LeetCode #2643 — EASY

Row With Maximum Ones

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

The Problem

Problem Statement

Given a m x n binary matrix mat, find the 0-indexed position of the row that contains the maximum count of ones, and the number of ones in that row.

In case there are multiple rows that have the maximum count of ones, the row with the smallest row number should be selected.

Return an array containing the index of the row, and the number of ones in it.

Example 1:

Input: mat = [[0,1],[1,0]]
Output: [0,1]
Explanation: Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1). So, the answer is [0,1].

Example 2:

Input: mat = [[0,0,0],[0,1,1]]
Output: [1,2]
Explanation: The row indexed 1 has the maximum count of ones (2). So we return its index, 1, and the count. So, the answer is [1,2].

Example 3:

Input: mat = [[0,0],[1,1],[0,0]]
Output: [1,2]
Explanation: The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 100
mat[i][j] is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: Given a m x n binary matrix mat, find the 0-indexed position of the row that contains the maximum count of ones, and the number of ones in that row. In case there are multiple rows that have the maximum count of ones, the row with the smallest row number should be selected. Return an array containing the index of the row, and the number of ones in it.

Baseline thinking

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

Pattern signal: Array

Example 1

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

Example 2

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

Example 3

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

Step 02

Core Insight

What unlocks the optimal approach

Iterate through each row and keep the count of ones.

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 #2643: Row With Maximum Ones
class Solution {
    public int[] rowAndMaximumOnes(int[][] mat) {
        int[] ans = new int[2];
        for (int i = 0; i < mat.length; ++i) {
            int cnt = 0;
            for (int x : mat[i]) {
                cnt += x;
            }
            if (ans[1] < cnt) {
                ans[0] = i;
                ans[1] = cnt;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2643: Row With Maximum Ones
func rowAndMaximumOnes(mat [][]int) []int {
	ans := []int{0, 0}
	for i, row := range mat {
		cnt := 0
		for _, x := range row {
			cnt += x
		}
		if ans[1] < cnt {
			ans = []int{i, cnt}
		}
	}
	return ans
}

# Accepted solution for LeetCode #2643: Row With Maximum Ones
class Solution:
    def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]:
        ans = [0, 0]
        for i, row in enumerate(mat):
            cnt = sum(row)
            if ans[1] < cnt:
                ans = [i, cnt]
        return ans

// Accepted solution for LeetCode #2643: Row With Maximum Ones
impl Solution {
    pub fn row_and_maximum_ones(mat: Vec<Vec<i32>>) -> Vec<i32> {
        let mut ans = vec![0, 0];
        for (i, row) in mat.iter().enumerate() {
            let cnt = row.iter().sum();
            if ans[1] < cnt {
                ans = vec![i as i32, cnt];
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2643: Row With Maximum Ones
function rowAndMaximumOnes(mat: number[][]): number[] {
    const ans: number[] = [0, 0];
    for (let i = 0; i < mat.length; i++) {
        const cnt = mat[i].reduce((sum, num) => sum + num, 0);
        if (ans[1] < cnt) {
            ans[0] = i;
            ans[1] = cnt;
        }
    }
    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(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.