1337. The K Weakest Rows in a Matrix

Q

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.

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]

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]

Constraints:

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

A

// vertical traversal

class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {        
        set<int> ss;// vertical traversal
        vector<int> res;
        // check by finding 0 vertically
        for(int j=0; j<mat[0].size() && res.size()<k; ++j )
        {
            for(int i=0; i< mat.size() && res.size() < k; ++i)
            {
                if(mat[i][j] ==0 &&  ss.find(i) == ss.end() )
                {
                    ss.insert(i);
                    res.push_back(i);
                }
            }
        }
        // now that all zeros have been found, but still not enough
        for(int i=0;i<mat.size() && (res.size() < k); ++i)
       {
            if(ss.find(i) ==ss.end()) // all 0 have been found
            {
                res.push_back(i);
            }
       }        
        return res;
    }
};