Leetcode #1337: The K Weakest Rows in a Matrix
In this guide, we solve Leetcode #1337 The K Weakest Rows in a 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 an m x n binary matrix mat of 1's (representing soldiers) and 0's (representing civilians). The soldiers are positioned in front of the civilians.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Array, Binary Search, Matrix, Sorting, Heap (Priority Queue)
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: mat =
[[1,1,0,0,0],
[1,1,1,1,0],
[1,0,0,0,0],
[1,1,0,0,0],
[1,1,1,1,1]],
k = 3
Output: [2,0,3]
Explanation:
The number of soldiers in each row is:
- Row 0: 2
- Row 1: 4
- Row 2: 1
- Row 3: 2
- Row 4: 5
The rows ordered from weakest to strongest are [2,0,3,1,4].
Python Solution
class Solution:
def kWeakestRows(self, mat: List[List[int]], k: int) -> List[int]:
m, n = len(mat), len(mat[0])
ans = [n - bisect_right(row[::-1], 0) for row in mat]
idx = list(range(m))
idx.sort(key=lambda i: ans[i])
return idx[:k]
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
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.