Leetcode #2123: Minimum Operations to Remove Adjacent Ones in Matrix
In this guide, we solve Leetcode #2123 Minimum Operations to Remove Adjacent Ones in 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
You are given a 0-indexed binary matrix grid. In one operation, you can flip any 1 in grid to be 0.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Graph, Array, Matrix
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: grid = [[1,1,0],[0,1,1],[1,1,1]]
Output: 3
Explanation: Use 3 operations to change grid[0][1], grid[1][2], and grid[2][1] to 0.
After, no more 1's are 4-directionally connected and grid is well-isolated.
Python Solution
class Solution:
def minimumOperations(self, grid: List[List[int]]) -> int:
def find(i: int) -> int:
for j in g[i]:
if j not in vis:
vis.add(j)
if match[j] == -1 or find(match[j]):
match[j] = i
return 1
return 0
g = defaultdict(list)
m, n = len(grid), len(grid[0])
for i, row in enumerate(grid):
for j, v in enumerate(row):
if (i + j) % 2 and v:
x = i * n + j
if i < m - 1 and grid[i + 1][j]:
g[x].append(x + n)
if i and grid[i - 1][j]:
g[x].append(x - n)
if j < n - 1 and grid[i][j + 1]:
g[x].append(x + 1)
if j and grid[i][j - 1]:
g[x].append(x - 1)
match = [-1] * (m * n)
ans = 0
for i in g.keys():
vis = set()
ans += find(i)
return ans
Complexity
The time complexity is , where and are the number of s in the matrix and the number of edges, respectively. The space complexity is O(V).
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.