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Implemented Palindrome Partitioning using Backtracking algorithm (TheAlgorithms#1591)
* Implemented Palindrome Partitioning using Backtracking algorithm * fix:Updated palindromePartition algorithm * code clean up * Rephrase doc comment & move to appropriate function --------- Co-authored-by: Lars Müller <34514239+appgurueu@users.noreply.github.com>
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Recursive/PalindromePartitioning.js

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import { palindrome } from './Palindrome'
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/*
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* Given a string s, return all possible palindrome partitionings of s.
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* A palindrome partitioning partitions a string into palindromic substrings.
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* @see https://www.cs.columbia.edu/~sedwards/classes/2021/4995-fall/proposals/Palindrome.pdf
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*/
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const partitionPalindrome = (s) => {
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const result = []
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backtrack(s, [], result)
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return result
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}
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const backtrack = (s, path, result) => {
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if (s.length === 0) {
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result.push([...path])
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return
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}
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for (let i = 0; i < s.length; i++) {
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const prefix = s.substring(0, i + 1)
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if (palindrome(prefix)) {
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path.push(prefix)
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backtrack(s.substring(i + 1), path, result)
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path.pop()
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}
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}
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}
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export default partitionPalindrome
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import partitionPalindrome from '../PalindromePartitioning'
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describe('Palindrome Partitioning', () => {
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it('should return all possible palindrome partitioning of s', () => {
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expect(partitionPalindrome('aab')).toEqual([
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['a', 'a', 'b'],
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['aa', 'b']
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])
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expect(partitionPalindrome('a')).toEqual([['a']])
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expect(partitionPalindrome('ab')).toEqual([['a', 'b']])
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})
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})

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