What is the time complexity of Maximum Product of Three Elements After One Replacement?

The optimal solution for Maximum Product of Three Elements After One Replacement runs in O(n log n) time and uses O(1) space. Sorting the array takes O(n log n), while finding the maximum product is O(1).

What is the brute force solution for Maximum Product of Three Elements After One Replacement?

The brute force approach for Maximum Product of Three Elements After One Replacement is Brute Force, with O(n²) time complexity and O(1) space complexity. Try replacing each element with the maximum possible value and calculate the product of the three largest elements. This approach checks all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Product of Three Elements After One Replacement?

Maximum Product of Three Elements After One Replacement uses the following concepts: Array, Math, Greedy, Sorting. The recommended patterns to study are: Sorting, Greedy.

What are the interview tips for Maximum Product of Three Elements After One Replacement?

Clarify the range of replacement values. Discuss edge cases like all negatives or zeros. Explain your thought process while coding.

What are common mistakes when solving Maximum Product of Three Elements After One Replacement?

Overlooking the case of negative numbers. Not considering the impact of replacing different elements.

#3732

Maximum Product of Three Elements After One Replacement

Medium

ArrayMathGreedySortingSortingGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

Identify the two largest and the two smallest numbers. Replace the smallest with 10^5 to maximize the product.

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Algorithm

3 steps

1Step 1: Find the two largest and two smallest numbers in the array.
2Step 2: Calculate the product of the two largest numbers with 10^5.
3Step 3: Calculate the product of the two smallest numbers with 10^5 (for negative scenarios) and return the maximum of both products.

solution.py3 lines

1def maxProduct(nums):
2    nums.sort()
3    return max(nums[-1] * nums[-2] * 100000, nums[0] * nums[1] * 100000)

ℹ

Complexity note: Sorting the array takes O(n log n), while finding the maximum product is O(1).

1Replacing with 10^5 can maximize product when combined with large negatives.
2Sorting helps quickly identify the largest and smallest values.

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