Leetcode – Maximum Product of Two Elements in an Array Solution

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One of the challenges you may encounter on Leetcode is the “Maximum Product of Two Elements in an Array” problem, a question focused on array manipulation and mathematical operations. This problem is categorized under array algorithms and tests your understanding of basic combinatorial mathematics and array operations.

In this comprehensive guide, we will delve into:

  1. Problem Statement
  2. Initial Thoughts and Conceptual Approach
  3. Optimized Approach
  4. Python Implementation
  5. Test Cases and Edge Cases
  6. Time and Space Complexity Analysis
  7. Conclusion

1. Problem Statement

You are given an integer array nums. You need to find two distinct indices i and j such that the product (nums[i]−1)×(nums[j]−1) is maximized. Return the maximum value of this expression.

Example 1:

Input: nums = [3, 4, 5, 2]
Output: 12
Explanation: (5−1)×(4−1)=12

Example 2:

Input: nums = [1, 5, 4, 5]
Output: 16
Explanation: (5−1)×(5−1)=16

2. Initial Thoughts and Conceptual Approach

A naive approach would be to calculate the product for every possible pair of numbers in the array and return the maximum product. This approach would involve nested loops and have a time complexity of O(n^2).

3. Optimized Approach

You can optimize this by sorting the array and then calculating the product of the largest two numbers, both of which should be at the end of the sorted array. The time complexity of sorting an array is O(n log ⁡n), which is better than the O(n^2) time complexity of the naive approach.

4. Python Implementation

Naive Implementation

def maxProduct(nums):
    max_prod = 0
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            max_prod = max(max_prod, (nums[i] - 1) * (nums[j] - 1))
    return max_prod

Optimized Implementation

def maxProduct(nums):
    nums.sort()
    return (nums[-1] - 1) * (nums[-2] - 1)

5. Test Cases and Edge Cases

Now, let’s consider some test cases and edge cases to ensure the implementation is robust:

# Test 1
assert maxProduct([3, 4, 5, 2]) == 12
# Test 2
assert maxProduct([1, 5, 4, 5]) == 16
# Test 3 (Edge Case)
assert maxProduct([10, 2, 5, 9]) == 72
# Test 4 (Edge Case)
assert maxProduct([10, 10]) == 81

6. Time and Space Complexity Analysis

Time Complexity

The naive implementation has a time complexity of O(n^2), whereas the optimized implementation reduces it to O(n log ⁡n) due to sorting.

Space Complexity

Both approaches have a constant space complexity of O(1), as we are not using any additional data structures that scale with the input.

7. Conclusion

The “Maximum Product of Two Elements in an Array” problem serves as an excellent example to demonstrate the importance of optimized algorithms. The naive O(n^2) approach may work well for small datasets but will struggle to scale with larger inputs. The optimized O(n log ⁡n) algorithm solves the problem efficiently by utilizing array sorting.

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