How do you solve Sum of GCD of Formed Pairs on LeetCode?

Sum of GCD of Formed Pairs (LeetCode #3867) 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: Prefix GCDs depend on the maximum value seen so far.

What is the time complexity of Sum of GCD of Formed Pairs?

The optimal solution for Sum of GCD of Formed Pairs runs in O(n log n) time and uses O(n) space. The dominant factor is sorting the prefixGcd array, which is O(n log n). The prefix GCD computation is O(n).

What is the brute force solution for Sum of GCD of Formed Pairs?

The brute force approach for Sum of GCD of Formed Pairs is Brute Force, with O(n²) time complexity and O(n) space complexity. This approach calculates the prefix GCD for each element and then sorts the array. It pairs elements from both ends of the sorted array to compute GCDs, leading to a high time complexity. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Sum of GCD of Formed Pairs?

Explain your thought process clearly. Discuss edge cases, such as empty arrays or single elements. Practice coding on a whiteboard or in a shared editor.

What are common mistakes when solving Sum of GCD of Formed Pairs?

Forgetting to handle odd-length arrays correctly. Not maintaining the maximum value during prefix GCD computation.

#3867

Sum of GCD of Formed Pairs

Medium

ArrayMathTwo PointersSimulationNumber TheoryPrefix SumSorting

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)

This approach leverages a single pass to compute prefix GCDs while maintaining the maximum value, followed by sorting and pairing, resulting in a more efficient solution.

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Algorithm

3 steps

1Step 1: Initialize mx and compute prefixGcd in a single pass over nums.
2Step 2: Sort the prefixGcd array.
3Step 3: Pair elements from both ends of the sorted array to compute GCDs.

solution.py10 lines

1def sumGCD(nums):
2    n = len(nums)
3    prefixGcd = []
4    mx = 0
5    for num in nums:
6        mx = max(mx, num)
7        prefixGcd.append(gcd(num, mx))
8    prefixGcd.sort()
9    total = sum(gcd(prefixGcd[i], prefixGcd[n - 1 - i]) for i in range(n // 2))
10    return total

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Complexity note: The dominant factor is sorting the prefixGcd array, which is O(n log n). The prefix GCD computation is O(n).

1Prefix GCDs depend on the maximum value seen so far.
2Sorting allows efficient pairing of elements.

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