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# 18. The K Weakest Rows in a Matrix

The problem can be found at the following link: [Question Link](https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/description/)

## My Approach

1. Create an empty vector of pairs `rowStrengths`. Each pair will store the row's strength (the sum of elements in the row) and the row index.
2. Iterate through each row of the matrix `mat`: a. Calculate the strength of the current row by summing all its elements using `accumulate`. b. Push a pair containing the row's strength and its index into the `rowStrengths` vector.
3. Sort the `rowStrengths` vector in ascending order based on the first element of each pair (i.e., row strength).
4. Create an empty vector `result` to store the indices of the `k` weakest rows.
5. Iterate from `i = 0` to `i < k`: a. Push the index of the `i`-th row from the sorted `rowStrengths` vector into the `result` vector.
6. Return the `result` vector, which contains the indices of the `k` weakest rows.

Here's the pointwise algorithm in a step-by-step format:

* Initialize an empty vector `rowStrengths`.
* Iterate through each row of the matrix `mat`:
  * Calculate the strength of the current row by summing all its elements.
  * Create a pair containing the row's strength and its index.
  * Push the pair into the `rowStrengths` vector.
* Sort the `rowStrengths` vector in ascending order based on the first element of each pair (i.e., row strength).
* Initialize an empty vector `result` to store the indices of the `k` weakest rows.
* Iterate from `i = 0` to `i < k`:
  * Push the index of the `i`-th row from the sorted `rowStrengths` vector into the `result` vector.
* Return the `result` vector, which contains the indices of the `k` weakest rows.

## Time and Auxiliary Space Complexity

* **Time Complexity**: `O(nlogn)`
* **Auxiliary Space Complexity**: `O(1)`

## Code (C++)

```cpp

class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        vector<pair<int, int>> rowStrengths;
        for (int i = 0; i < mat.size(); ++i) {
            int strength = accumulate(mat[i].begin(), mat[i].end(), 0);
            rowStrengths.push_back({strength, i});
        }
        
        sort(rowStrengths.begin(), rowStrengths.end());
        
        vector<int> result;
        for (int i = 0; i < k; ++i) {
            result.push_back(rowStrengths[i].second);
        }
        
        return result;
    }
};

```

## Contribution and Support

For discussions, questions, or doubts related to this solution, please visit our [discussion section](https://leetcode.com/discuss/general-discussion). We welcome your input and aim to foster a collaborative learning environment.

If you find this solution helpful, consider supporting us by giving a `⭐ star` to the [rishabhv12/Daily-Leetcode-Solution](https://github.com/rishabhv12/Daily-Leetcode-Solution) repository.


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