# 2536. Increment Submatrices by One

## Description

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.

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 <= row1i <= row2i < n
• 0 <= col1i <= col2i < n

## Solutions

• 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;
}
}

• class Solution {
public:
vector<vector<int>> rangeAddQueries(int n, vector<vector<int>>& queries) {
vector<vector<int>> mat(n, vector<int>(n));
for (auto& 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;
}
};

• 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


• 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
}