LeetCode #1337 — EASY

The K Weakest Rows in a Matrix

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[[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]]
3

Example 2

[[1,0,0,0],[1,1,1,1],[1,0,0,0],[1,0,0,0]]
2

Step 02

Core Insight

What unlocks the optimal approach

Sort the matrix row indexes by the number of soldiers and then row indexes.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1337: The K Weakest Rows in a Matrix
class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {
        int m = mat.length, n = mat[0].length;
        int[] res = new int[m];
        List<Integer> idx = new ArrayList<>();
        for (int i = 0; i < m; ++i) {
            idx.add(i);
            int[] row = mat[i];
            int left = 0, right = n;
            while (left < right) {
                int mid = (left + right) >> 1;
                if (row[mid] == 0) {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }
            res[i] = left;
        }
        idx.sort(Comparator.comparingInt(a -> res[a]));
        int[] ans = new int[k];
        for (int i = 0; i < k; ++i) {
            ans[i] = idx.get(i);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1337: The K Weakest Rows in a Matrix
func kWeakestRows(mat [][]int, k int) []int {
	m, n := len(mat), len(mat[0])
	res := make([]int, m)
	var idx []int
	for i, row := range mat {
		idx = append(idx, i)
		left, right := 0, n
		for left < right {
			mid := (left + right) >> 1
			if row[mid] == 0 {
				right = mid
			} else {
				left = mid + 1
			}
		}
		res[i] = left
	}
	sort.Slice(idx, func(i, j int) bool {
		return res[idx[i]] < res[idx[j]] || (res[idx[i]] == res[idx[j]] && idx[i] < idx[j])
	})
	return idx[:k]
}

# Accepted solution for LeetCode #1337: The K Weakest Rows in a Matrix
class Solution:
    def kWeakestRows(self, mat: List[List[int]], k: int) -> List[int]:
        m, n = len(mat), len(mat[0])
        ans = [n - bisect_right(row[::-1], 0) for row in mat]
        idx = list(range(m))
        idx.sort(key=lambda i: ans[i])
        return idx[:k]

// Accepted solution for LeetCode #1337: The K Weakest Rows in a Matrix
struct Solution;
use std::collections::BinaryHeap;

type Pair = (i32, usize);

impl Solution {
    fn k_weakest_rows(mat: Vec<Vec<i32>>, k: i32) -> Vec<i32> {
        let mut pq: BinaryHeap<Pair> = BinaryHeap::new();
        let n = mat.len();
        for i in 0..n {
            let sum = mat[i].iter().sum::<i32>();
            let pair = (sum, i);
            pq.push(pair);
            if pq.len() > k as usize {
                pq.pop();
            }
        }
        let mut res: Vec<i32> = vec![];
        while !pq.is_empty() {
            let biggest = pq.pop().unwrap();
            res.push(biggest.1 as i32);
        }
        res.reverse();
        res
    }
}

#[test]
fn test() {
    let mat: Vec<Vec<i32>> = [
        [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],
    ]
    .iter()
    .map(|v| v.to_vec())
    .collect();
    let k = 3;
    let res = vec![2, 0, 3];
    assert_eq!(Solution::k_weakest_rows(mat, k), res);
}

// Accepted solution for LeetCode #1337: The K Weakest Rows in a Matrix
function kWeakestRows(mat: number[][], k: number): number[] {
    let n = mat.length;
    let sumMap = mat.map((d, i) => [d.reduce((a, c) => a + c, 0), i]);
    let ans = [];
    // 冒泡排序
    for (let i = 0; i < k; i++) {
        for (let j = i; j < n; j++) {
            if (
                sumMap[j][0] < sumMap[i][0] ||
                (sumMap[j][0] == sumMap[i][0] && sumMap[i][1] > sumMap[j][1])
            ) {
                [sumMap[i], sumMap[j]] = [sumMap[j], sumMap[i]];
            }
        }
        ans.push(sumMap[i][1]);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.