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LeetCode - 1337. The K Weakest Rows in a Matrix

1337. The K Weakest Rows in a Matrix - easy

문제

Given a m * n matrix mat of ones (representing soldiers) and zeros (representing civilians), return the indexes of the k weakest rows in the matrix ordered from the weakest to the strongest.

A row i is weaker than row j, if the number of soldiers in row i is less than the number of soldiers in row j, or they have the same number of soldiers but i is less than j. Soldiers are always stand in the frontier of a row, that is, always ones may appear first and then zeros.

제한사항

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

입출력 예

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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 for each row is: 
row 0 -> 2 
row 1 -> 4 
row 2 -> 1 
row 3 -> 2 
row 4 -> 5 
Rows ordered from the weakest to the strongest are [2,0,3,1,4]
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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 for each row is: 
row 0 -> 1 
row 1 -> 4 
row 2 -> 1 
row 3 -> 1 
Rows ordered from the weakest to the strongest are [0,2,3,1]

풀이

  • Arary
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class Solution {
public:    
    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        std::vector<std::pair<int, int>> items;
        std::vector<int> result;
        
        int row_index = 0;
        for (const auto& row : mat) {
            int sum = std::accumulate(row.begin(), row.end(), 0);
            items.push_back({sum, row_index});
            ++row_index;
        }
        
        std::sort(items.begin(), items.end(), 
            [](const std::pair<int, int>& a, const std::pair<int, int>& b){
                if (a.first == b.first) {
                    return a.second < b.second;
                }
                return a.first < b.first;
            }
        );

        for (int i = 0 ; i < k ; ++i) {
            result.push_back(items[i].second);
        }
        
        return result;
    }
};
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