LeetCode 1913. Maximum Product Difference Between Two Pairs Solution in Java, C++, Python & More | Explanation + Code

Description

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

<li>For example, the product difference between <code>(5, 6)</code> and <code>(2, 7)</code> is <code>(5 * 6) - (2 * 7) = 16</code>.</li>

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:

<li><code>4 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>

<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>

Solutions

Solution 1

PythonJavaC++GoJavaScript

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

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]; } }(code-box)

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]; } };(code-box)

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

/** * @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; };(code-box)