What is the primary algorithmic approach for this problem?

The primary approach is to use binary search to count soldiers in each row and then sort the rows based on soldier count and index.

How does sorting affect the time complexity of the solution?

Sorting the rows by soldier count and index contributes to an O(m log m) time complexity, where m is the number of rows in the matrix.

What is the significance of using binary search in this problem?

Binary search is used to count soldiers efficiently by exploiting the sorted nature of each row, reducing the need to scan each row completely.

Can the solution be optimized further for larger matrices?

While the current approach is optimal in terms of time complexity for this problem, further optimizations may be explored based on the matrix size and constraints.

What happens if multiple rows have the same number of soldiers?

In the case of a tie, the rows are ordered by their index in the matrix, ensuring a stable and deterministic output.

#1337

Easy

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LeetCode 题解工作台

矩阵中战斗力最弱的 K 行

给你一个大小为 m * n 的矩阵 mat ，矩阵由若干军人和平民组成，分别用 1 和 0 表示。请你返回矩阵中战斗力最弱的 k 行的索引，按从最弱到最强排序。如果第 i 行的军人数量少于第 j 行，或者两行军人数量相同但 i 小于 j ，那么我们认为第 i 行的战斗力比第 j 行弱。军人总…

数组二分查找排序堆（优先队列）矩阵

题目描述

给你一个大小为 m * n 的矩阵 mat，矩阵由若干军人和平民组成，分别用 1 和 0 表示。

请你返回矩阵中战斗力最弱的 k 行的索引，按从最弱到最强排序。

如果第 i 行的军人数量少于第 j 行，或者两行军人数量相同但 i 小于 j，那么我们认为第 i 行的战斗力比第 j 行弱。

军人总是排在一行中的靠前位置，也就是说 1 总是出现在 0 之前。

示例 1：

输入：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
输出：[2,0,3]
解释：
每行中的军人数目：
行 0 -> 2 
行 1 -> 4 
行 2 -> 1 
行 3 -> 2 
行 4 -> 5 
从最弱到最强对这些行排序后得到 [2,0,3,1,4]

示例 2：

输入：mat = 
[[1,0,0,0],
 [1,1,1,1],
 [1,0,0,0],
 [1,0,0,0]], 
k = 2
输出：[0,2]
解释： 
每行中的军人数目：
行 0 -> 1 
行 1 -> 4 
行 2 -> 1 
行 3 -> 1 
从最弱到最强对这些行排序后得到 [0,2,3,1]

提示：

m == mat.length
n == mat[i].length
2 <= n, m <= 100
1 <= k <= m
matrix[i][j] 不是 0 就是 1

lightbulb

解题思路

方法一

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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]

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Check if the candidate efficiently applies binary search to count soldiers in each row.
question_mark
Evaluate how the candidate handles sorting and tie-breaking when soldier counts are equal.
question_mark
Assess the candidate's ability to optimize for both time and space complexity within the problem's constraints.

warning

常见陷阱

外企场景

error
Misunderstanding the requirement for sorting rows based on both soldier count and row index in case of ties.
error
Using a brute-force approach to count soldiers in each row instead of binary search, leading to inefficiency.
error
Not properly handling edge cases, such as when multiple rows have the same number of soldiers.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Handling larger matrices where time and space complexity become more significant.
arrow_right_alt
Modifying the problem to return the indices of the k strongest rows instead of the weakest.
arrow_right_alt
Allowing for different sorting criteria, such as sorting by the number of civilians instead of soldiers.

help

常见问题

外企场景

继续练习

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