2711. Difference of Number of Distinct Values on Diagonals

Problem

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

Solution (Java)

class Solution {
    public int[][] differenceOfDistinctValues(int[][] grid) {
               int m = grid.length;
        int n = grid[0].length;
        int[][] res=new int[m][n];
        for(int i=0;i<grid.length;i++){
            for(int j=0;j<grid[0].length;j++){
                res[i][j]=calculate(grid,i,j);
            }
        }
        return res;
    }

    int calculate(int[][] arr,int i,int j){
        int r1=i;
        int r2=i;
        int c1=j;
        int c2=j;

        HashSet<Integer> hs1=new HashSet(); // to store distinct values of upper diagonal
        HashSet<Integer> hs2=new HashSet(); // to store distinct values of lower diagonal

        // to store distinct values of upper diagonal
        while(r1>0 && c1>0){
            hs1.add(arr[--r1][--c1]);
        }

        // to store distinct values of lower diagonal
        while(r2<arr.length-1 && c2<arr[0].length-1){
            hs2.add(arr[++r2][++c2]);
        }
        return Math.abs(hs1.size()-hs2.size());
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).