1337. The K Weakest Rows in a Matrix LeetCode Solution

In this guide, you will get 1337. The K Weakest Rows in a Matrix LeetCode Solution with the best time and space complexity. The solution to The K Weakest Rows in a Matrix problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. The K Weakest Rows in a Matrix solution in C++
4. The K Weakest Rows in a Matrix solution in Java
5. The K Weakest Rows in a Matrix solution in Python
6. Additional Resources

Problem Statement of 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.

Complexity Analysis

• Time Complexity:
• Space Complexity:

1337. The K Weakest Rows in a Matrix LeetCode Solution in C++

class Solution {
 public:
  vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
    vector<int> ans;
    vector<pair<int, int>> rowSums;  // (sum(mat[i]), i)

    for (int i = 0; i < mat.size(); ++i)
      rowSums.emplace_back(accumulate(mat[i].begin(), mat[i].end(), 0), i);

    ranges::sort(rowSums, ranges::less{},
                 [](const pair<int, int>& rowSum) { return rowSum; });

    for (int i = 0; i < k; ++i)
      ans.push_back(rowSums[i].second);

    return ans;
  }
};
/* code provided by PROGIEZ */

1337. The K Weakest Rows in a Matrix LeetCode Solution in Java

class Solution {
  public int[] kWeakestRows(int[][] mat, int k) {
    int[] ans = new int[k];
    Pair<Integer, Integer>[] rowSums = new Pair[mat.length];

    for (int i = 0; i < mat.length; ++i)
      rowSums[i] = new Pair<>(Arrays.stream(mat[i]).sum(), i);

    Arrays.sort(rowSums, Comparator.comparing(Pair<Integer, Integer>::getKey)
                             .thenComparing(Pair<Integer, Integer>::getValue));

    for (int i = 0; i < k; ++i)
      ans[i] = rowSums[i].getValue();

    return ans;
  }
}
// code provided by PROGIEZ

1337. The K Weakest Rows in a Matrix LeetCode Solution in Python

class Solution:
  def kWeakestRows(self, mat: list[list[int]], k: int) -> list[int]:
    rowSums = [(sum(row), i) for i, row in enumerate(mat)]
    return [i for _, i in sorted(rowSums)[:k]]
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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