• Nov 19, 2022 •CodeCatch
0 likes • 2 views
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"))
• Feb 21, 2025 •leafboo
console.log("hello world")
• Mar 22, 2022 •LeifMessinger
0 likes • 1 view
//QM Helper by Leif Messinger //Groups numbers by number of bits and shows their binary representations. //To be used on x.com const minterms = prompt("Enter your minterms separated by commas").split(",").map(x => parseInt(x.trim())); const maxNumBits = minterms.reduce(function(a, b) { return Math.max(a, Math.log2(b)); }); const bitGroups = []; for(let i = 0; i < maxNumBits; ++i){ bitGroups.push([]); } for(const minterm of minterms){ let outputString = (minterm+" "); //Count the bits let count = 0; for (var i = maxNumBits; i >= 0; --i) { if((minterm >> i) & 1){ ++count; outputString += "1"; }else{ outputString += "0"; } } bitGroups[count].push(outputString); } document.body.textContent = ""; document.body.style.setProperty("white-space", "pre"); for(const group of bitGroups){ for(const outputString of group){ document.body.textContent += outputString + "\r\n"; } }
const union = (...arr) => [...new Set(arr.flat())]; // Example union([1, 2], [2, 3], [3]); // [1, 2, 3]
// There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9. // You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where: // The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B'). // The second character of the ith pair denotes the rod that the ith ring is placed on ('0' to '9'). // For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1. // Return the number of rods that have all three colors of rings on them. let rings = "B0B6G0R6R0R6G9"; var countPoints = function(rings) { let sum = 0; // Always 10 Rods for (let i = 0; i < 10; i++) { if (rings.includes(`B${i}`) && rings.includes(`G${i}`) && rings.includes(`R${i}`)) { sum+=1; } } return sum; }; console.log(countPoints(rings));
• Apr 10, 2025 •hasnaoui1
0 likes • 11 views
const express = require('express') const mongoose = require('mongoose') const cors = require('cors') const dotenv = require('dotenv') const server = express() dotenv.config() server.use(express.json()) server.get('/' , (req , res)=>{ res.send('Hello') }) //4.lancement du serveur server.listen(process.env.PORT , ()=>{ console.log('server run on http://localhost:'+process.env.PORT) })