Increment Submatrices by One
Problem Statement - link #
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:
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 <= query.length <= 10000
0 <= row1i <= row2i < n
0 <= col1i <= col2i < n
Solutions #
class Solution {
public:
vector<vector<int>> rangeAddQueries(int n, vector<vector<int>>& queries) {
vector<vector<int>> arr(n, vector<int> (n, 0));
int r1, r2, c1, c2;
for(auto &it: queries) {
r1 = it[0], r2 = it[2], c1 = it[1], c2 = it[3];
for(int i = r1; i <= r2; i++) {
arr[i][c1] += 1;
if((c2 + 1) >= n) {
if(i + 1 < n) {
arr[i + 1][0] -= 1;
}
else {
continue;
}
}
else {
arr[i][c2 + 1] -= 1;
}
}
}
int count = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
count += arr[i][j];
arr[i][j] = count;
}
}
return arr;
}
};