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Leetcode 1913. Maximum Product Difference Between Two Pairs

Problem Statement

You are given an integer array nums, where the length of nums is 4 or more. You need to find the maximum product difference between two pairs (a, b) and (c, d) where a and b are elements from the array nums as well as c and d, and their positions are distinct. The product difference is defined and computed as:

Product Difference = (a * b) - (c * d)

Return the maximum such product difference you can achieve.

Clarifying Questions

  1. What is the expected input size?
    • The array nums will have a length of at least 4 and can be considerably large.
  2. Are there any constraints on the values of the integers in nums?
    • The problem does not specify this, but we can assume based on typical constraints that the integers can be both negative and positive.
  3. Can the pairs (a, b) and (c, d) include the same elements?
    • No, the problem specifies that the elements have to be in distinct positions.

Strategy

  1. Sorting Approach:
    • Sort the array nums in ascending order.
    • The two largest elements (which will produce the maximum product) will be the last two elements of the sorted array.
    • The two smallest elements (which will produce the minimum product) will be the first two elements of the sorted array.
    • Calculate the product difference using these four elements.

Code

The following Java code implements the sorting approach described:

import java.util.Arrays;

public class MaxProductDifference {
    public int maxProductDifference(int[] nums) {
        // Step 1: Sort the array
        Arrays.sort(nums);
        
        // Step 2: Identify the two largest and two smallest elements
        int n = nums.length;
        int max1 = nums[n - 1];
        int max2 = nums[n - 2];
        int min1 = nums[0];
        int min2 = nums[1];
        
        // Step 3: Calculate the product differences
        int productDifference = (max1 * max2) - (min1 * min2);
        
        return productDifference;
    }

    public static void main(String[] args) {
        MaxProductDifference solution = new MaxProductDifference();
        int[] nums = {5, 6, 2, 7, 4};
        System.out.println(solution.maxProductDifference(nums));  // Output: 34
    }
}

Time Complexity

The overall time complexity is dominated by the sorting step, so the time complexity is O(n log n).

In this solution, we efficiently derive the maximum product difference by sorting the array and then using the appropriate elements from the sorted array to perform the calculation.

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