //abcdefe
console.log(longestChain(["a", "ab", "abc", "abcde", "abcdefe", "axde"])); //axde
console.log(longestChain(["a", "ab", "abc", "abcde", "abcdefe", "axde"])); //none
I believe this is the problem that the OP is trying to solve using JavaScript:
Longest Character Removal Chain
and
Interview Questions - String Chain
Anyone please feel welcome to edit this answer to provide a solution for the question asked.
var StackOverFlow;
(function(StackOverFlow) {
var LongestChain = (function() {
function LongestChain() {}
LongestChain.main = function(args) {
// Array of words
var words = ["a", "ab", "abc", "abcdefe", "axde"];
console.info(
"Longest Chain Length : " + LongestChain.longest_chain(words)
);
};
LongestChain.longest_chain = function(w) {
if (null == w || w.length < 1) {
return 0;
}
var maxChainLen = 0;
var words = w.slice(0).slice(0);
var wordToLongestChain = {};
for (var index7809 = 0; index7809 < w.length; index7809++) {
var word = w[index7809];
{
if (maxChainLen > word.length) {
continue;
}
var curChainLen =
LongestChain.find_chain_len(word, words, wordToLongestChain) + 1;
/* put */ wordToLongestChain[word] = curChainLen;
maxChainLen = Math.max(maxChainLen, curChainLen);
}
}
return maxChainLen;
};
LongestChain.find_chain_len = function(word, words, wordToLongestChain) {
var curChainLen = 0;
for (var i = 0; i < word.length; i++) {
var nextWord = word.substring(0, i) + word.substring(i + 1);
if (words.indexOf(nextWord) >= 0) {
if (wordToLongestChain.hasOwnProperty(nextWord)) {
curChainLen = Math.max(
curChainLen,
/* get */ (function(m, k) {
return m[k] ? m[k] : null;
})(wordToLongestChain, nextWord)
);
} else {
var nextWordChainLen = LongestChain.find_chain_len(
nextWord,
words,
wordToLongestChain
);
curChainLen = Math.max(curChainLen, nextWordChainLen + 1);
}
}
}
return curChainLen;
};
return LongestChain;
})();
StackOverFlow.LongestChain = LongestChain;
LongestChain["__class"] = "StackOverFlow.LongestChain";
})(StackOverFlow || (StackOverFlow = {}));
StackOverFlow.LongestChain.main(null);
This question already has answers here:
Palindrome check in Javascript
(45 answers)
Closed 4 years ago.
The question I have been given is this;
create a function that takes an array of words and returns an array containing only the palindromes.
A palindrome is a word that is spelled the same way backwards.
E.g. ['foo', 'racecar', 'pineapple', 'porcupine', 'pineenip'] => ['racecar', 'pineenip']
This is the code that I create;
let arr = []
let str = words.slice(0)
let pal = str.toString().split("").reverse().join("")
console.log(pal);
for (let i = 0; i < words.length; i++) {
for (let k = 0; k < pal.length; k++) {
if (words[i] == pal[k]) {
arr.push(words[i])
}
}
}
return arr
}
This is the test that my code is run against;
describe("findPalindromes", () => {
it("returns [] when passed []", () => {
expect(findPalindromes([])).to.eql([]);
});
it("identifies a palindrom", () => {
expect(findPalindromes(["racecar"])).to.eql(["racecar"]);
});
it("ignores non-palindromes", () => {
expect(findPalindromes(["pineapple", "racecar", "pony"])).to.eql([
"racecar"
]);
});
it("returns [] when passed no palindromes", () => {
expect(findPalindromes(["pineapple", "watermelon", "pony"])).to.eql([]);
});
});
Does anyone have any any suggestion of how to make my code work?
This is the simplest function that returns true or false if the str is a palindrome or not.
I would use this in combination with the filter function to filter on all palindromes. Like this
function checkPalindrom(str) { //function that checks if palindrome or not
return str == str.split('').reverse().join('');
}
const result = words.filter(word => checkPalindrom(word)); //filter function that filters array to only keep palindromes
Without giving spoilers to the answer (this is a common interview question) a clean approach would be as follows:
Define a function isPalindrome(string): boolean
Use the filter property available on the Array prototype to return an array of only palindromes e.g. inputArray.filter(isPalindrome)
Both can be unit tested separately, for example:
You could define an array of inputs and expected outputs for isPalindrome [{ input: "racecar", expectedOutput: true}, {input: "pineapple", expectedOutput: false}, ...] and loop over each test case.
