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Subsets of an array

0 likes • Nov 19, 2022
JavaScript
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count substrings

CodeCatch
0 likes • Nov 19, 2022
JavaScript
const countSubstrings = (str, searchValue) => {
let count = 0,
i = 0;
while (true) {
const r = str.indexOf(searchValue, i);
if (r !== -1) [count, i] = [count + 1, r + 1];
else return count;
}
};
countSubstrings('tiktok tok tok tik tok tik', 'tik'); // 3
countSubstrings('tutut tut tut', 'tut'); // 4

permutations

CodeCatch
0 likes • Nov 19, 2022
JavaScript
const permutations = arr => {
if (arr.length <= 2) return arr.length === 2 ? [arr, [arr[1], arr[0]]] : arr;
return arr.reduce(
(acc, item, i) =>
acc.concat(
permutations([...arr.slice(0, i), ...arr.slice(i + 1)]).map(val => [
item,
...val,
])
),
[]
);
};
permutations([1, 33, 5]);
// [ [1, 33, 5], [1, 5, 33], [33, 1, 5], [33, 5, 1], [5, 1, 33], [5, 33, 1] ]

Longest Palindromic Substring

CodeCatch
0 likes • Nov 19, 2022
JavaScript
const longestPalindrome = (s) => {
if(s.length === 1) return s
const odd = longestOddPalindrome(s)
const even = longestEvenPalindrome(s)
if(odd.length >= even.length) return odd
if(even.length > odd.length) return even
};
const longestOddPalindrome = (s) => {
let longest = 0
let longestSubStr = ""
for(let i = 1; i < s.length; i++) {
let currentLongest = 1
let currentLongestSubStr = ""
let left = i - 1
let right = i + 1
while(left >= 0 && right < s.length) {
const leftLetter = s[left]
const rightLetter = s[right]
left--
right++
if(leftLetter === rightLetter) {
currentLongest += 2
currentLongestSubStr = s.slice(left+1, right)
} else break
}
if(currentLongest > longest) {
if(currentLongestSubStr) longestSubStr = currentLongestSubStr
else longestSubStr = s[i]
longest = currentLongest
}
}
return longestSubStr
}
const longestEvenPalindrome = (s) => {
let longest = 0
let longestSubStr = ""
for(let i = 0; i < s.length - 1; i++) {
if(s[i] !== s[i+1]) continue;
let currentLongest = 2
let currentLongestSubStr = ""
let left = i - 1
let right = i + 2
while(left >= 0 && right < s.length) {
const leftLetter = s[left]
const rightLetter = s[right]
left--
right++
if(leftLetter === rightLetter) {
currentLongest += 2
currentLongestSubStr = s.slice(left+1, right)
} else break
}
if(currentLongest > longest) {
if(currentLongestSubStr) longestSubStr = currentLongestSubStr
else longestSubStr = s[i] + s[i+1]
longest = currentLongest
}
}
return longestSubStr
}
console.log(longestPalindrome("babad"))

Sequential Queue

ThiccDaddyLOAF
0 likes • Feb 6, 2021
JavaScript
class SequentialQueue{ //if you want it to go backwards, too bad
next(){
return this.i++;
}
constructor(start = 0){
this.i = start;
}
}
const que = new SequentialQueue(0);
for(let i = 0; i < 10; i++){
console.log(que.next());
}

K-means clustering

CodeCatch
0 likes • Nov 19, 2022
JavaScript
const kMeans = (data, k = 1) => {
const centroids = data.slice(0, k);
const distances = Array.from({ length: data.length }, () =>
Array.from({ length: k }, () => 0)
);
const classes = Array.from({ length: data.length }, () => -1);
let itr = true;
while (itr) {
itr = false;
for (let d in data) {
for (let c = 0; c < k; c++) {
distances[d][c] = Math.hypot(
...Object.keys(data[0]).map(key => data[d][key] - centroids[c][key])
);
}
const m = distances[d].indexOf(Math.min(...distances[d]));
if (classes[d] !== m) itr = true;
classes[d] = m;
}
for (let c = 0; c < k; c++) {
centroids[c] = Array.from({ length: data[0].length }, () => 0);
const size = data.reduce((acc, _, d) => {
if (classes[d] === c) {
acc++;
for (let i in data[0]) centroids[c][i] += data[d][i];
}
return acc;
}, 0);
for (let i in data[0]) {
centroids[c][i] = parseFloat(Number(centroids[c][i] / size).toFixed(2));
}
}
}
return classes;
};
kMeans([[0, 0], [0, 1], [1, 3], [2, 0]], 2); // [0, 1, 1, 0]

Generate a random HEX color

AustinLeath
0 likes • Nov 18, 2022
JavaScript
// Generate a random HEX color
randomColor = () => `#${Math.random().toString(16).slice(2, 8).padStart(6, '0')}`;
// Or
const randomColor = () => `#${(~~(Math.random()*(1<<24))).toString(16)}`;