Welcome to Subscribe On Youtube
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, add1
tomat[x][y]
for allrow1i <= x <= row2i
andcol1i <= 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 }