LeetCode 2711: Difference of Number of Distinct Values on Diagonals Solution

Master LeetCode problem 2711 (Difference of Number of Distinct Values on Diagonals), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2711. Difference of Number of Distinct Values on Diagonals

Medium

Array Hash Table Matrix

Problem Explanation

Explanation:

To solve this problem, we need to iterate through each cell of the given grid and calculate the number of distinct values on the diagonals to the left and above the cell (leftAbove) and on the diagonals to the right and below the cell (rightBelow). We then compute the absolute difference between these two values to get the answer for each cell.

Create a new matrix answer of the same size as the input grid.
For each cell in the grid:
- Calculate the distinct values on the left-above diagonal (leftAbove) and right-below diagonal (rightBelow).
- Compute the absolute difference between leftAbove and rightBelow to get the answer for that cell.
Return the answer matrix.

Time Complexity: O(mn) where m is the number of rows and n is the number of columns in the grid. Space Complexity: O(mn) for the answer matrix.

Solution Code

class Solution {
    public int[][] diagonalDifference(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] answer = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                Set<Integer> leftAbove = new HashSet<>();
                Set<Integer> rightBelow = new HashSet<>();
                
                int r = i - 1, c = j - 1;
                while (r >= 0 && c >= 0) {
                    leftAbove.add(grid[r][c]);
                    r--;
                    c--;
                }
                
                r = i + 1;
                c = j + 1;
                while (r < m && c < n) {
                    rightBelow.add(grid[r][c]);
                    r++;
                    c++;
                }
                
                answer[i][j] = Math.abs(leftAbove.size() - rightBelow.size());
            }
        }
        
        return answer;
    }
}

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