How do you solve The K Weakest Rows in a Matrix on LeetCode?

The K Weakest Rows in a Matrix (LeetCode #1337) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: The soldiers are always positioned to the left, allowing for efficient counting.

What is the brute force solution for The K Weakest Rows in a Matrix?

The brute force approach for The K Weakest Rows in a Matrix is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute force approach, we will count the number of soldiers in each row and then sort the rows based on the count and their indices. This is straightforward but not the most efficient way to solve the problem. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for The K Weakest Rows in a Matrix?

Clearly explain your thought process and the reasoning behind each step. Be prepared to discuss time and space complexity for your solutions. Practice writing clean and efficient code, as readability matters in interviews.

What are common mistakes when solving The K Weakest Rows in a Matrix?

Not considering the order of indices when counts are equal. Overlooking the binary search optimization for counting soldiers.

#1337

The K Weakest Rows in a Matrix

Easy

ArrayBinary SearchSortingHeap (Priority Queue)MatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n)	O(n)

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Intuition

Time O(n log n)Space O(n)

In the optimal approach, we can use a more efficient counting method combined with sorting. Since the soldiers are always on the left, we can use binary search to count them, making the solution faster.

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Algorithm

5 steps

1Step 1: Initialize an empty list to store the count of soldiers and their respective row indices.
2Step 2: For each row, use binary search to find the number of soldiers (1's) in the row.
3Step 3: Store the count along with the row index in the list.
4Step 4: Sort the list based on the number of soldiers and then by row index.
5Step 5: Extract the first k indices from the sorted list and return them.

solution.py23 lines

1# Full working Python code
2mat = [[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]]
3k = 3
4
5def kWeakestRows(mat, k):
6    soldiers = []
7    for i in range(len(mat)):
8        count = binarySearch(mat[i])
9        soldiers.append((count, i))
10    soldiers.sort()  # Sort by count and then by index
11    return [soldiers[i][1] for i in range(k)]
12
13def binarySearch(row):
14    left, right = 0, len(row)
15    while left < right:
16        mid = (left + right) // 2
17        if row[mid] == 1:
18            left = mid + 1
19        else:
20            right = mid
21    return left
22
23print(kWeakestRows(mat, k))

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the soldiers array, while the binary search for counting soldiers is O(log n) for each row. The space complexity remains O(n) for storing the counts and indices.

1The soldiers are always positioned to the left, allowing for efficient counting.
2Sorting is essential for determining the weakest rows based on multiple criteria.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.