对角线上不同值的数量差

Given a 0-indexed 2D grid of size m x n, you should find the matrix answer of size m x n.

The value of each cell (r, c) of the matrix answer is calculated in the following way:

• Let topLeft[r][c] be the number of distinct values in the top-left diagonal of the cell (r, c) in the matrix grid.
• Let bottomRight[r][c] be the number of distinct values in the bottom-right diagonal of the cell (r, c) in the matrix grid.

Then answer[r][c] = |topLeft[r][c] - bottomRight[r][c]|.

Return the matrix answer.

A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end.

A cell (r1, c1) belongs to the top-left diagonal of the cell (r, c), if both belong to the same diagonal and r1 < r. Similarly is defined bottom-right diagonal.

 

Example 1:

Input: grid = [[1,2,3],[3,1,5],[3,2,1]]
Output: [[1,1,0],[1,0,1],[0,1,1]]
Explanation: The 1st diagram denotes the initial grid. 
The 2nd diagram denotes a grid for cell (0,0), where blue-colored cells are cells on its bottom-right diagonal.
The 3rd diagram denotes a grid for cell (1,2), where red-colored cells are cells on its top-left diagonal.
The 4th diagram denotes a grid for cell (1,1), where blue-colored cells are cells on its bottom-right diagonal and red-colored cells are cells on its top-left diagonal.
- The cell (0,0) contains [1,1] on its bottom-right diagonal and [] on its top-left diagonal. The answer is |1 - 0| = 1.
- The cell (1,2) contains [] on its bottom-right diagonal and [2] on its top-left diagonal. The answer is |0 - 1| = 1.
- The cell (1,1) contains [1] on its bottom-right diagonal and [1] on its top-left diagonal. The answer is |1 - 1| = 0.
The answers of other cells are similarly calculated.

Example 2:

Input: grid = [[1]]
Output: [[0]]
Explanation: - The cell (0,0) contains [] on its bottom-right diagonal and [] on its top-left diagonal. The answer is |0 - 0| = 0.

 

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n, grid[i][j] <= 50

/**
 * Problem description:
 * Given a 2D matrix `grid` of size m x n, return a matrix `answer` of the same size.
 * For each cell (r, c) in `answer`, the value is defined as the absolute difference between
 * the number of distinct values in the top-left diagonal and the bottom-right diagonal in `grid`.
 *
 * Solution (using Java Streams):
 * 1. Stream through each cell and calculate the distinct values in the top-left and bottom-right diagonals.
 * 2. Store the results in the `answer` matrix by taking the absolute difference.
 */
import java.util.*;
import java.util.stream.*;

public class Solution {
    public int[][] diagonalDifference(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] answer = new int[m][n];

        IntStream.range(0, m).forEach(r ->
            IntStream.range(0, n).forEach(c -> {
                // Calculate distinct count in top-left diagonal
                long topLeft = IntStream.iterate(r, i -> i - 1)
                    .limit(Math.min(r, c) + 1)
                    .mapToObj(i -> grid[i][c - (r - i)])
                    .distinct()
                    .count();

                // Calculate distinct count in bottom-right diagonal
                long bottomRight = IntStream.iterate(r, i -> i + 1)
                    .limit(Math.min(m - r, n - c))
                    .mapToObj(i -> grid[i][c + (i - r)])
                    .distinct()
                    .count();

                // Calculate the answer for cell (r, c)
                answer[r][c] = (int) Math.abs(topLeft - bottomRight);
            })
        );

        return answer;
    }
}

此题解方法通过遍历矩阵中的所有对角线来计算左上角和右下角的对角线上的不同值数量。首先，使用四个独立的循环来分别处理矩阵的四个边界开始的对角线：首先是从上边界开始的左上到右下的对角线（两个循环），接着是从右边界和底边界开始的右下到左上的对角线（两个循环）。每个循环中使用一个集合来存储遍历过程中遇到的不同的值，并在每个单元格更新answer矩阵中的对应值。最后，根据左上和右下的值计算其差的绝对值。

O((m + n) * min(m, n))

O(min(m, n))