#2711

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Difference of Number of Distinct Values on Diagonals

Find the difference in the number of distinct diagonal values in a matrix, returning results in a new matrix.

Problem Statement

You are given a 2D matrix of size m x n. For each cell in the matrix, you need to calculate the difference between the number of distinct values on its diagonal compared to the main diagonal. A diagonal is defined as a line of cells starting from any cell in the topmost row or the leftmost column, extending bottom-right until the end of the matrix.

Return a new matrix of the same size, where each cell contains the calculated difference of distinct diagonal values for that position. You must focus on efficient array scanning and hash lookups to solve the problem.

Examples

Example 1

Input: grid = [[1,2,3],[3,1,5],[3,2,1]]

Output: Output: [[1,1,0],[1,0,1],[0,1,1]]

To calculate the answer cells:

Example 2

Input: grid = [[1]]

Output: Output: [[0]]

Example details omitted.

Constraints

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

Solution Approach

Array Scanning

Iterate over the matrix and for each element, scan its diagonal starting from that element. This ensures you capture all relevant diagonal values for each cell.

Hash Set for Distinct Values

Use a hash set to keep track of the distinct values encountered on the diagonal. This enables efficient counting and ensures uniqueness.

Efficient Calculation

For each cell, calculate the number of distinct diagonal values by comparing the hash set size and store the result in the output matrix.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach chosen for scanning the diagonals. If each diagonal is scanned independently, the time complexity is O(m * n). The space complexity is also O(m * n) because of the additional storage required for the hash set and the output matrix.

What Interviewers Usually Probe

Ability to optimize matrix traversal with hash sets for distinct values.
Understanding of how to use efficient data structures like hash sets in matrix-related problems.
Experience with solving problems involving matrix diagonals and distinct value counting.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the problem by failing to recognize the need for efficient scanning.
Misunderstanding the concept of diagonals, leading to incorrect traversal logic.
Failing to handle edge cases such as single-row or single-column matrices.

Follow-up variants

Modify the problem to handle larger matrices efficiently.
Consider cases where the diagonals can be in different directions.
Optimize the approach for matrices with larger values or constraints.

FAQ

How do I calculate the difference of distinct values on diagonals?

You can calculate the difference by scanning each cell's diagonal, storing values in a hash set to count distinct elements, and comparing the results.

What pattern does the 'Difference of Number of Distinct Values on Diagonals' problem follow?

The problem follows the array scanning plus hash lookup pattern, where you traverse the matrix and use hash sets to count distinct values efficiently.

How do I optimize this problem?

Optimize by using hash sets to track distinct values on each diagonal and by minimizing redundant scans of the matrix.

What if the matrix is very large?

For larger matrices, optimize the diagonal scanning logic and consider using more advanced data structures to handle larger input sizes efficiently.

Are there any edge cases for this problem?

Yes, edge cases include matrices with only one row or column, which should be handled with care to avoid incorrect diagonal scanning.

terminal

Solution

Solution 1: Simulation

We can simulate the process described in the problem statement, calculating the number of distinct values on the top-left diagonal $tl$ and the bottom-right diagonal $br$ for each cell, then compute their difference $|tl - br|$.

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class Solution:
    def differenceOfDistinctValues(self, grid: List[List[int]]) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        ans = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                x, y = i, j
                s = set()
                while x and y:
                    x, y = x - 1, y - 1
                    s.add(grid[x][y])
                tl = len(s)
                x, y = i, j
                s = set()
                while x + 1 < m and y + 1 < n:
                    x, y = x + 1, y + 1
                    s.add(grid[x][y])
                br = len(s)
                ans[i][j] = abs(tl - br)
        return ans

Continue Practicing

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