LeetCode 2123: Minimum Operations to Remove Adjacent Ones in Matrix Solution

Master LeetCode problem 2123 (Minimum Operations to Remove Adjacent Ones in Matrix), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2123. Minimum Operations to Remove Adjacent Ones in Matrix

Hard

Array Graph Matrix

Problem Explanation

Explanation

To solve this problem, we can use a dynamic programming approach. We will iterate through the matrix row by row and keep track of the minimum number of operations needed to remove adjacent ones up to that point. We will maintain two arrays to store the minimum operations for having the current row's elements as 0 or 1. The final answer will be the minimum of the last row's minimum operations for having 0 or 1.

Let's break down the steps:

Initialize two arrays dp0 and dp1 to store the minimum operations for having the current row's elements as 0 or 1 respectively.
Iterate through the matrix row by row.
For each row, update the dp0 and dp1 arrays based on the current row's elements and the previous row's minimum operations.
The final answer will be the minimum of the last row's minimum operations for having 0 or 1.

Time complexity: O(m*n) where m is the number of rows and n is the number of columns in the matrix. Space complexity: O(n) where n is the number of columns in the matrix.

Solution Code

class Solution {
    public int minOperationsToRemoveAdjacentOnes(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        
        int[] dp0 = new int[n];
        int[] dp1 = new int[n];
        
        for (int i = 0; i < m; i++) {
            int[] newDp0 = new int[n];
            int[] newDp1 = new int[n];
            
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 1) {
                    newDp1[j] = Math.min(dp0[j], dp1[j]);
                    newDp0[j] = Math.min(dp0[j], dp1[j] + 1);
                } else {
                    newDp0[j] = dp0[j];
                    newDp1[j] = dp1[j] + 1;
                }
                
                if (j > 0) {
                    newDp0[j] = Math.min(newDp0[j], newDp0[j - 1]);
                    newDp1[j] = Math.min(newDp1[j], newDp1[j - 1]);
                }
            }
            
            dp0 = newDp0;
            dp1 = newDp1;
        }
        
        return Math.min(dp0[n - 1], dp1[n - 1]);
    }
}

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