The Minimum Number of Flips to Convert Binary Matrix to Zero Matrix interview question involves a grid of 0s and 1s. Flipping a cell (changing 0 to 1 or vice versa) also flips its four neighbors (up, down, left, right). The goal is to find the minimum number of flips to turn the entire matrix into all 0s. If it's impossible, return -1.
This problem, popular at Airbnb and Google, tests a candidate's ability to model a matrix state as a graph node. Since the matrix is small (at most 3x3), the total number of possible states is only 2^9 = 512. This allows the interviewer to see if the candidate can apply Breadth-First Search (BFS) to find the shortest path in a state-space graph, as well as their skill in using bitmasking to represent those states.
This problem utilizes the Array, Matrix, BFS, Hash Table, Bit Manipulation interview pattern. You treat each configuration of the matrix as a state. Using BFS, you start from the initial matrix and explore all possible flips. To efficiently store and compare these states, you convert the 2D matrix into a single integer (a bitmask). A set or hash table keeps track of visited states to prevent infinite loops and ensure we find the minimum steps.
Initial 3x3 Matrix: 0 0 0 1 (Let's use 2x2 for simplicity) Initial: [0,0,0,1] -> bitmask: 0001 (binary) Flip (1,1): It flips itself and neighbors (0,1) and (1,0). New state: [0,1,1,0] -> bitmask: 0110 We continue this process until we reach state 0000. Each layer of BFS represents one additional flip.
Candidates often try to use complex recursion or greedy strategies, which don't work here because a single flip has far-reaching consequences. Another mistake is not recognizing that flipping the same cell twice is useless (it returns the matrix to its previous state), which is why BFS naturally finds the shortest path without redundant flips. Failing to handle the "impossible" case (where BFS ends without reaching the zero matrix) is also common.
For problems with small "state spaces" (like 3x3 grids or 10-element arrays), think about representing the entire state as an integer bitmask. This makes your code faster and allows you to use standard graph algorithms like BFS or Dijkstra's very easily.
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