Maximum Product Difference: NLogN

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Problem can be solved in NLogN with 2 lines: sort and take the extremes. That's it. Problem and code are below, cheers, ACC.

Maximum Product Difference Between Two Pairs - LeetCode

1913. Maximum Product Difference Between Two Pairs
Easy

The product difference between two pairs (a, b) and (c, d) is defined as (a * b) - (c * d).

  • For example, the product difference between (5, 6) and (2, 7) is (5 * 6) - (2 * 7) = 16.

Given an integer array nums, choose four distinct indices w, x, y, and z such that the product difference between pairs (nums[w], nums[x]) and (nums[y], nums[z]) is maximized.

Return the maximum such product difference.

 

Example 1:

Input: nums = [5,6,2,7,4]
Output: 34
Explanation: We can choose indices 1 and 3 for the first pair (6, 7) and indices 2 and 4 for the second pair (2, 4).
The product difference is (6 * 7) - (2 * 4) = 34.

Example 2:

Input: nums = [4,2,5,9,7,4,8]
Output: 64
Explanation: We can choose indices 3 and 6 for the first pair (9, 8) and indices 1 and 5 for the second pair (2, 4).
The product difference is (9 * 8) - (2 * 4) = 64.

 

Constraints:

  • 4 <= nums.length <= 104
  • 1 <= nums[i] <= 104

public int MaxProductDifference(int[] nums)
{
    Array.Sort(nums);
    return nums[nums.Length - 1] * nums[nums.Length - 2] - nums[0] * nums[1];
}

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