2711

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 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

Related Topics:
Array, Hash Table, Matrix

Hints:

• Use the set to count the number of distinct elements on diagonals.

Solution 1.

// OJ: https://leetcode.com/problems/difference-of-number-of-distinct-values-on-diagonals
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(max(M, N)) extra space
class Solution {
public:
    vector<vector<int>> differenceOfDistinctValues(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size();
        vector<vector<int>> ans(M, vector<int>(N));
        auto fill = [&](int i, int j) {
            int x = i, y = j;
            unordered_map<int, int> m;
            for (; x < M && y < N; ++x, ++y) m[A[x][y]]++;
            x = i, y = j;
            unordered_set<int> s;
            for (; x < M && y < N; ++x, ++y) {
                if (--m[A[x][y]] == 0) m.erase(A[x][y]);
                ans[x][y] = abs((int)s.size() - (int)m.size());
                s.insert(A[x][y]);
            }
        };
        for (int j = 0; j < N; ++j) fill(0, j);
        for (int i = 1; i < M; ++i) fill(i, 0);
        return ans;
    }
};