How do you solve Divide Array Into Equal Pairs on LeetCode?

Divide Array Into Equal Pairs (LeetCode #2206) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Counting occurrences allows for a quick check on pairing feasibility.

What is the time complexity of Divide Array Into Equal Pairs?

The optimal solution for Divide Array Into Equal Pairs runs in O(n) time and uses O(n) space. This complexity is linear because we only traverse the array a couple of times: once for counting and once for checking counts.

What is the brute force solution for Divide Array Into Equal Pairs?

The brute force approach for Divide Array Into Equal Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible pairs to see if they can be formed. This is straightforward but inefficient, as it may require checking many combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Divide Array Into Equal Pairs?

Divide Array Into Equal Pairs uses the following concepts: Array, Hash Table, Bit Manipulation, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Divide Array Into Equal Pairs?

Always think about counting elements when dealing with pairs or groups. Check if the problem can be simplified by using data structures like HashMaps. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Divide Array Into Equal Pairs?

Not considering edge cases where numbers appear an odd number of times. Overcomplicating the problem by trying to form pairs directly instead of counting.

#2206

Divide Array Into Equal Pairs

Easy

ArrayHash TableBit ManipulationCountingHash MapArray

LeetCode ↗

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)

The optimal solution leverages counting the occurrences of each number. If every number appears an even number of times, we can form pairs. This is efficient and straightforward.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency count of each number in the array.
2Step 2: Iterate through the frequency count and check if each count is even.
3Step 3: If all counts are even, return true; otherwise, return false.

solution.py9 lines

1# Full working Python code
2
3def can_form_pairs(nums):
4    from collections import Counter
5    count = Counter(nums)
6    for freq in count.values():
7        if freq % 2 != 0:
8            return False
9    return True

ℹ

Complexity note: This complexity is linear because we only traverse the array a couple of times: once for counting and once for checking counts.

1Counting occurrences allows for a quick check on pairing feasibility.
2Even counts are essential for forming pairs.

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