Solved Problems

Author: gouthampradhan

README.md

@@ -265,6 +265,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - ![#f03c15](https://placehold.it/15/f03c15/000000?text=+) [Number of Ways to Stay in the Same Place After Some Steps](problems/src/dynamic_programming/NumberOfWaysToStayInTheSamePlace.java) (Hard)
 - [Toss Strange Coins](problems/src/dynamic_programming/TossStrangeCoins.java) (Medium)
 - [Knight Dialer](problems/src/dynamic_programming/KnightDialer.java) (Medium)
+- ![#f03c15](https://placehold.it/15/f03c15/000000?text=+) [Palindrome Removal](problems/src/dynamic_programming/PalindromeRemoval.java) (Hard)
 
 
 #### [Greedy](problems/src/greedy)

problems/src/dynamic_programming/PalindromeRemoval.java

@@ -0,0 +1,49 @@
+package dynamic_programming;
+
+/**
+ * Created by gouthamvidyapradhan on 12/05/2020 Given an integer array arr, in one move you can
+ * select a palindromic subarray arr[i], arr[i+1], ..., arr[j] where i <= j, and remove that
+ * subarray from the given array. Note that after removing a subarray, the elements on the left and
+ * on the right of that subarray move to fill the gap left by the removal.
+ *
+ * <p>Return the minimum number of moves needed to remove all numbers from the array.
+ *
+ * <p>Example 1:
+ *
+ * <p>Input: arr = [1,2] Output: 2 Example 2:
+ *
+ * <p>Input: arr = [1,3,4,1,5] Output: 3 Explanation: Remove [4] then remove [1,3,1] then remove
+ * [5].
+ *
+ * <p>Constraints:
+ *
+ * <p>1 <= arr.length <= 100 1 <= arr[i] <= 20
+ */
+public class PalindromeRemoval {
+  public static void main(String[] args) {
+    int[] A = {1, 3, 1, 2, 4, 2};
+    System.out.println(new PalindromeRemoval().minimumMoves(A));
+  }
+
+  int[][] DP;
+
+  public int minimumMoves(int[] arr) {
+    DP = new int[arr.length][arr.length];
+    return dp(0, arr.length - 1, arr);
+  }
+
+  private int dp(int i, int j, int[] arr) {
+    if (i > j) return 1;
+    else if (DP[i][j] != 0) return DP[i][j];
+    else {
+      int min = Integer.MAX_VALUE;
+      for (int t = j; t >= i; t--) {
+        if (arr[i] == arr[t]) {
+          min = Math.min(min, dp(i + 1, t - 1, arr) + ((t + 1 > j) ? 0 : dp(t + 1, j, arr)));
+        }
+      }
+      DP[i][j] = min;
+      return min;
+    }
+  }
+}