What is the time complexity of Minimum Number of Flips to Convert Binary Matrix to Zero Matrix?

The optimal solution for Minimum Number of Flips to Convert Binary Matrix to Zero Matrix runs in O(2^(m*n)) time and uses O(2^(m*n)) space. The BFS explores all possible states, leading to an exponential number of states. The space complexity is also exponential due to the storage of visited states.

What is the brute force solution for Minimum Number of Flips to Convert Binary Matrix to Zero Matrix?

The brute force approach for Minimum Number of Flips to Convert Binary Matrix to Zero Matrix is Brute Force, with O(2^(m*n)) time complexity and O(1) space complexity. The brute force approach tries every possible combination of flips to see if it can turn the matrix into a zero matrix. Given the small size of the matrix (maximum 3x3), this approach is feasible despite being inefficient. The optimal Optimal Solution approach improves this to O(2^(m*n)).

What are the interview tips for Minimum Number of Flips to Convert Binary Matrix to Zero Matrix?

Clearly explain your thought process and how you approach generating states. Discuss edge cases, such as already being in a zero matrix. Practice BFS problems to become comfortable with state exploration.

What are common mistakes when solving Minimum Number of Flips to Convert Binary Matrix to Zero Matrix?

Not considering all neighbors when flipping a cell. Failing to track visited states, leading to infinite loops.

#1284

Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

Hard

ArrayHash TableBit ManipulationBreadth-First SearchMatrixBreadth-First SearchState Transition

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^(m*n))	O(2^(m*n))
Space	O(1)	O(2^(m*n))

💡

Intuition

Time O(2^(m*n))Space O(2^(m*n))

The optimal solution uses a breadth-first search (BFS) to explore the matrix states efficiently. By treating each state as a node and each flip as an edge, we can find the shortest path to the zero matrix without exploring all combinations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a queue for BFS and a set to track visited states.
2Step 2: Start from the initial matrix state and enqueue it with a step count of 0.
3Step 3: For each state, generate new states by flipping each cell and its neighbors, enqueueing new states if they haven't been visited.
4Step 4: If a new state is the zero matrix, return the step count.

solution.py28 lines

1from collections import deque
2
3def minFlips(mat):
4    m, n = len(mat), len(mat[0])
5    target = tuple([0] * (m * n))
6    start = tuple(cell for row in mat for cell in row)
7    if start == target:
8        return 0
9    queue = deque([(start, 0)])
10    visited = set([start])
11    directions = [(0, 0), (0, 1), (1, 0), (0, -1), (-1, 0)]
12
13    while queue:
14        current, steps = queue.popleft()
15        for i in range(m):
16            for j in range(n):
17                new_state = list(current)
18                for dx, dy in directions:
19                    x, y = i + dx, j + dy
20                    if 0 <= x < m and 0 <= y < n:
21                        new_state[x * n + y] ^= 1
22                new_state = tuple(new_state)
23                if new_state == target:
24                    return steps + 1
25                if new_state not in visited:
26                    visited.add(new_state)
27                    queue.append((new_state, steps + 1))
28    return -1

ℹ

Complexity note: The BFS explores all possible states, leading to an exponential number of states. The space complexity is also exponential due to the storage of visited states.

1Flipping a cell affects its neighbors, creating complex interactions.
2The problem can be viewed as a state transition problem, suitable for BFS.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.