How do you solve Diagonal Traverse II on LeetCode?

Diagonal Traverse II (LeetCode #1424) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Elements on the same diagonal share the same sum of their row and column indices.

What is the brute force solution for Diagonal Traverse II?

The brute force approach for Diagonal Traverse II is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves iterating through each element in the 2D array and grouping them based on their diagonal indices. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Diagonal Traverse II?

Diagonal Traverse II uses the following concepts: Array, Sorting, Heap (Priority Queue). The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Diagonal Traverse II?

Always clarify the problem constraints and edge cases before diving into coding. Think about how you can optimize your initial solution as you code. Practice explaining your thought process clearly as you work through the problem.

What are common mistakes when solving Diagonal Traverse II?

Not considering varying row lengths in the 2D array. Failing to handle the sorting of diagonal elements correctly.

#1424

Diagonal Traverse II

Medium

ArraySortingHeap (Priority Queue)Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

The optimal solution leverages a priority queue to efficiently group and sort the diagonal elements in one pass, reducing the need for multiple iterations.

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Algorithm

3 steps

1Step 1: Initialize a priority queue to store tuples of (sum, row, value).
2Step 2: Iterate through the 2D array and for each element, calculate its diagonal index and push it into the priority queue.
3Step 3: Extract elements from the priority queue in order of their diagonal indices and append them to the result list.

solution.py12 lines

1import heapq
2
3def findDiagonalOrder(nums):
4    pq = []
5    for r in range(len(nums)):
6        for c in range(len(nums[r])):
7            heapq.heappush(pq, (r + c, r, nums[r][c]))
8    result = []
9    while pq:
10        _, r, val = heapq.heappop(pq)
11        result.append(val)
12    return result

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of the priority queue, where n is the total number of elements. The space complexity remains O(n) for storing the elements.

1Elements on the same diagonal share the same sum of their row and column indices.
2Using a priority queue helps in efficiently sorting elements based on their diagonal indices.

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