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

Problem Link

https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/

Problem Statement

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.

Approach 1: Brute Force

Iterate each row to find out the number of soldiers, store the count with the row index. Sort it and take the first $k$ counts.

C++

Written by @wkw

class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        vector<int> ans;
        vector<pair<int, int>> a;
        for (int i = 0; i < mat.size(); i++) {
            a.push_back({
                // number of soldiers
                accumulate(mat[i].begin(), mat[i].end(), 0),
                // row index
                i
            });
        }
        sort(a.begin(), a.end());
        // or use partial_sort
        // partial_sort(a.begin(), a.begin() + k, a.end());
        for (int i = 0; i < k; i++) {
            // add the count to answer
            ans.push_back(a[i].second);
        }
        return ans;
    }
};

Approach 2: Heap

Python

Written by @heiheihang

class Solution:
    def kWeakestRows(self, mat: List[List[int]], k: int) -> List[int]:
        rows = []
        for i in range(len(mat)):
            cnt = 0
            for j in range(len(mat[i])):
                if(mat[i][j] == 1):
                    cnt += 1
                else:
                    break
            heappush(rows, [-cnt, -i])
            if len(rows) > k:
                heappop(rows)
        ans = []
        while rows:
            cnt, row = heappop(rows)
            ans.append(-row)
        ans.reverse()
        return ans

Approach 3: Binary Search

Instead of searching linearly, we can use binary search to find out the number of soldiers. The rest is same as approach 1.

C++

Written by @wkw

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size(), l = 0, r = n;
        while (l < r) {
            int m = l + (r - l) / 2;
            if (nums[m] == target) l = m + 1;
            else r = m;
        }
        return l;
    }

    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        vector<int> ans;
        vector<pair<int, int>> a;
        for (int i = 0; i < mat.size(); i++) {
            a.push_back({
                search(mat[i], 1),
                // accumulate(mat[i].begin(), mat[i].end(), 0),
                i
            });
        }
        sort(a.begin(), a.end());
        // or use partial_sort
        // partial_sort(a.begin(), a.begin() + k, a.end());
        for (int i = 0; i < k; i++) {
            ans.push_back(a[i].second);
        }
        return ans;
    }
};

Approach 4: Binary Search + Priority Queue

Instead of using a vector to store and sort, we use priority queue to handle the order.

C++

Written by @wkw

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size(), l = 0, r = n;
        while (l < r) {
            int m = l + (r - l) / 2;
            if (nums[m] == target) l = m + 1;
            else r = m;
        }
        return l;
    }

    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        vector<int> ans;
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
        for (int i = 0; i < mat.size(); i++) {
            pq.push({
                search(mat[i], 1),
                i
            });
        }
        for (int i = 0; i < k; i++) {
            ans.push_back(pq.top().second);
            pq.pop();
        }
        return ans;
    }
};

Kotlin

Written by @wkw

class Solution {
    fun kWeakestRows(mat: Array<IntArray>, k: Int): IntArray {
        var ans = IntArray(k)
        var pq = PriorityQueue<Pair<Int, Int>>(compareBy( {it.first }, { it.second }))
        fun search(nums: IntArray, target: Int): Int {
            val n = nums.size
            var l = 0
            var r = n
            while (l < r) {
                var m = (l + r) / 2
                if (nums[m] == target) l = m + 1
                else r = m
            }
            return l
        }
        for (i in mat.indices) {
            pq.add(Pair(search(mat[i], 1), i))
        }
        for (i in 0 until k) {
            ans[i] = pq.poll().second
        }
        return ans
    }
}