LeetCode 1337. The K Weakest Rows in a Matrix

Description

https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/

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.

Explanation

Count 1 at each row and use a priority queue to help sort.

Python Solution

class Solution:
    def kWeakestRows(self, mat: List[List[int]], k: int) -> List[int]:
        results = []
        
        counter = {}
        for i, row in enumerate(mat):
            counter[i] = row.count(1)
        

        pq = []
        for key, value in counter.items():
            heapq.heappush(pq, (value, key))
        
        for j in range(k):
            results.append(heapq.heappop(pq)[1])
                      
        
        return results
        
  • Time Complexity: O(N).
  • Space Complexity: O(N).

Leave a Reply

Your email address will not be published.