How do you solve Array of Doubled Pairs on LeetCode?

Array of Doubled Pairs (LeetCode #954) 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: The problem requires pairing numbers with their doubles, which can be efficiently managed using counting.

What is the brute force solution for Array of Doubled Pairs?

The brute force approach for Array of Doubled Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible reorderings of the array and checking if any of them satisfy the condition. This is simple but inefficient, especially for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Array of Doubled Pairs?

Array of Doubled Pairs uses the following concepts: Array, Hash Table, Greedy, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Array of Doubled Pairs?

Always think about how you can reduce the problem size, like using counting or sorting. Clarify the problem statement and constraints before jumping into coding. Discuss your thought process and approach with the interviewer to demonstrate your problem-solving skills.

What are common mistakes when solving Array of Doubled Pairs?

Assuming that the order of elements in the original array matters, while it does not. Not considering negative numbers and their doubles correctly.

#954

Array of Doubled Pairs

Medium

ArrayHash TableGreedySortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution involves using a hash map to count occurrences of each number and then checking if we can form pairs of numbers and their doubles. This is efficient and avoids the need for generating permutations.

⚙️

Algorithm

4 steps

1Step 1: Count the occurrences of each number in a hash map.
2Step 2: Sort the unique numbers in the hash map.
3Step 3: For each number, check if the double exists in the hash map and if there are enough occurrences to form pairs.
4Step 4: If all numbers can be paired successfully, return true; otherwise, return false.

solution.py10 lines

1# Full working Python code
2from collections import Counter
3
4def canReorderDoubled(arr):
5    count = Counter(arr)
6    for x in sorted(count.keys()):
7        if count[x] > count[2 * x]:
8            return False
9        count[2 * x] -= count[x]
10    return True

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the unique keys of the hash map, while the space complexity is O(n) for storing the counts.

1The problem requires pairing numbers with their doubles, which can be efficiently managed using counting.
2Sorting the unique numbers helps ensure that we handle smaller numbers first, which is crucial for pairing.

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