Implemented Palindrome Partitioning using Backtracking algorithm

Trikcode · Trikcode · commit 4bdaa5cff51b · 2023-11-25T00:28:10.000+03:00
diff --git a/Backtracking/PalindromePartitioning.js b/Backtracking/PalindromePartitioning.js
@@ -0,0 +1,46 @@
+/*
+ * Problem Statement: Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.
+ * what is palindrome partitioning?
+ * - Palindrome partitioning means, partitioning a string into substrings such that every substring is a palindrome.
+ * Reference to know more about palindrome partitioning:
+ * - https://www.cs.columbia.edu/~sedwards/classes/2021/4995-fall/proposals/Palindrome.pdf
+ */
+
+class PalindromePartitioning {
+  partition(s) {
+    const result = []
+    this.backtrack(s, [], result)
+    return result
+  }
+
+  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 (this.isPalindrome(prefix)) {
+        path.push(prefix)
+        this.backtrack(s.substring(i + 1), path, result)
+        path.pop()
+      }
+    }
+  }
+
+  isPalindrome(s) {
+    let start = 0
+    let end = s.length - 1
+    while (start < end) {
+      if (s.charAt(start) !== s.charAt(end)) {
+        return false
+      }
+      start++
+      end--
+    }
+    return true
+  }
+}
+
+export default PalindromePartitioning
diff --git a/Backtracking/tests/PalindromePartitioning.test.js b/Backtracking/tests/PalindromePartitioning.test.js
@@ -0,0 +1,29 @@
+import PalindromePartitioning from '../PalindromePartitioning'
+
+describe('PalindromePartitioning', () => {
+  it('should partition a string into palindromes', () => {
+    const pp = new PalindromePartitioning()
+    const result = pp.partition('aab')
+
+    expect(result).toEqual(
+      expect.arrayContaining([
+        ['a', 'a', 'b'],
+        ['aa', 'b']
+      ])
+    )
+  })
+
+  it('should handle empty string', () => {
+    const pp = new PalindromePartitioning()
+    const result = pp.partition('')
+
+    expect(result).toEqual([[]])
+  })
+
+  it('should handle a single character string', () => {
+    const pp = new PalindromePartitioning()
+    const result = pp.partition('c')
+
+    expect(result).toEqual([['c']])
+  })
+})
