Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

class Solution {
    int chooseBest(int first, int second) { return first < 0 || (second >= 0 && second < first) ? second : first; }

    int minimizeFlips(const vector<vector<int>>& grid, vector<int>& flips) {
        if (flips.size() == grid[0].size()) {
            vector<int> currentState = vector<int>(grid[0].size());
            vector<int> nextState = flips;
            int currentAttempt = 0;
            for (const vector<int>& row : grid) {
                vector<int> tempState = currentState;
                for (int index = 0; index < row.size(); ++index) {
                    tempState[index] ^= row[index];
                    if (nextState[index]) {
                        tempState[index] ^= 1;
                        if (index > 0) {
                            tempState[index - 1] ^= 1;
                        }
                        if (index + 1 < row.size()) {
                            tempState[index + 1] ^= 1;
                        }
                        ++currentAttempt;
                    }
                }
                currentState = nextState;
                nextState = tempState;
            }
            for (int value : nextState) {
                if (value) {
                    return -1;
                }
            }
            return currentAttempt;
        }
        flips.push_back(0);
        const int attemptOne = minimizeFlips(grid, flips);
        flips.back() = 1;
        const int attemptTwo = minimizeFlips(grid, flips);
        flips.pop_back();
        return chooseBest(attemptOne, attemptTwo);
    }

public:
    int minFlips(vector<vector<int>>& grid) {
        vector<int> flips;
        return minimizeFlips(grid, flips);
    }
};

The solution provided solves the problem of determining the minimum number of flips required to transform a binary matrix to a matrix consisting only of zeros. The code is implemented in C++ and utilizes a recursive approach backed by backtracking logic.

• The function minimizeFlips performs the recusive computation, checking every possible combination of flips on each row of the matrix and determines the minimum flips needed for the entire grid.
• The program employs a helper function chooseBest to decide the minimum number of valid flips between two given scenarios.
• It uses a depth-first search strategy to try all possible flips for the matrix cells.
• Performance-wise, the approach maintains states of flips for each column in vectors and evaluates the outcome row by row.
• The recursion explores flipping a specific column and not flipping it, hence exhausting possible transformations.
• If flipping the current cell results in achieving a state with more rows containing zeros, the count is updated and the algorithm continues to the next possibility.
• If a row is reached where no sequence of flips leads to zeros across, the function returns -1, indicating the impossibility of achieving the desired state.
• The condition checking is performed every time a row transformation is attempted, ensuring that all configurations are tested for their ability to contribute to the zero matrix.

This solution provides a clear implementation for addressing combinatorial optimization problems involving grid transformations, leveraging recursive checks coupled with strategic flips per column. Through careful state management, the recursive function minimizeFlips ensures comprehensive checks for achieving a matrix of zeros, which is then leveraged in the minFlips public method to start the process using an empty initial flip configuration.

class Solution {

    private int improvedMin(int p, int q) {
        return p < 0 || (q >= 0 && q < p) ? q : p;
    }

    private int depthFirstSearch(int[][] matrix, List<Integer> flips) {
        if (flips.size() == matrix[0].length) {
            int[] tempState = new int[matrix[0].length];
            int[] prevState = flips.stream().mapToInt(Integer::intValue).toArray();
            int countFlips = 0;
            for (int[] row : matrix) {
                int[] currentState = tempState;
                for (int idx = 0; idx < row.length; ++idx) {
                    currentState[idx] ^= row[idx];
                    if (prevState[idx] == 1) {
                        currentState[idx] ^= 1;
                        if (idx > 0) {
                            currentState[idx - 1] ^= 1;
                        }
                        if (idx + 1 < row.length) {
                            currentState[idx + 1] ^= 1;
                        }
                        ++countFlips;
                    }
                }
                tempState = prevState;
                prevState = currentState;
            }
            for (int element : prevState) {
                if (element == 1) {
                    return -1;
                }
            }
            return countFlips;
        }
        flips.add(0);
        final int recurse1 = depthFirstSearch(matrix, flips);
        flips.set(flips.size() - 1, 1);
        final int recurse2 = depthFirstSearch(matrix, flips);
        flips.remove(flips.size() - 1);
        return improvedMin(recurse1, recurse2);
    }

    public int minFlips(int[][] matrix) {
        return depthFirstSearch(matrix, new ArrayList<>());  
    }
}

The Java solution revolves around determining the minimum number of flips required to convert a given binary matrix to a zero matrix. The strategy employed involves depth-first search (DFS) combined with a backtracking approach, making use of both bitwise operations and dynamic programming techniques.

The code defines a Solution class containing several methods:

• improvedMin(p, q): Helper method to determine the smaller value between p and q, considering possible negative values.
• depthFirstSearch(matrix, flips): Main recursive DFS function that explores flipping each bit's possibility in the matrix. This function uses a local state representation to track flips applied to every element for each row of the matrix. If flipping the current bit leads to all bits becoming zero (a favorable scenario), it then recursively proceeds to the next configuration, either flipping or not flipping the current bit, maintaining a count of total flips.
  • The function processes each bit in the current row, adjusting adjacent bits in accordance with the flipping operation and increasing the flip count.
  • After processing all rows, it checks if the final state is fully converted to a zero matrix. If any bit is not zero, the function returns -1, indicating failure for the particular combination.
  • If the matrix is successfully flipped to zeroes, it returns the count of flips made.
• minFlips(matrix): Initiates the DFS by providing the initial empty list for flips and passing the matrix to depthFirstSearch.

Key operational insights are:

• Initially, the DFS is invoked without any flips.
• For every bit, it recursively evaluates the consequences of flipping and not flipping, optimizing to find the minimal number of flips required using improvedMin.
• Bitwise operations are used to simulate the effects of flips on the matrix.
• The solution iteratively improves upon itself by backtracking, ensuring that every combination of flips is evaluated.

Summarily, the algorithm efficiently searches through all potential combinations of flips for converting the matrix to a zero matrix, utilizing backtracking to identify the minimum required flips while conducting bitwise checks to ensure accuracy after each potential flip. This combination of strategies ensures a thorough yet optimized approach to solving the problem.