Problem

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.

Examples

Example 1

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![](https://assets.leetcode.com/uploads/2022/11/24/p2example11.png)

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

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![](https://assets.leetcode.com/uploads/2022/11/24/p2example22.png)

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 <= 10^4
  • 0 <= row1i <= row2i < n
  • 0 <= col1i <= col2i < n

Solution

Method 1 – 2D Prefix Sum (Imos Method)

Intuition

Directly incrementing all elements in each submatrix for every query is too slow. Instead, we use a 2D prefix sum trick (Imos method) to efficiently apply all increments, then compute the final matrix in one pass.

Approach

  1. Create a 2D difference matrix diff of size (n+1) x (n+1), initialized to 0.
  2. For each query [r1, c1, r2, c2], update the corners of diff to mark the increment region:
    • diff[r1][c1] += 1
    • diff[r1][c2+1] -= 1
    • diff[r2+1][c1] -= 1
    • diff[r2+1][c2+1] += 1
  3. After all queries, compute the prefix sums row-wise and then column-wise to get the final matrix.
  4. Return the resulting matrix.

Code

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vector<vector<int>> rangeAddQueries(int n, vector<vector<int>>& queries) {
    vector<vector<int>> diff(n+1, vector<int>(n+1));
    for (auto& q : queries) {
        int r1 = q[0], c1 = q[1], r2 = q[2], c2 = q[3];
        diff[r1][c1]++;
        diff[r1][c2+1]--;
        diff[r2+1][c1]--;
        diff[r2+1][c2+1]++;
    }
    vector<vector<int>> ans(n, vector<int>(n));
    for (int i = 0; i < n; ++i)
        for (int j = 1; j < n; ++j)
            diff[i][j] += diff[i][j-1];
    for (int j = 0; j < n; ++j)
        for (int i = 1; i < n; ++i)
            diff[i][j] += diff[i-1][j];
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
            ans[i][j] = diff[i][j];
    return ans;
}
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func rangeAddQueries(n int, queries [][]int) [][]int {
    diff := make([][]int, n+1)
    for i := range diff {
        diff[i] = make([]int, n+1)
    }
    for _, q := range queries {
        r1, c1, r2, c2 := q[0], q[1], q[2], q[3]
        diff[r1][c1]++
        diff[r1][c2+1]--
        diff[r2+1][c1]--
        diff[r2+1][c2+1]++
    }
    for i := 0; i < n; i++ {
        for j := 1; j < n; j++ {
            diff[i][j] += diff[i][j-1]
        }
    }
    for j := 0; j < n; j++ {
        for i := 1; i < n; i++ {
            diff[i][j] += diff[i-1][j]
        }
    }
    ans := make([][]int, n)
    for i := 0; i < n; i++ {
        ans[i] = make([]int, n)
        for j := 0; j < n; j++ {
            ans[i][j] = diff[i][j]
        }
    }
    return ans
}
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class Solution {
    public int[][] rangeAddQueries(int n, int[][] queries) {
        int[][] diff = new int[n+1][n+1];
        for (int[] q : queries) {
            int r1 = q[0], c1 = q[1], r2 = q[2], c2 = q[3];
            diff[r1][c1]++;
            diff[r1][c2+1]--;
            diff[r2+1][c1]--;
            diff[r2+1][c2+1]++;
        }
        for (int i = 0; i < n; ++i)
            for (int j = 1; j < n; ++j)
                diff[i][j] += diff[i][j-1];
        for (int j = 0; j < n; ++j)
            for (int i = 1; i < n; ++i)
                diff[i][j] += diff[i-1][j];
        int[][] ans = new int[n][n];
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n; ++j)
                ans[i][j] = diff[i][j];
        return ans;
    }
}
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fun rangeAddQueries(n: Int, queries: Array<IntArray>): Array<IntArray> {
    val diff = Array(n+1) { IntArray(n+1) }
    for (q in queries) {
        val (r1, c1, r2, c2) = q
        diff[r1][c1]++
        diff[r1][c2+1]--
        diff[r2+1][c1]--
        diff[r2+1][c2+1]++
    }
    for (i in 0 until n)
        for (j in 1 until n)
            diff[i][j] += diff[i][j-1]
    for (j in 0 until n)
        for (i in 1 until n)
            diff[i][j] += diff[i-1][j]
    val ans = Array(n) { IntArray(n) }
    for (i in 0 until n)
        for (j in 0 until n)
            ans[i][j] = diff[i][j]
    return ans
}
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def rangeAddQueries(n: int, queries: list[list[int]]) -> list[list[int]]:
    diff = [[0]*(n+1) for _ in range(n+1)]
    for r1, c1, r2, c2 in queries:
        diff[r1][c1] += 1
        diff[r1][c2+1] -= 1
        diff[r2+1][c1] -= 1
        diff[r2+1][c2+1] += 1
    for i in range(n):
        for j in range(1, n):
            diff[i][j] += diff[i][j-1]
    for j in range(n):
        for i in range(1, n):
            diff[i][j] += diff[i-1][j]
    ans = [[diff[i][j] for j in range(n)] for i in range(n)]
    return ans
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fn range_add_queries(n: i32, queries: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
    let n = n as usize;
    let mut diff = vec![vec![0; n+1]; n+1];
    for q in queries {
        let (r1, c1, r2, c2) = (q[0] as usize, q[1] as usize, q[2] as usize, q[3] as usize);
        diff[r1][c1] += 1;
        diff[r1][c2+1] -= 1;
        diff[r2+1][c1] -= 1;
        diff[r2+1][c2+1] += 1;
    }
    for i in 0..n {
        for j in 1..n {
            diff[i][j] += diff[i][j-1];
        }
    }
    for j in 0..n {
        for i in 1..n {
            diff[i][j] += diff[i-1][j];
        }
    }
    let mut ans = vec![vec![0; n]; n];
    for i in 0..n {
        for j in 0..n {
            ans[i][j] = diff[i][j];
        }
    }
    ans
}
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function rangeAddQueries(n: number, queries: number[][]): number[][] {
  const diff = Array.from({length: n+1}, () => Array(n+1).fill(0));
  for (const [r1, c1, r2, c2] of queries) {
    diff[r1][c1]++;
    diff[r1][c2+1]--;
    diff[r2+1][c1]--;
    diff[r2+1][c2+1]++;
  }
  for (let i = 0; i < n; ++i)
    for (let j = 1; j < n; ++j)
      diff[i][j] += diff[i][j-1];
  for (let j = 0; j < n; ++j)
    for (let i = 1; i < n; ++i)
      diff[i][j] += diff[i-1][j];
  const ans = Array.from({length: n}, () => Array(n));
  for (let i = 0; i < n; ++i)
    for (let j = 0; j < n; ++j)
      ans[i][j] = diff[i][j];
  return ans;
}

Complexity

  • ⏰ Time complexity: O(q + n^2) — q is the number of queries, n is the matrix size. Each query is O(1), and prefix sum is O(n^2).
  • 🧺 Space complexity: O(n^2) — For the difference and answer matrices.