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 <[email protected]>

Author: Trikcode

Recursive/PalindromePartitioning.js

@@ -0,0 +1,30 @@
+import { palindrome } from './Palindrome'
+
+/*
+ * Given a string s, return all possible palindrome partitionings of s.
+ * A palindrome partitioning partitions a string into palindromic substrings.
+ * @see https://www.cs.columbia.edu/~sedwards/classes/2021/4995-fall/proposals/Palindrome.pdf
+ */
+const partitionPalindrome = (s) => {
+  const result = []
+  backtrack(s, [], result)
+  return result
+}
+
+const backtrack = (s, path, result) => {
+  if (s.length === 0) {
+    result.push([...path])
+    return
+  }
+
+  for (let i = 0; i < s.length; i++) {
+    const prefix = s.substring(0, i + 1)
+    if (palindrome(prefix)) {
+      path.push(prefix)
+      backtrack(s.substring(i + 1), path, result)
+      path.pop()
+    }
+  }
+}
+
+export default partitionPalindrome

Recursive/test/PalindromePartitioning.test.js

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+import partitionPalindrome from '../PalindromePartitioning'
+
+describe('Palindrome Partitioning', () => {
+  it('should return all possible palindrome partitioning of s', () => {
+    expect(partitionPalindrome('aab')).toEqual([
+      ['a', 'a', 'b'],
+      ['aa', 'b']
+    ])
+    expect(partitionPalindrome('a')).toEqual([['a']])
+    expect(partitionPalindrome('ab')).toEqual([['a', 'b']])
+  })
+})