Introduction

In this blog post, we will explore how to solve the LeetCode problem "1284" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are called neighbors if they share one edge. Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot. A binary matrix is a matrix with all cells equal to 0 or 1 only. A zero matrix is a matrix with all cells equal to 0. Example 1: Input: mat = [[0,0],[0,1]] Output: 3 Explanation: One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown. Example 2: Input: mat = [[0]] Output: 0 Explanation: Given matrix is a zero matrix. We do not need to change it. Example 3: Input: mat = [[1,0,0],[1,0,0]] Output: -1 Explanation: Given matrix cannot be a zero matrix. Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 3 mat[i][j] is either 0 or 1.

Explanation

Here's the breakdown of the problem and the Python code:

• High-Level Approach:

  • Since the constraints on m and n are small (<= 3), we can explore all possible combinations of flipping the first row. Each combination will lead to a unique solution (if one exists).
  • For each combination of flips in the first row, we propagate the flips to the subsequent rows to make the upper rows all zeros.
  • After fixing all rows based on the first row flips, we check if the entire matrix becomes zero. If so, we update the minimum steps found so far.

• Complexity:

  • Runtime: O(2ⁿ m n), where n is the number of columns and m is the number of rows.
  • Storage: O(m * n)

Code

    def minFlips(mat):
    m = len(mat)
    n = len(mat[0])
    min_steps = float('inf')

    def flip(arr, i, j):
        if 0 <= i < m and 0 <= j < n:
            arr[i][j] ^= 1

    def solve(initial_flips):
        temp_mat = [row[:] for row in mat]
        steps = 0

        # Apply initial flips to the first row
        for j in range(n):
            if (initial_flips >> j) & 1:
                flip(temp_mat, 0, j)
                flip(temp_mat, 0, j - 1)
                flip(temp_mat, 0, j + 1)
                flip(temp_mat, 1, j)
                steps += 1

        # Propagate flips to the subsequent rows
        for i in range(1, m):
            for j in range(n):
                if temp_mat[i - 1][j] == 1:
                    flip(temp_mat, i, j)
                    flip(temp_mat, i - 1, j)
                    flip(temp_mat, i + 1, j)
                    flip(temp_mat, i, j - 1)
                    flip(temp_mat, i, j + 1)
                    steps += 1

        # Check if all elements are zero
        for i in range(m):
            for j in range(n):
                if temp_mat[i][j] == 1:
                    return float('inf')  # Not a valid solution

        return steps

    # Iterate through all possible combinations of flips in the first row
    for i in range(2**n):
        min_steps = min(min_steps, solve(i))

    return min_steps if min_steps != float('inf') else -1