1337. The K Weakest Rows in a Matrix

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.

class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> (a[0] == b[0]) ? b[1] - a[1] : b[0] - a[0]);
        
        int n = mat.length;
        for(int i = 0; i < n; i++) {
            pq.add(new int[]{help(mat[i]), i});
            if(pq.size() > k) pq.poll();
        }
        
        int[] res = new int[k];
        
        while(k > 0) res[--k] = pq.poll()[1];
        return res;
    }
    
    public int help(int[] nums) {
        int hi = nums.length, lo = 0;
        
        while(lo < hi) {
            int mid = lo + (hi - lo) / 2;
            if(nums[mid] == 1) {
                lo = mid + 1;
            }
            else hi = mid;
        }
        
        return lo;
    }
}

https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/discuss/496555/Java-Best-Solution-100-TimeSpace-Binary-Search-%2B-Heap

很多细节，首先是pq，因为要返回最weak的（最小）rows，必须得inverse order排列，并且维持k size，才能保证pq里是需要的rows。

然后返回也是逆序返回。

help method是计算这一行的1个数，用binary search，这样能更快，判断表达式是nums[mid] == 1, 如果是，就 lo = mid + 1，否则hi = mid