• 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"))
• Sep 13, 2023 •C S
0 likes • 3 views
/** * @param {number[]} nums * @param {number} target * @return {number[]} */ var twoSum = function(nums, target) { for(let i = 0; i < nums.length; i++) { for(let j = 0; j < nums.length; j++) { if(nums[i] + nums[j] === target && i !== j) { return [i, j] } } } };
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// 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));
const geometricProgression = (end, start = 1, step = 2) => Array.from({ length: Math.floor(Math.log(end / start) / Math.log(step)) + 1, }).map((_, i) => start * step ** i); geometricProgression(256); // [1, 2, 4, 8, 16, 32, 64, 128, 256] geometricProgression(256, 3); // [3, 6, 12, 24, 48, 96, 192] geometricProgression(256, 1, 4); // [1, 4, 16, 64, 256]
• Mar 9, 2021 •LeifMessinger
alert("bruh")
const toFixed = (n, fixed) => ~~(Math.pow(10, fixed) * n) / Math.pow(10, fixed); // Examples toFixed(25.198726354, 1); // 25.1 toFixed(25.198726354, 2); // 25.19 toFixed(25.198726354, 3); // 25.198 toFixed(25.198726354, 4); // 25.1987 toFixed(25.198726354, 5); // 25.19872 toFixed(25.198726354, 6); // 25.198726