What is the brute force solution for Difference of Number of Distinct Values on Diagonals?

The brute force approach for Difference of Number of Distinct Values on Diagonals is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will iterate through each cell in the grid and calculate the distinct values on the diagonals by checking all the relevant cells. This is straightforward but inefficient as it requires multiple passes for each cell. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Difference of Number of Distinct Values on Diagonals?

Difference of Number of Distinct Values on Diagonals uses the following concepts: Array, Hash Table, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Difference of Number of Distinct Values on Diagonals?

Clarify the problem statement and ask questions about edge cases. Explain your thought process while coding to demonstrate your understanding. Practice similar problems to become familiar with diagonal traversal techniques.

What are common mistakes when solving Difference of Number of Distinct Values on Diagonals?

Not considering the boundaries of the grid when traversing diagonals. Confusing the leftAbove and rightBelow definitions.

#2711

Difference of Number of Distinct Values on Diagonals

Medium

ArrayHash TableMatrixHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

In the optimal approach, we will traverse the grid only once and maintain two arrays to track the distinct counts of leftAbove and rightBelow values dynamically. This avoids redundant calculations and significantly improves efficiency.

⚙️

Algorithm

4 steps

1Step 1: Create two arrays to store the distinct counts for leftAbove and rightBelow diagonals.
2Step 2: Traverse the grid to fill the leftAbove counts in one pass.
3Step 3: Traverse the grid in reverse to fill the rightBelow counts.
4Step 4: Calculate the answer matrix using the counts from both arrays.

solution.py30 lines

1# Full working Python code
2
3def differenceOfDistinctValues(grid):
4    m, n = len(grid), len(grid[0])
5    leftAbove = [{} for _ in range(m + n - 1)]
6    rightBelow = [{} for _ in range(m + n - 1)]
7    answer = [[0] * n for _ in range(m)]
8    
9    # Fill leftAbove counts
10    for r in range(m):
11        for c in range(n):
12            diag = r - c + (n - 1)
13            leftAbove[diag][grid[r][c]] = leftAbove[diag].get(grid[r][c], 0) + 1
14    
15    # Fill rightBelow counts
16    for r in range(m - 1, -1, -1):
17        for c in range(n - 1, -1, -1):
18            diag = r - c + (n - 1)
19            rightBelow[diag][grid[r][c]] = rightBelow[diag].get(grid[r][c], 0) + 1
20    
21    # Calculate answer
22    for r in range(m):
23        for c in range(n):
24            diag = r - c + (n - 1)
25            leftCount = len(leftAbove[diag]) - (1 if grid[r][c] in leftAbove[diag] else 0)
26            rightCount = len(rightBelow[diag]) - (1 if grid[r][c] in rightBelow[diag] else 0)
27            answer[r][c] = abs(leftCount - rightCount)
28    
29    return answer
30

ℹ

Complexity note: This complexity arises because we traverse the grid a constant number of times (twice) and use additional space for storing counts of distinct elements, which is proportional to the number of diagonals.

1Understanding diagonal traversal is key to solving this problem.
2Using hash maps can help efficiently count distinct elements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.