function isPalindrome(word) {
const firstHalf = word.slice(0, Math.ceil(word.length/2));
const secondHalfReversed = word.slice(Math.floor(word.length/2)).split('').reverse().join('');
return firstHalf === secondHalfReversed;
}
function getPalindromesFromArray(arr) {
return arr.filter(isPalindrome);
}
const wordsArr = ['foo', 'racecar', 'pineapple', 'porcupine', 'pineenip'];
console.log(getPalindromesFromArray(wordsArr));
using for loop and filter
let arr = ["foo", "racecar", "pineapple", "porcupine", "pineenip",'pap','aaaa'];
let palindromes = arr.filter(w => {
let len = w.length;
for (let i = 0; i < len / 2; i++) {
if (w[i] == w[len - i - 1]) {
return true;
} else {
return false;
}
}
});
console.log(palindromes)
To solve that first I would create an isPalindrome function like this:
function isPalindrome(word) {
palindromeWord = ''
for(var i = word.length - 1; i >= 0; i--) {
palindromeWord += word.charAt(i)
}
return palindromeWord === word
}
and then I would check for each word inside the array like this:
let arr = ['foo', 'racecar', 'pineapple', 'porcupine', 'pineenip']
let palindromeArr = []
arr.forEach(word => {
if (isPalindrome(word)) {
palindromeArr.push(word)
}
})
console.log(palindromeArr)
What you have is good, however when you did
var pal = str.toString().split("").reverse().join("")
You changed from an array to a string, then you went into the loop with the string, so pal[k] gave a character and not a word.
To change pal back to an array of strings, split it again, use
var pal = str.toString().split("").reverse().join("").split(",");
var words = ['foo', 'racecar', 'pineapple', 'porcupine', 'pineenip'];
var arr = [];
var str = words.slice(0);
var pal = str.toString().split("").reverse().join("").split(",");
console.log(pal);
for (let i = 0; i < words.length; i++) {
for (let k = 0; k < pal.length; k++) {
if (words[i] == pal[k]) {
arr.push(words[i])
}
}
}
console.log(arr);
This question already has answers here:
Cartesian product of multiple arrays in JavaScript
(35 answers)
Closed 1 year ago.
I'm having trouble coming up with code to generate combinations from n number of arrays with m number of elements in them, in JavaScript. I've seen similar questions about this for other languages, but the answers incorporate syntactic or library magic that I'm unsure how to translate.
Consider this data:
[[0,1], [0,1,2,3], [0,1,2]]
3 arrays, with a different number of elements in them. What I want to do is get all combinations by combining an item from each array.
For example:
0,0,0 // item 0 from array 0, item 0 from array 1, item 0 from array 2
0,0,1
0,0,2
0,1,0
0,1,1
0,1,2
0,2,0
0,2,1
0,2,2
And so on.
If the number of arrays were fixed, it would be easy to make a hard coded implementation. But the number of arrays may vary:
[[0,1], [0,1]]
[[0,1,3,4], [0,1], [0], [0,1]]
Any help would be much appreciated.
Here is a quite simple and short one using a recursive helper function:
function cartesian(...args) {
var r = [], max = args.length-1;
function helper(arr, i) {
for (var j=0, l=args[i].length; j<l; j++) {
var a = arr.slice(0); // clone arr
a.push(args[i][j]);
if (i==max)
r.push(a);
else
helper(a, i+1);
}
}
helper([], 0);
return r;
}
Usage:
cartesian([0,1], [0,1,2,3], [0,1,2]);
To make the function take an array of arrays, just change the signature to function cartesian(args) instead of using rest parameter syntax.
I suggest a simple recursive generator function:
// JS
function* cartesianIterator(head, ...tail) {
const remainder = tail.length ? cartesianIterator(...tail) : [[]];
for (let r of remainder) for (let h of head) yield [h, ...r];
}
// get values:
const cartesian = items => [...cartesianIterator(items)];
console.log(cartesian(input));
// TS
function* cartesianIterator<T>(items: T[][]): Generator<T[]> {
const remainder = items.length > 1 ? cartesianIterator(items.slice(1)) : [[]];
for (let r of remainder) for (let h of items.at(0)!) yield [h, ...r];
}
// get values:
const cartesian = <T>(items: T[][]) => [...cartesianIterator(items)];
console.log(cartesian(input));
You could take an iterative approach by building sub arrays.
var parts = [[0, 1], [0, 1, 2, 3], [0, 1, 2]],
result = parts.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));
console.log(result.map(a => a.join(', ')));
.as-console-wrapper { max-height: 100% !important; top: 0; }
After doing a little research I discovered a previous related question:
Finding All Combinations of JavaScript array values
I've adapted some of the code from there so that it returns an array of arrays containing all of the permutations:
function(arraysToCombine) {
var divisors = [];
for (var i = arraysToCombine.length - 1; i >= 0; i--) {
divisors[i] = divisors[i + 1] ? divisors[i + 1] * arraysToCombine[i + 1].length : 1;
}
function getPermutation(n, arraysToCombine) {
var result = [],
curArray;
for (var i = 0; i < arraysToCombine.length; i++) {
curArray = arraysToCombine[i];
result.push(curArray[Math.floor(n / divisors[i]) % curArray.length]);
}
return result;
}
var numPerms = arraysToCombine[0].length;
for(var i = 1; i < arraysToCombine.length; i++) {
numPerms *= arraysToCombine[i].length;
}
var combinations = [];
for(var i = 0; i < numPerms; i++) {
combinations.push(getPermutation(i, arraysToCombine));
}
return combinations;
}
I've put a working copy at http://jsfiddle.net/7EakX/ that takes the array you gave earlier ([[0,1], [0,1,2,3], [0,1,2]]) and outputs the result to the browser console.
