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1913. Maximum Product Difference Between Two Pairs

Description

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

Solutions

Solution 1

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class Solution:
    def maxProductDifference(self, nums: List[int]) -> int:
        nums.sort()
        return nums[-1] * nums[-2] - nums[0] * nums[1]

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class Solution {
    public int maxProductDifference(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        return nums[n - 1] * nums[n - 2] - nums[0] * nums[1];
    }
}

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class Solution {
public:
    int maxProductDifference(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        return nums[n - 1] * nums[n - 2] - nums[0] * nums[1];
    }
};

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func maxProductDifference(nums []int) int {
    sort.Ints(nums)
    n := len(nums)
    return nums[n-1]*nums[n-2] - nums[0]*nums[1]
}

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/**
 * @param {number[]} nums
 * @return {number}
 */
var maxProductDifference = function (nums) {
    nums.sort((a, b) => a - b);
    let n = nums.length;
    let ans = nums[n - 1] * nums[n - 2] - nums[0] * nums[1];
    return ans;
};

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