solutions: 1464 - Maximum Product of Two Elements in an Array (Easy)

solutions/1400-1499/1464-maximum-product-of-two-elements-in-an-array-easy.md

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+---
+description: 'Author: @wingkwong, @jaffar, @jit | https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/'
+tags: [Array, Sorting, Heap (Priority Queue)]
+---
+
+# 1464 - Maximum Product of Two Elements in an Array (Easy) 
+
+## Problem Link
+
+https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/
+
+## Problem Statement
+
+Given the array of integers `nums`, you will choose two different indices `i` and `j` of that array. *Return the maximum value of* `(nums[i]-1)*(nums[j]-1)`.
+
+**Example 1:**
+
+```
+Input: nums = [3,4,5,2]
+Output: 12 
+Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,5,4,5]
+Output: 16
+Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.
+```
+
+**Example 3:**
+
+```
+Input: nums = [3,7]
+Output: 12
+```
+
+**Constraints:**
+
+- $2 <= nums.length <= 500$
+- $1 <= nums[i] <= 10^3$
+
+## Approach 1: Sorting
+
+We can just choose the two largest values to maximum the product. 
+
+<Tabs>
+
+<TabItem value="py" label="Python">
+<SolutionAuthor name="@jaffar"/>
+
+```py
+class Solution:
+    def maxProduct(self, nums: List[int]) -> int:
+        sorted_nums = sorted(nums, reverse=True)
+        first_max = sorted_nums[0] - 1
+        second_max = sorted_nums[1] - 1
+        return first_max * second_max
+```
+
+</TabItem>
+
+<TabItem value="erlang" label="Erlang">
+<SolutionAuthor name="@jit"/>
+
+```cpp
+%% Lazy but whatever
+max_product(Nums) ->
+  [A, B | _] = lists:sort(fun erlang:'>='/2, Nums),
+  (A - 1) * (B - 1).
+```
+
+</TabItem>
+
+<TabItem value="java" label="Java">
+<SolutionAuthor name="@wingkwong"/>
+
+```java
+class Solution {
+    public int maxProduct(int[] nums) {
+        Arrays.sort(nums);
+        int n = nums.length;
+        return (nums[n - 1] - 1) * (nums[n - 2] - 1);
+    }
+}
+```
+
+</TabItem>
+</Tabs>