What is the time complexity of Maximum Product of First and Last Elements of a Subsequence?

The optimal solution for Maximum Product of First and Last Elements of a Subsequence runs in O(n) time and uses O(1) space. This approach runs in linear time by iterating through the array and calculating products dynamically.

What is the brute force solution for Maximum Product of First and Last Elements of a Subsequence?

The brute force approach for Maximum Product of First and Last Elements of a Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subsequences of size m and calculate the product of the first and last elements. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Product of First and Last Elements of a Subsequence?

Maximum Product of First and Last Elements of a Subsequence uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Two Pointers, Dynamic Programming.

What are the interview tips for Maximum Product of First and Last Elements of a Subsequence?

Explain your thought process clearly. Discuss edge cases, like all negative numbers. Practice similar problems to build intuition.

What are common mistakes when solving Maximum Product of First and Last Elements of a Subsequence?

Forgetting to check bounds when selecting subsequences. Not considering negative numbers affecting the product.

#3584

Maximum Product of First and Last Elements of a Subsequence

Medium

ArrayTwo PointersTwo PointersDynamic Programming

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Instead of generating all subsequences, we can find the maximum product by iterating through the array and maintaining the maximum and minimum values dynamically.

⚙️

Algorithm

3 steps

1Step 1: Initialize maxProduct to a very small value.
2Step 2: Iterate through the array, considering each element as a potential first element of a subsequence.
3Step 3: For each valid first element, calculate the product with the maximum last element from the remaining elements.

solution.py7 lines

1def maxProduct(nums, m):
2    maxProduct = float('-inf')
3    n = len(nums)
4    for i in range(n - m + 1):
5        maxLast = max(nums[i + m - 1:])
6        maxProduct = max(maxProduct, nums[i] * maxLast)
7    return maxProduct

ℹ

Complexity note: This approach runs in linear time by iterating through the array and calculating products dynamically.

1Maximize the first element and find the best possible last element.
2Consider both negative and positive values for maximum product.

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