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; } };
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