Skip to content

Longest Palindromic Substring #1489

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Open

ZainabMalik5 wants to merge 6 commits into TheAlgorithms:master

from ZainabMalik5:newbranch3

Open

Longest Palindromic Substring #1489

Changes from 5 commits

Commits

Show all changes

Select commit Hold shift + click to select a range

Add files via upload

ZainabMalik5 Oct 5, 2023

Delete Maths/Divisors.js

ZainabMalik5 Oct 8, 2023

Add files via upload

ZainabMalik5 Oct 9, 2023

Delete Dynamic-Programming/LongestPalindromicSubstring.js

ZainabMalik5 Oct 9, 2023

Add files via upload

ZainabMalik5 Oct 9, 2023

Update LongestPalindromicSubstring.js

ZainabMalik5 Oct 9, 2023

File filter

Filter by extension

Conversations

Jump to

Jump to file

Failed to load files.

Loading

Diff view

Diff view

42 changes: 42 additions & 0 deletions Dynamic-Programming/LongestPalindromicSubstring.js

Show comments View file Open in desktop

Original file line number	Diff line number	Diff line change
		@@ -0,0 +1,42 @@
	/*
	LeetCode -> https://leetcode.com/problems/longest-palindromic-substring

	Given a string s, find the longest palindromic substring in s.
	You may assume that the maximum length of s is 1000.

	*/
	function longestPalindrome(s) {
	const n = s.length;
	const dp = new Array(n + 1).fill(0).map(() => new Array(n + 1).fill(0));
	const len = new Array(n + 1).fill(0).map(() => new Array(n + 1).fill(0));
	let str = "";
	let mx = 0;
	//fill for single character
	for (let i = 0; i < n; i++) {
	dp[i][i] = 1;
	len[i][i] = 1;
	}

	for (let i = n - 2; i >= 0; i--) {
	for (let j = i + 1; j < n; j++) {
	if (s[i] === s[j] && j - i === 1) {
	len[i][j] = 2;
	dp[i][j] = 1;
	} else if (s[i] === s[j] && dp[i + 1][j - 1]) {
	dp[i][j] = 1;
	len[i][j] = j - i + 1;
	}
	}
	}

	for (let i = 0; i < n; i++) {
	for (let j = 0; j < n; j++) {
	if (len[i][j] > mx && dp[i][j]) {
	mx = len[i][j];
	str = s.substring(i, j + 1);
	}
	}
	}

	return str;
	}