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

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 \(\textit{diff}\), initially with all elements set to \(0\). For each query \([\textit{row1}, \textit{col1}, \textit{row2}, \textit{col2}]\), we can update the difference matrix through the following steps:

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

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:

\[ \textit{mat}[i][j] = \textit{diff}[i][j] + (\textit{mat}[i-1][j] \text{ if } i > 0 \text{ else } 0) + (\textit{mat}[i][j-1] \text{ if } j > 0 \text{ else } 0) - (\textit{mat}[i-1][j-1] \text{ if } i > 0 \text{ and } j > 0 \text{ else } 0) \]

The time complexity is \(O(m + n^2)\), where \(m\) and \(n\) are the length of \(\textit{queries}\) and the given \(n\), respectively. Ignoring the space consumed by the answer, the space complexity is \(O(1)\).

Python3JavaC++GoTypeScriptRust

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

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;
    }
};

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
}

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;
}

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
    }
}

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