LeetCode-in-Java

2711. Difference of Number of Distinct Values on Diagonals

Medium

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 1^st diagram denotes the initial grid.

The 2^nd diagram denotes a grid for cell (0,0), where blue-colored cells are cells on its bottom-right diagonal.

The 3^rd diagram denotes a grid for cell (1,2), where red-colored cells are cells on its top-left diagonal.

The 4^th 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

Solution

import java.util.HashSet;
import java.util.Set;

public class Solution {
    public int[][] differenceOfDistinctValues(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] arrTopLeft = new int[m][n];
        int[][] arrBotRight = new int[m][n];

        for (int i = m - 1; i >= 0; i--) {
            int c = 0;
            int r = i;
            Set<Integer> set = new HashSet<>();
            while (cellExists(r, c, grid)) {
                arrTopLeft[r][c] = set.size();
                set.add(grid[r++][c++]);
            }
        }

        for (int i = 1; i < n; i++) {
            int r = 0;
            int c = i;
            Set<Integer> set = new HashSet<>();
            while (cellExists(r, c, grid)) {
                arrTopLeft[r][c] = set.size();
                set.add(grid[r++][c++]);
            }
        }

        for (int i = 0; i < n; i++) {
            int r = m - 1;
            int c = i;
            Set<Integer> set = new HashSet<>();
            while (cellExists(r, c, grid)) {
                arrBotRight[r][c] = set.size();
                set.add(grid[r--][c--]);
            }
        }

        for (int i = m - 1; i >= 0; i--) {
            int c = n - 1;
            int r = i;
            Set<Integer> set = new HashSet<>();
            while (cellExists(r, c, grid)) {
                arrBotRight[r][c] = set.size();
                set.add(grid[r--][c--]);
            }
        }

        int[][] result = new int[m][n];
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                result[r][c] = Math.abs(arrTopLeft[r][c] - arrBotRight[r][c]);
            }
        }

        return result;
    }

    private boolean cellExists(int r, int c, int[][] grid) {
        return r >= 0 && r < grid.length && c >= 0 && c < grid[0].length;
    }
}

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