LeetCode 1284: Minimum Number of Flips to Convert Binary Matrix to Zero Matrix Solution

Master LeetCode problem 1284 (Minimum Number of Flips to Convert Binary Matrix to Zero 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.

1284. Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

Hard

Array Hash Table Bit Manipulation Breadth-First Search Matrix

Problem Explanation

Explanation:

To solve this problem, we can use a breadth-first search (BFS) approach. We will start by converting the given binary matrix to a string representation to keep track of visited states. We will use a queue to store the current state of the matrix and the number of flips required to reach that state. At each step, we will flip a cell and its neighbors, generate a new state, and check if it matches the zero matrix. If it does, we return the number of flips needed. If not, we continue exploring all possible states until we have exhausted all options.

Algorithm:
1. Convert the binary matrix to a string representation and push it to the queue along with the number of flips = 0.
2. While the queue is not empty, pop the front element.
3. Iterate over all cells in the matrix, flip the cell and its neighbors, generate a new state.
4. If the new state matches the zero matrix, return the number of flips.
5. If the new state is not visited before, add it to the queue with the updated number of flips.
6. If all possible states are explored and the zero matrix is not reached, return -1.
Time Complexity: O(2^(m*n)), where m and n are the dimensions of the binary matrix.
Space Complexity: O(2^(m*n)) for the visited set and queue.

Solution Code

import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Set;

class Solution {
    public int minFlips(int[][] mat) {
        int m = mat.length, n = mat[0].length;
        String start = matrixToString(mat);
        Queue<String> queue = new LinkedList<>();
        queue.add(start);
        Set<String> visited = new HashSet<>();
        visited.add(start);
        int flips = 0;

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                String curr = queue.poll();
                if (isZeroMatrix(curr, m, n)) return flips;

                int[][] matrix = stringToMatrix(curr, m, n);
                for (int r = 0; r < m; r++) {
                    for (int c = 0; c < n; c++) {
                        flip(matrix, r, c, m, n);
                        String nextState = matrixToString(matrix);
                        if (!visited.contains(nextState)) {
                            queue.add(nextState);
                            visited.add(nextState);
                        }
                        flip(matrix, r, c, m, n); // Flip back for the next iteration
                    }
                }
            }
            flips++;
        }

        return -1;
    }

    private String matrixToString(int[][] mat) {
        StringBuilder sb = new StringBuilder();
        for (int[] row : mat) {
            for (int cell : row) {
                sb.append(cell);
            }
        }
        return sb.toString();
    }

    private int[][] stringToMatrix(String str, int m, int n) {
        int[][] mat = new int[m][n];
        int k = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                mat[i][j] = str.charAt(k++) - '0';
            }
        }
        return mat;
    }

    private void flip(int[][] mat, int r, int c, int m, int n) {
        int[][] dirs = {{0, 0}, {0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        for (int[] dir : dirs) {
            int nr = r + dir[0], nc = c + dir[1];
            if (nr >= 0 && nr < m && nc >= 0 && nc < n) {
                mat[nr][nc] ^= 1; // Flip the cell
            }
        }
    }

    private boolean isZeroMatrix(String str, int m, int n) {
        for (int i = 0; i < m * n; i++) {
            if (str.charAt(i) == '1') return false;
        }
        return true;
    }
}

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