1337

1337. The K Weakest Rows in a Matrix (Easy)

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.

Companies: Amazon, Apple

Related Topics:
Array, Binary Search, Sorting, Heap (Priority Queue), Matrix

Solution 1. Binary Search (L <= R) + Sorting

// OJ: https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/
// Author: github.com/lzl124631x
// Time: O(MlogN + MlogM)
// Space: O(M)
class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& A, int k) {
        vector<pair<int, int>> B;
        int M = A.size(), N = A[0].size();
        for (int i = 0; i < M; ++i) {
            int L = 0, R = N - 1;
            while (L <= R) {
                int M = (L + R) / 2;
                if (A[i][M] == 1) L = M + 1;
                else R = M - 1;
            }
            B.emplace_back(R, i);
        }
        sort(begin(B), end(B));
        vector<int> ans;
        for (int i = 0; i < k; ++i) ans.push_back(B[i].second);
        return ans;
    }
};

Or

// OJ: https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix
// Author: github.com/lzl124631x
// Time: O(MlogN + MlogM)
// Space: O(M)
class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& A, int k) {
        int M = A.size(), N = A[0].size();
        vector<int> cnt(M), id(M), ans(k);
        for (int i = 0; i < M; ++i) {
            int L = 0, R = N - 1;
            while (L <= R) {
                int M = (L + R) / 2;
                if (A[i][M] == 1) L = M + 1;
                else R = M - 1;
            }
            cnt[i] = L;
        }
        iota(begin(id), end(id), 0);
        sort(begin(id), end(id), [&](int a, int b) { return cnt[a] != cnt[b] ? cnt[a] < cnt[b] : a < b; });
        for (int i = 0; i < k; ++i) ans[i] = id[i];
        return ans;
    }
};

Solution 2. Binary Search (L < R) + Sorting

// OJ: https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/
// Author: github.com/lzl124631x
// Time: O(MlogN + MlogM)
// Space: O(M)
class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& A, int k) {
        vector<pair<int, int>> B;
        int M = A.size(), N = A[0].size();
        for (int i = 0; i < M; ++i) {
            int L = 0, R = N; // Note that since we are looking for the first index of `0`, the `R` should be initialized as `N`.
            while (L < R) {
                int M = (L + R) / 2;
                if (A[i][M] == 1) L = M + 1;
                else R = M;
            }
            B.emplace_back(R, i);
        }
        sort(begin(B), end(B));
        vector<int> ans;
        for (int i = 0; i < k; ++i) ans.push_back(B[i].second);
        return ans;
    }
};

Solution 3. Binary Search (L <= R) + Heap

// OJ: https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix
// Author: github.com/lzl124631x
// Time: O(M * (logN + logK))
// Space: O(K)
class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& A, int k) {
        int M = A.size(), N = A[0].size();
        vector<int> ans;
        typedef pair<int, int> PII; // cnt, index
        priority_queue<PII> pq;
        for (int i = 0; i < M; ++i) {
            int L = 0, R = N - 1;
            while (L <= R) {
                int M = (L + R) / 2;
                if (A[i][M] == 1) L = M + 1;
                else R = M - 1;
            }
            pq.emplace(L, i);
            if (pq.size() > k) pq.pop();
        }
        while (pq.size()) {
            ans.push_back(pq.top().second);
            pq.pop();
        }
        reverse(begin(ans), end(ans));
        return ans;
    }
};