LeetCode #2536 — MEDIUM

Increment Submatrices by One

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

The Problem

Problem Statement

You are given a positive integer n, indicating that we initially have an n x n 0-indexed integer matrix mat filled with zeroes.

You are also given a 2D integer array query. For each query[i] = [row1_i, col1_i, row2_i, col2_i], you should do the following operation:

Add 1 to every element in the submatrix with the top left corner (row1_i, col1_i) and the bottom right corner (row2_i, col2_i). That is, add 1 to mat[x][y] for all row1_i <= x <= row2_i and col1_i <= y <= col2_i.

Return the matrix mat after performing every query.

Example 1:

Input: n = 3, queries = [[1,1,2,2],[0,0,1,1]]
Output: [[1,1,0],[1,2,1],[0,1,1]]
Explanation: The diagram above shows the initial matrix, the matrix after the first query, and the matrix after the second query.
- In the first query, we add 1 to every element in the submatrix with the top left corner (1, 1) and bottom right corner (2, 2).
- In the second query, we add 1 to every element in the submatrix with the top left corner (0, 0) and bottom right corner (1, 1).

Example 2:

Input: n = 2, queries = [[0,0,1,1]]
Output: [[1,1],[1,1]]
Explanation: The diagram above shows the initial matrix and the matrix after the first query.
- In the first query we add 1 to every element in the matrix.

Constraints:

1 <= n <= 500
1 <= queries.length <= 10⁴
0 <= row1_i <= row2_i < n
0 <= col1_i <= col2_i < n

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer n, indicating that we initially have an n x n 0-indexed integer matrix mat filled with zeroes. You are also given a 2D integer array query. For each query[i] = [row1i, col1i, row2i, col2i], you should do the following operation: Add 1 to every element in the submatrix with the top left corner (row1i, col1i) and the bottom right corner (row2i, col2i). That is, add 1 to mat[x][y] for all row1i <= x <= row2i and col1i <= y <= col2i. Return the matrix mat after performing every query.

Baseline thinking

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

Pattern signal: Array

Example 1

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

Example 2

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

Core Insight

What unlocks the optimal approach

Imagine each row as a separate array. Instead of updating the whole submatrix together, we can use prefix sum to update each row separately.
For each query, iterate over the rows i in the range [row1, row2] and add 1 to prefix sum S[i][col1], and subtract 1 from S[i][col2 + 1].
After doing this operation for all the queries, update each row separately with S[i][j] = S[i][j] + S[i][j - 1].

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 #2536: Increment Submatrices by One
class Solution {
    public int[][] rangeAddQueries(int n, int[][] queries) {
        int[][] mat = new int[n][n];
        for (var q : queries) {
            int x1 = q[0], y1 = q[1], x2 = q[2], y2 = q[3];
            mat[x1][y1]++;
            if (x2 + 1 < n) {
                mat[x2 + 1][y1]--;
            }
            if (y2 + 1 < n) {
                mat[x1][y2 + 1]--;
            }
            if (x2 + 1 < n && y2 + 1 < n) {
                mat[x2 + 1][y2 + 1]++;
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i > 0) {
                    mat[i][j] += mat[i - 1][j];
                }
                if (j > 0) {
                    mat[i][j] += mat[i][j - 1];
                }
                if (i > 0 && j > 0) {
                    mat[i][j] -= mat[i - 1][j - 1];
                }
            }
        }
        return mat;
    }
}

// Accepted solution for LeetCode #2536: Increment Submatrices by One
func rangeAddQueries(n int, queries [][]int) [][]int {
	mat := make([][]int, n)
	for i := range mat {
		mat[i] = make([]int, n)
	}
	for _, q := range queries {
		x1, y1, x2, y2 := q[0], q[1], q[2], q[3]
		mat[x1][y1]++
		if x2+1 < n {
			mat[x2+1][y1]--
		}
		if y2+1 < n {
			mat[x1][y2+1]--
		}
		if x2+1 < n && y2+1 < n {
			mat[x2+1][y2+1]++
		}
	}
	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			if i > 0 {
				mat[i][j] += mat[i-1][j]
			}
			if j > 0 {
				mat[i][j] += mat[i][j-1]
			}
			if i > 0 && j > 0 {
				mat[i][j] -= mat[i-1][j-1]
			}
		}
	}
	return mat
}

# Accepted solution for LeetCode #2536: Increment Submatrices by One
class Solution:
    def rangeAddQueries(self, n: int, queries: List[List[int]]) -> List[List[int]]:
        mat = [[0] * n for _ in range(n)]
        for x1, y1, x2, y2 in queries:
            mat[x1][y1] += 1
            if x2 + 1 < n:
                mat[x2 + 1][y1] -= 1
            if y2 + 1 < n:
                mat[x1][y2 + 1] -= 1
            if x2 + 1 < n and y2 + 1 < n:
                mat[x2 + 1][y2 + 1] += 1

        for i in range(n):
            for j in range(n):
                if i:
                    mat[i][j] += mat[i - 1][j]
                if j:
                    mat[i][j] += mat[i][j - 1]
                if i and j:
                    mat[i][j] -= mat[i - 1][j - 1]
        return mat

// Accepted solution for LeetCode #2536: Increment Submatrices by One
impl Solution {
    pub fn range_add_queries(n: i32, queries: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let n = n as usize;
        let mut mat = vec![vec![0; n]; n];

        for q in queries {
            let (x1, y1, x2, y2) = (q[0] as usize, q[1] as usize, q[2] as usize, q[3] as usize);
            mat[x1][y1] += 1;
            if x2 + 1 < n {
                mat[x2 + 1][y1] -= 1;
            }
            if y2 + 1 < n {
                mat[x1][y2 + 1] -= 1;
            }
            if x2 + 1 < n && y2 + 1 < n {
                mat[x2 + 1][y2 + 1] += 1;
            }
        }

        for i in 0..n {
            for j in 0..n {
                if i > 0 {
                    mat[i][j] += mat[i - 1][j];
                }
                if j > 0 {
                    mat[i][j] += mat[i][j - 1];
                }
                if i > 0 && j > 0 {
                    mat[i][j] -= mat[i - 1][j - 1];
                }
            }
        }

        mat
    }
}

// Accepted solution for LeetCode #2536: Increment Submatrices by One
function rangeAddQueries(n: number, queries: number[][]): number[][] {
    const mat: number[][] = Array.from({ length: n }, () => Array(n).fill(0));

    for (const [x1, y1, x2, y2] of queries) {
        mat[x1][y1] += 1;
        if (x2 + 1 < n) mat[x2 + 1][y1] -= 1;
        if (y2 + 1 < n) mat[x1][y2 + 1] -= 1;
        if (x2 + 1 < n && y2 + 1 < n) mat[x2 + 1][y2 + 1] += 1;
    }

    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            if (i > 0) mat[i][j] += mat[i - 1][j];
            if (j > 0) mat[i][j] += mat[i][j - 1];
            if (i > 0 && j > 0) mat[i][j] -= mat[i - 1][j - 1];
        }
    }

    return mat;
}

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.