diff --git a/src/main/java/com/thealgorithm/backtracking/PalindromePartitioning.java b/src/main/java/com/thealgorithm/backtracking/PalindromePartitioning.java
new file mode 100644
index 0000000..004b662
--- /dev/null
+++ b/src/main/java/com/thealgorithm/backtracking/PalindromePartitioning.java
@@ -0,0 +1,57 @@
+package com.thealgorithm.backtracking;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ * @author: Subham Santra
+ * @implNote
+ * This is a pure back-tracking approach This problem can be solved using dynamic programming
+ * also, I will add DP code in DP section
+ */
+public class PalindromePartitioning {
+
+ Boolean[][] memo;
+
+ public List> partition(String s) {
+ memo = new Boolean[s.length()][s.length()];
+ List> possiblePalindromes = new ArrayList<>();
+ partition(s, 0, new ArrayList<>(), possiblePalindromes);
+ return possiblePalindromes;
+ }
+
+ void partition(String s, int lo, List current, List> possiblePalindromes) {
+ if (lo == s.length()) {
+ possiblePalindromes.addFirst(new ArrayList<>(current));
+ return;
+ }
+
+ for (int p = lo; p < s.length(); ++p) {
+ if (isPalindrome(s, lo, p)) {
+ current.addLast(s.substring(lo, p + 1));
+ partition(s, p + 1, current, possiblePalindromes);
+ current.removeLast();
+ }
+ }
+ }
+
+ boolean isPalindrome(String s, int l, int h) {
+ if (memo[l][h] == null) {
+ for (int i = l, j = h; i <= j; ++i, --j) {
+ if (s.charAt(i) != s.charAt(j)) {
+ return memo[l][h] = false;
+ }
+ }
+ return memo[l][h] = true;
+ }
+ return memo[l][h];
+ }
+
+ public static void main(String[] args) {
+ System.out.println(new PalindromePartitioning().partition("baab"));
+ System.out.println(new PalindromePartitioning().partition("A"));
+ System.out.println(new PalindromePartitioning().partition("aab"));
+ System.out.println(new PalindromePartitioning().partition(""));
+ System.out.println(new PalindromePartitioning().partition("aababbababbabbabba"));
+ }
+}