Leetcode #1284: Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
In this guide, we solve Leetcode #1284 Minimum Number of Flips to Convert Binary Matrix to Zero Matrix in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Breadth-First Search, Array, Hash Table, Matrix
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Solution:
def minFlips(self, mat: List[List[int]]) -> int:
m, n = len(mat), len(mat[0])
state = sum(1 << (i * n + j) for i in range(m) for j in range(n) if mat[i][j])
q = deque([state])
vis = {state}
ans = 0
dirs = [0, -1, 0, 1, 0, 0]
while q:
for _ in range(len(q)):
state = q.popleft()
if state == 0:
return ans
for i in range(m):
for j in range(n):
nxt = state
for k in range(5):
x, y = i + dirs[k], j + dirs[k + 1]
if not 0 <= x < m or not 0 <= y < n:
continue
if nxt & (1 << (x * n + y)):
nxt -= 1 << (x * n + y)
else:
nxt |= 1 << (x * n + y)
if nxt not in vis:
vis.add(nxt)
q.append(nxt)
ans += 1
return -1
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.