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+# Leetcode 1332. Remove Palindromic Subsequences
+
+## Problem: 
+The description of the problem is given [here](https://leetcode.com/problems/remove-palindromic-subsequences/).
+
+In fact we are given a string. In a single step one can remove one `palindromic subsequence` from s. We want to compute the `minimum number of steps` needed to make the string empty.
+One constraint is that we can only use two characters in the string `a` and `b`.
+Example 1:
+```
+Input: s = "ababa"
+Output: 1
+Explanation: s is already a palindrome, so its entirety can be removed in a single step.
+```
+
+Example 2:
+```
+Input: s = "abb"
+Output: 2
+Explanation: "abb" -> "bb" -> "". 
+Remove palindromic subsequence "a" then "bb".
+```
+
+Example 3:
+```
+Input: s = "baabb"
+Output: 2
+Explanation: "baabb" -> "b" -> "". 
+Remove palindromic subsequence "baab" then "b".
+```
+
+## Thinking process
+This problem is categorized as an easy problem and i think so too. we only need to see that if the string is empty, that means the number of steps to make it empty is zero, if the string is a palindrome, we just have to remove the whole string and that can be done in one step. It becomes tricky when the string is not a palindrome.
+When the string is not a palindrome, we have two steps to make the string empty, either we can regroup the `a` together and the `b` together or we can remove a part of the string and the rest is also a palindrome. we ar just leveraging the fact that we only have two type of characters `a` and `b`.
+
+The time complexity of this approch is `O(n)`.
+
+## Code
+```
+public int removePalindromeSub(String s) {
+  if (s == null || s.length() == 0) {
+    return 0;
+  }
+  
+  StringBuilder sb = new StringBuilder(s);
+  sb.reverse();
+  if (s.equals(sb.toString())) {
+    return 1;
+  }
+  
+  return 2;
+}
+```
+
+