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
console.log(charSet.reduce((a,b)=>a.flatMap(x=>b.map(y=>x+y)),['']))
Just for fun, here's a more functional variant of the solution in my first answer:
function cartesian() {
var r = [], args = Array.from(arguments);
args.reduceRight(function(cont, factor, i) {
return function(arr) {
for (var j=0, l=factor.length; j<l; j++) {
var a = arr.slice(); // clone arr
a[i] = factor[j];
cont(a);
}
};
}, Array.prototype.push.bind(r))(new Array(args.length));
return r;
}
Alternative, for full speed we can dynamically compile our own loops:
function cartesian() {
return (cartesian.cache[arguments.length] || cartesian.compile(arguments.length)).apply(null, arguments);
}
cartesian.cache = [];
cartesian.compile = function compile(n) {
var args = [],
indent = "",
up = "",
down = "";
for (var i=0; i<n; i++) {
var arr = "$"+String.fromCharCode(97+i),
ind = String.fromCharCode(105+i);
args.push(arr);
up += indent+"for (var "+ind+"=0, l"+arr+"="+arr+".length; "+ind+"<l"+arr+"; "+ind+"++) {\n";
down = indent+"}\n"+down;
indent += " ";
up += indent+"arr["+i+"] = "+arr+"["+ind+"];\n";
}
var body = "var res=[],\n arr=[];\n"+up+indent+"res.push(arr.slice());\n"+down+"return res;";
return cartesian.cache[n] = new Function(args, body);
}
var f = function(arr){
if(typeof arr !== 'object'){
return false;
}
arr = arr.filter(function(elem){ return (elem !== null); }); // remove empty elements - make sure length is correct
var len = arr.length;
var nextPerm = function(){ // increase the counter(s)
var i = 0;
while(i < len)
{
arr[i].counter++;
if(arr[i].counter >= arr[i].length){
arr[i].counter = 0;
i++;
}else{
return false;
}
}
return true;
};
var getPerm = function(){ // get the current permutation
var perm_arr = [];
for(var i = 0; i < len; i++)
{
perm_arr.push(arr[i][arr[i].counter]);
}
return perm_arr;
};
var new_arr = [];
for(var i = 0; i < len; i++) // set up a counter property inside the arrays
{
arr[i].counter = 0;
}
while(true)
{
new_arr.push(getPerm()); // add current permutation to the new array
if(nextPerm() === true){ // get next permutation, if returns true, we got them all
break;
}
}
return new_arr;
};
Here's another way of doing it. I treat the indices of all of the arrays like a number whose digits are all different bases (like time and dates), using the length of the array as the radix.
So, using your first set of data, the first digit is base 2, the second is base 4, and the third is base 3. The counter starts 000, then goes 001, 002, then 010. The digits correspond to indices in the arrays, and since order is preserved, this is no problem.
I have a fiddle with it working here: http://jsfiddle.net/Rykus0/DS9Ea/1/
and here is the code:
// Arbitrary base x number class
var BaseX = function(initRadix){
this.radix = initRadix ? initRadix : 1;
this.value = 0;
this.increment = function(){
return( (this.value = (this.value + 1) % this.radix) === 0);
}
}
function combinations(input){
var output = [], // Array containing the resulting combinations
counters = [], // Array of counters corresponding to our input arrays
remainder = false, // Did adding one cause the previous digit to rollover?
temp; // Holds one combination to be pushed into the output array
// Initialize the counters
for( var i = input.length-1; i >= 0; i-- ){
counters.unshift(new BaseX(input[i].length));
}
// Get all possible combinations
// Loop through until the first counter rolls over
while( !remainder ){
temp = []; // Reset the temporary value collection array
remainder = true; // Always increment the last array counter
// Process each of the arrays
for( i = input.length-1; i >= 0; i-- ){
temp.unshift(input[i][counters[i].value]); // Add this array's value to the result
// If the counter to the right rolled over, increment this one.
if( remainder ){
remainder = counters[i].increment();
}
}
output.push(temp); // Collect the results.
}
return output;
}
// Input is an array of arrays
console.log(combinations([[0,1], [0,1,2,3], [0,1,2]]));
You can use a recursive function to get all combinations
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
let loopOver = (arr, str = '', final = []) => {
if (arr.length > 1) {
arr[0].forEach(v => loopOver(arr.slice(1), str + v, final))
} else {
arr[0].forEach(v => final.push(str + v))
}
return final
}
console.log(loopOver(charSet))
This code can still be shorten using ternary but i prefer the first version for readability 😊
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
let loopOver = (arr, str = '') => arr[0].map(v => arr.length > 1 ? loopOver(arr.slice(1), str + v) : str + v).flat()
console.log(loopOver(charSet))
Another implementation with ES6 recursive style
Array.prototype.cartesian = function(a,...as){
return a ? this.reduce((p,c) => (p.push(...a.cartesian(...as).map(e => as.length ? [c,...e] : [c,e])),p),[])
: this;
};
console.log(JSON.stringify([0,1].cartesian([0,1,2,3], [[0],[1],[2]])));