Is The K Weakest Rows in a Matrix easy or hard?

The K Weakest Rows in a Matrix is classified as an Easy problem on LeetCode with a high acceptance rate. The core idea involves counting soldiers in each row and sorting, while binary search provides a small optimization for larger inputs.

The K Weakest Rows in a Matrix Python/Java solution

Python and Java solutions typically compute soldier counts using either a loop or binary search, store pairs of (count, index), and then sort them. Python often uses tuples and the built-in sort, while Java commonly uses arrays with a custom comparator or a priority queue.

How to solve The K Weakest Rows in a Matrix in O(m log n)?

Use binary search on every row to locate the first 0, which directly gives the soldier count. This takes O(log n) per row, so counting across m rows costs O(m log n). Then use a heap of size k or sorting to determine the k weakest rows.

What is the best approach for The K Weakest Rows in a Matrix?

Binary search combined with sorting or a heap is the most efficient approach. Because each row contains soldiers (1s) followed by civilians (0s), binary search can find the soldier count in O(log n) time. After computing counts for all rows, sort them or maintain a heap to extract the k weakest rows.

Is The K Weakest Rows in a Matrix asked at Google/Amazon/Meta?

This problem appears frequently in coding interview practice sets and is commonly associated with companies like Amazon and Google. It tests basic matrix processing, sorting, and binary search optimization, which are common interview topics.

What data structure is used in The K Weakest Rows in a Matrix?

Typical solutions use arrays to store (soldierCount, rowIndex) pairs and apply sorting. Optimized implementations may use binary search for counting and a heap or priority queue to track the k weakest rows efficiently.

What is the time complexity of The K Weakest Rows in a Matrix?

The straightforward solution scans each row and sorts the results, giving O(mn + m log m) time. Using binary search reduces counting to O(m log n), and sorting adds O(m log m). A heap optimization can reduce sorting overhead to O(m log k).

Description Hints 1 Solution Video 8:06

1337. The K Weakest Rows in a Matrix

Easy74.3% acceptance5 companies

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. That is, all the 1's will appear to the left of all the 0's in each row.

A row i is weaker than a row j if one of the following is true:

The number of soldiers in row i is less than the number of soldiers in row j.
Both rows have the same number of soldiers and i < j.

Return the indices of the k weakest rows in the matrix ordered from weakest to strongest.

Example 1:

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].

Example 2:

Input: mat = 
[[1,0,0,0],
 [1,1,1,1],
 [1,0,0,0],
 [1,0,0,0]], 
k = 2
Output: [0,2]
Explanation: 
The number of soldiers in each row is: 
- Row 0: 1 
- Row 1: 4 
- Row 2: 1 
- Row 3: 1 
The rows ordered from weakest to strongest are [0,2,3,1].

Constraints:

m == mat.length
n == mat[i].length
2 <= n, m <= 100
1 <= k <= m
matrix[i][j] is either 0 or 1.

Asked at 5 companies

Amazon

Google Microsoft Meta Bloomberg

Topics

Hints open one at a time

Hint 1

Sort the matrix row indexes by the number of soldiers and then row indexes.

8:06

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
2    def kWeakestRows(self, mat: List[List[int]], k: int) -> List[int]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[[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]]
3

Input

[[1,0,0,0],[1,1,1,1],[1,0,0,0],[1,0,0,0]]
2

Your code is saved on this device.Solution