LeetCode 1246: Palindrome Removal

Explanation

The problem "Palindrome Removal" can be solved using dynamic programming. We can define a DP array that stores the minimum cost to remove a subarray from i to j such that the remaining array is a palindrome. The goal is to find the minimum cost to remove all elements in the given array such that the remaining array is a palindrome.

The key idea is to break down the problem into subproblems and build the solution incrementally. We can consider all possible partitions of the array and recursively find the minimum cost for each partition. We can optimize this using dynamic programming to avoid recomputing overlapping subproblems.

The algorithm involves iterating over all possible subarrays and calculating the minimum cost to make the remaining subarray a palindrome. We need to consider different cases depending on the length of the subarray and whether the endpoints match or not.

Time complexity: O(n^3) where n is the length of the input array. Space complexity: O(n^2) for the DP array.

class Solution {
    public int minimumMoves(int[] arr) {
        int n = arr.length;
        int[][] dp = new int[n][n];

        for (int len = 1; len <= n; len++) {
            for (int i = 0; i + len - 1 < n; i++) {
                int j = i + len - 1;
                if (len == 1) {
                    dp[i][j] = 1;
                } else {
                    dp[i][j] = 1 + dp[i + 1][j];
                    if (arr[i] == arr[i + 1]) {
                        dp[i][j] = Math.min(dp[i][j], 1 + dp[i + 2][j]);
                    }
                    for (int k = i + 2; k <= j; k++) {
                        if (arr[i] == arr[k]) {
                            dp[i][j] = Math.min(dp[i][j], dp[i + 1][k - 1] + dp[k + 1][j]);
                        }
                    }
                }
            }
        }

        return dp[0][n - 1];
    }
}