LeetCode #2482 — MEDIUM

Difference Between Ones and Zeros in Row and Column

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed m x n binary matrix grid.

A 0-indexed m x n difference matrix diff is created with the following procedure:

Let the number of ones in the i^th row be onesRow_i.
Let the number of ones in the j^th column be onesCol_j.
Let the number of zeros in the i^th row be zerosRow_i.
Let the number of zeros in the j^th column be zerosCol_j.
diff[i][j] = onesRow_i + onesCol_j - zerosRow_i - zerosCol_j

Return the difference matrix diff.

Example 1:

Input: grid = [[0,1,1],[1,0,1],[0,0,1]]
Output: [[0,0,4],[0,0,4],[-2,-2,2]]
Explanation:
- diff[0][0] = onesRow₀ + onesCol₀ - zerosRow₀ - zerosCol₀ = 2 + 1 - 1 - 2 = 0 
- diff[0][1] = onesRow₀ + onesCol₁ - zerosRow₀ - zerosCol₁ = 2 + 1 - 1 - 2 = 0 
- diff[0][2] = onesRow₀ + onesCol₂ - zerosRow₀ - zerosCol₂ = 2 + 3 - 1 - 0 = 4 
- diff[1][0] = onesRow₁ + onesCol₀ - zerosRow₁ - zerosCol₀ = 2 + 1 - 1 - 2 = 0 
- diff[1][1] = onesRow₁ + onesCol₁ - zerosRow₁ - zerosCol₁ = 2 + 1 - 1 - 2 = 0 
- diff[1][2] = onesRow₁ + onesCol₂ - zerosRow₁ - zerosCol₂ = 2 + 3 - 1 - 0 = 4 
- diff[2][0] = onesRow₂ + onesCol₀ - zerosRow₂ - zerosCol₀ = 1 + 1 - 2 - 2 = -2
- diff[2][1] = onesRow₂ + onesCol₁ - zerosRow₂ - zerosCol₁ = 1 + 1 - 2 - 2 = -2
- diff[2][2] = onesRow₂ + onesCol₂ - zerosRow₂ - zerosCol₂ = 1 + 3 - 2 - 0 = 2

Example 2:

Input: grid = [[1,1,1],[1,1,1]]
Output: [[5,5,5],[5,5,5]]
Explanation:
- diff[0][0] = onesRow₀ + onesCol₀ - zerosRow₀ - zerosCol₀ = 3 + 2 - 0 - 0 = 5
- diff[0][1] = onesRow₀ + onesCol₁ - zerosRow₀ - zerosCol₁ = 3 + 2 - 0 - 0 = 5
- diff[0][2] = onesRow₀ + onesCol₂ - zerosRow₀ - zerosCol₂ = 3 + 2 - 0 - 0 = 5
- diff[1][0] = onesRow₁ + onesCol₀ - zerosRow₁ - zerosCol₀ = 3 + 2 - 0 - 0 = 5
- diff[1][1] = onesRow₁ + onesCol₁ - zerosRow₁ - zerosCol₁ = 3 + 2 - 0 - 0 = 5
- diff[1][2] = onesRow₁ + onesCol₂ - zerosRow₁ - zerosCol₂ = 3 + 2 - 0 - 0 = 5

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 10⁵
1 <= m * n <= 10⁵
grid[i][j] is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed m x n binary matrix grid. A 0-indexed m x n difference matrix diff is created with the following procedure: Let the number of ones in the ith row be onesRowi. Let the number of ones in the jth column be onesColj. Let the number of zeros in the ith row be zerosRowi. Let the number of zeros in the jth column be zerosColj. diff[i][j] = onesRowi + onesColj - zerosRowi - zerosColj Return the difference matrix diff.

Baseline thinking

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

Pattern signal: Array

Example 1

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

Example 2

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

Core Insight

What unlocks the optimal approach

You need to reuse information about a row or a column many times. Try storing it to avoid computing it multiple times.
Use an array to store the number of 1’s in each row and another array to store the number of 1’s in each column. Once you know the number of 1’s in each row or column, you can also easily calculate the number of 0’s.

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 #2482: Difference Between Ones and Zeros in Row and Column
class Solution {
    public int[][] onesMinusZeros(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[] rows = new int[m];
        int[] cols = new int[n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int v = grid[i][j];
                rows[i] += v;
                cols[j] += v;
            }
        }
        int[][] diff = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                diff[i][j] = rows[i] + cols[j] - (n - rows[i]) - (m - cols[j]);
            }
        }
        return diff;
    }
}

// Accepted solution for LeetCode #2482: Difference Between Ones and Zeros in Row and Column
func onesMinusZeros(grid [][]int) [][]int {
	m, n := len(grid), len(grid[0])
	rows := make([]int, m)
	cols := make([]int, n)
	diff := make([][]int, m)
	for i, row := range grid {
		diff[i] = make([]int, n)
		for j, v := range row {
			rows[i] += v
			cols[j] += v
		}
	}
	for i, r := range rows {
		for j, c := range cols {
			diff[i][j] = r + c - (n - r) - (m - c)
		}
	}
	return diff
}

# Accepted solution for LeetCode #2482: Difference Between Ones and Zeros in Row and Column
class Solution:
    def onesMinusZeros(self, grid: List[List[int]]) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        rows = [0] * m
        cols = [0] * n
        for i, row in enumerate(grid):
            for j, v in enumerate(row):
                rows[i] += v
                cols[j] += v
        diff = [[0] * n for _ in range(m)]
        for i, r in enumerate(rows):
            for j, c in enumerate(cols):
                diff[i][j] = r + c - (n - r) - (m - c)
        return diff

// Accepted solution for LeetCode #2482: Difference Between Ones and Zeros in Row and Column
impl Solution {
    pub fn ones_minus_zeros(grid: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let m = grid.len();
        let n = grid[0].len();
        let mut rows = vec![0; m];
        let mut cols = vec![0; n];
        for i in 0..m {
            for j in 0..n {
                if grid[i][j] == 1 {
                    rows[i] += 1;
                    cols[j] += 1;
                }
            }
        }
        let mut ans = vec![vec![0; n]; m];
        for i in 0..m {
            for j in 0..n {
                ans[i][j] = (rows[i] + cols[j] - (m - rows[i]) - (n - cols[j])) as i32;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2482: Difference Between Ones and Zeros in Row and Column
function onesMinusZeros(grid: number[][]): number[][] {
    const m = grid.length;
    const n = grid[0].length;
    const rows = new Array(m).fill(0);
    const cols = new Array(n).fill(0);
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (grid[i][j]) {
                rows[i]++;
                cols[j]++;
            }
        }
    }
    const ans = Array.from({ length: m }, () => new Array(n).fill(0));
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            ans[i][j] = rows[i] + cols[j] - (m - rows[i]) - (n - cols[j]);
        }
    }
    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(m × n)

Space

O(m + n)

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.