The K Weakest Rows in a Matrix

Description

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.

Code

Max Heap

Time Complexity: $O(m\log (\max (n,k))$, Space Complexity: $O(k)$

cmp 因為是 max heap 所以沒用到，不過還是寫出來，代表我會寫。

class Solution {
    struct cmp {
        bool operator() (const pair<int, int>& p1, const pair<int, int>& p2) {
            if(p1.first < p2.first) 
                return true;
            else if (p1.first == p2.first && p1.second < p2.second)
                return true;
            else
                return false;
        }
    };
 
public:
    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        priority_queue<pair<int, int>, vector<pair<int, int>>> max_heap;
        for(int i = 0; i < mat.size(); i++) {
            int s = getNumberOfSoilder(mat[i]);
            max_heap.push({s, i});
            if(max_heap.size() > k) {
                max_heap.pop();
            }
        }
 
        vector<int> res;
        while(max_heap.size()) {
            res.push_back(max_heap.top().second);
            max_heap.pop();
        }
 
        reverse(res.begin(), res.end());
        return res;
 
    }
 
    int getNumberOfSoilder(vector<int>& row) {
        int l = 0, r = row.size() - 1;
        while(l < r) {
            int m = l + (r - l + 1) / 2;
            if(row[m] == 0) {
                r = m - 1;
            } else {
                l = m;
            }
        }
        return row[l] == 0 ? 0 : l + 1;
    }
};

Sorting

Time Complexity: $O(mn\log m)$, Space Complexity: $O(k)$

class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        int n = mat[0].size();
        for(int i = 0; i < mat.size(); i++) {
            mat[i].push_back(i);
        }
 
        sort(mat.begin(), mat.end());
        vector<int> res(k);
        for(int i = 0; i < k; i++) {
            res[i] = mat[i][n];
        }
        return res;
 
    }
};

Source

• The K Weakest Rows in a Matrix - LeetCode