How do you solve Number of Perfect Pairs on LeetCode?

Number of Perfect Pairs (LeetCode #3649) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting helps in efficiently finding valid pairs.

What is the time complexity of Number of Perfect Pairs?

The optimal solution for Number of Perfect Pairs runs in O(n log n) time and uses O(1) space. The sorting step takes O(n log n), and the two-pointer traversal is O(n), making it efficient overall.

What is the brute force solution for Number of Perfect Pairs?

The brute force approach for Number of Perfect Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible pair of indices to see if they satisfy the perfect pair conditions. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Number of Perfect Pairs?

Number of Perfect Pairs uses the following concepts: Array, Math, Two Pointers, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Perfect Pairs?

Explain your thought process clearly. Consider edge cases like empty arrays or single elements. Practice similar problems to build confidence.

What are common mistakes when solving Number of Perfect Pairs?

Not considering negative values correctly. Failing to sort the array before applying conditions.

#3649

Number of Perfect Pairs

Medium

ArrayMathTwo PointersSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(1)

Sort the array by absolute values and use a two-pointer technique to efficiently count valid pairs that meet the conditions.

⚙️

Algorithm

3 steps

1Step 1: Sort the array based on absolute values.
2Step 2: Use two pointers to find pairs (i, j) such that the conditions are satisfied.
3Step 3: Count valid pairs while moving the pointers appropriately.

solution.py10 lines

1def perfectPairs(nums):
2    nums.sort(key=abs)
3    count = 0
4    n = len(nums)
5    j = 0
6    for i in range(n):
7        while j < n and abs(nums[j]) <= 2 * abs(nums[i]):
8            j += 1
9        count += j - i - 1
10    return count

ℹ

Complexity note: The sorting step takes O(n log n), and the two-pointer traversal is O(n), making it efficient overall.

1Sorting helps in efficiently finding valid pairs.
2Using two pointers reduces the need for nested loops.

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