LeetCode #1252 — EASY

Cells with Odd Values in a Matrix

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

The Problem

Problem Statement

There is an m x n matrix that is initialized to all 0's. There is also a 2D array indices where each indices[i] = [r_i, c_i] represents a 0-indexed location to perform some increment operations on the matrix.

For each location indices[i], do both of the following:

Increment all the cells on row r_i.
Increment all the cells on column c_i.

Given m, n, and indices, return the number of odd-valued cells in the matrix after applying the increment to all locations in indices.

Example 1:

Input: m = 2, n = 3, indices = [[0,1],[1,1]]
Output: 6
Explanation: Initial matrix = [[0,0,0],[0,0,0]].
After applying first increment it becomes [[1,2,1],[0,1,0]].
The final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers.

Example 2:

Input: m = 2, n = 2, indices = [[1,1],[0,0]]
Output: 0
Explanation: Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix.

Constraints:

1 <= m, n <= 50
1 <= indices.length <= 100
0 <= r_i < m
0 <= c_i < n

Follow up: Could you solve this in O(n + m + indices.length) time with only O(n + m) extra space?

Step 01

Brute Force Baseline

Problem summary: There is an m x n matrix that is initialized to all 0's. There is also a 2D array indices where each indices[i] = [ri, ci] represents a 0-indexed location to perform some increment operations on the matrix. For each location indices[i], do both of the following: Increment all the cells on row ri. Increment all the cells on column ci. Given m, n, and indices, return the number of odd-valued cells in the matrix after applying the increment to all locations in indices.

Baseline thinking

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

Pattern signal: Array · Math

Example 1

2
3
[[0,1],[1,1]]

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

Simulation : With small constraints, it is possible to apply changes to each row and column and count odd cells after applying it.
You can accumulate the number you should add to each row and column and then you can count the number of odd cells.

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 #1252: Cells with Odd Values in a Matrix
class Solution {
    public int oddCells(int m, int n, int[][] indices) {
        int[][] g = new int[m][n];
        for (int[] e : indices) {
            int r = e[0], c = e[1];
            for (int i = 0; i < m; ++i) {
                g[i][c]++;
            }
            for (int j = 0; j < n; ++j) {
                g[r][j]++;
            }
        }
        int ans = 0;
        for (int[] row : g) {
            for (int v : row) {
                ans += v % 2;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1252: Cells with Odd Values in a Matrix
func oddCells(m int, n int, indices [][]int) int {
	g := make([][]int, m)
	for i := range g {
		g[i] = make([]int, n)
	}
	for _, e := range indices {
		r, c := e[0], e[1]
		for i := 0; i < m; i++ {
			g[i][c]++
		}
		for j := 0; j < n; j++ {
			g[r][j]++
		}
	}
	ans := 0
	for _, row := range g {
		for _, v := range row {
			ans += v % 2
		}
	}
	return ans
}

# Accepted solution for LeetCode #1252: Cells with Odd Values in a Matrix
class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        g = [[0] * n for _ in range(m)]
        for r, c in indices:
            for i in range(m):
                g[i][c] += 1
            for j in range(n):
                g[r][j] += 1
        return sum(v % 2 for row in g for v in row)

// Accepted solution for LeetCode #1252: Cells with Odd Values in a Matrix
struct Solution;

impl Solution {
    fn odd_cells(n: i32, m: i32, indices: Vec<Vec<i32>>) -> i32 {
        let n = n as usize;
        let m = m as usize;
        let mut a = vec![vec![0; m]; n];
        let mut res = 0;
        for p in indices {
            let r = p[0] as usize;
            let c = p[1] as usize;
            for j in 0..m {
                a[r][j] += 1;
                if a[r][j] % 2 == 1 {
                    res += 1;
                } else {
                    res -= 1;
                }
            }
            for i in 0..n {
                a[i][c] += 1;
                if a[i][c] % 2 == 1 {
                    res += 1;
                } else {
                    res -= 1;
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let n = 3;
    let m = 3;
    let indices: Vec<Vec<i32>> = vec_vec_i32![[0, 1], [1, 1]];
    assert_eq!(Solution::odd_cells(n, m, indices), 6);
    let n = 2;
    let m = 2;
    let indices: Vec<Vec<i32>> = vec_vec_i32![[1, 1], [0, 0]];
    assert_eq!(Solution::odd_cells(n, m, indices), 0);
}

// Accepted solution for LeetCode #1252: Cells with Odd Values in a Matrix
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1252: Cells with Odd Values in a Matrix
// class Solution {
//     public int oddCells(int m, int n, int[][] indices) {
//         int[][] g = new int[m][n];
//         for (int[] e : indices) {
//             int r = e[0], c = e[1];
//             for (int i = 0; i < m; ++i) {
//                 g[i][c]++;
//             }
//             for (int j = 0; j < n; ++j) {
//                 g[r][j]++;
//             }
//         }
//         int ans = 0;
//         for (int[] row : g) {
//             for (int v : row) {
//                 ans += v % 2;
//             }
//         }
//         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(k × (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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.