Solution Explanation: Palindrome Removal

This problem asks for the minimum number of moves to remove all elements from an array, where a move consists of removing a palindromic subarray. The optimal solution utilizes dynamic programming with an interval approach.

Approach:

The core idea is to build a solution bottom-up using dynamic programming. We define a 2D array f where f[i][j] represents the minimum number of moves needed to remove all elements from index i to j (inclusive).

1. Base Case: f[i][i] = 1 for all i. Removing a single element takes one move.

2. Two-Element Case: If i + 1 == j, we have two elements. If they are equal (forming a palindrome), f[i][j] = 1; otherwise, f[i][j] = 2.

3. Recursive Relation (More than Two Elements): For j > i + 1, we consider two possibilities:

   • Palindrome: If arr[i] == arr[j], the subarray is potentially a palindrome. We can remove the outer elements in one move, and the remaining subarray arr[i+1...j-1] requires f[i+1][j-1] moves.
   • Non-Palindrome or Partial Palindromes: We can split the array at any index k between i and j and recursively solve the problem for arr[i...k] and arr[k+1...j]. The total moves will be f[i][k] + f[k+1][j].

   We choose the minimum of these possibilities:

   f[i][j] = min( (arr[i] == arr[j] ? f[i+1][j-1] : Infinity), min(f[i][k] + f[k+1][j] for k in range(i, j)) )

4. Result: The final answer is f[0][n-1], where n is the length of the array.

Time and Space Complexity:

• Time Complexity: O(n³). The three nested loops in the dynamic programming solution lead to a cubic time complexity.
• Space Complexity: O(n²). We use a 2D array f of size n x n to store the results of subproblems.

Code Implementation (Python):

class Solution:
    def minimumMoves(self, arr: List[int]) -> int:
        n = len(arr)
        f = [[float('inf')] * n for _ in range(n)]  # Initialize with infinity
 
        for i in range(n):
            f[i][i] = 1  # Base case: one element requires one move
 
        for length in range(2, n + 1):  # Iterate through subarray lengths
            for i in range(n - length + 1):
                j = i + length - 1
                if arr[i] == arr[j]:
                    f[i][j] = min(f[i][j], f[i + 1][j - 1] if i + 1 <= j - 1 else 1)
                for k in range(i, j):
                    f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j])
 
        return f[0][n - 1]

The other languages (Java, C++, Go, TypeScript) follow the same logic, differing only in syntax and data structure handling. The core dynamic programming approach remains consistent.