[LeetCode] 1337. The K Weakest Rows in a Matrix

[frdmu]

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.

 
解法：
根据矩阵的性质，1总是排在0前面，对于力量的定义，可以用第一个0出现的位置表示，所以可以用二分法。然后将力量和索引放入小顶堆，取前k个即可。代码如下：

class Solution {
public:
    struct node {
        int index;
        int power;
        friend bool operator < (node a, node b) {
            if (a.power > b.power) return true;
            if (a.power == b.power) return a.index > b.index;
            return false;
        }
    };
    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        vector<int> res;
        priority_queue<node> heap; 
        int m = mat.size(), n = mat[0].size();

        for (int i = 0; i < m; i++) {
            int left = 0, right = n;
            while (left < right) {
                int mid = left + (right - left) / 2; 
                if (mat[i][mid] >= 1) {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
            heap.push({i, right});
        }
        for (int i = 0; i < k; i++) {
            res.push_back(heap.top().index);
            heap.pop();
        }

        return res;
    }
};