LeetCode 2536. Increment Submatrices by One Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: 2D Difference Array

A 2D difference array is a technique used to efficiently handle range updates on 2D arrays. We can implement fast updates on submatrices by maintaining a difference matrix of the same size as the original matrix.

Suppose we have a 2D difference matrix diff, initially with all elements set to 0. For each query [row1, col1, row2, col2], we can update the difference matrix through the following steps:

1. Increment position (row1, col1) by 1.
2. Decrement position (row2 + 1, col1) by 1, provided that row2 + 1 < n.
3. Decrement position (row1, col2 + 1) by 1, provided that col2 + 1 < n.
4. Increment position (row2 + 1, col2 + 1) by 1, provided that row2 + 1 < n and col2 + 1 < n.

After completing all queries, we need to convert the difference matrix back to the original matrix using prefix sums. That is, for each position (i, j), we calculate:

mat[i][j] = diff[i][j] + (mat[i-1][j] if i > 0 else 0) + (mat[i][j-1] if j > 0 else 0) - (mat[i-1][j-1] if i > 0 and j > 0 else 0)

The time complexity is O(m + n²), where m and n are the length of queries and the given n, respectively. Ignoring the space consumed by the answer, the space complexity is O(1).

PythonJavaC++GoTypeScriptRustC#

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

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 } }(code-box)

public class Solution { public int[][] RangeAddQueries(int n, int[][] queries) { int[][] mat = new int[n][]; for (int i = 0; i < n; i++) { mat[i] = new int[n]; } foreach (var q in 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; } }(code-box)