# 1464. Maximum Product of Two Elements in an Array

## Description

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

## Solutions

• class Solution {
public int maxProduct(int[] nums) {
int ans = 0;
int n = nums.length;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
ans = Math.max(ans, (nums[i] - 1) * (nums[j] - 1));
}
}
return ans;
}
}

• class Solution {
public:
int maxProduct(vector<int>& nums) {
int ans = 0;
int n = nums.size();
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
ans = max(ans, (nums[i] - 1) * (nums[j] - 1));
}
}
return ans;
}
};

• class Solution:
def maxProduct(self, nums: List[int]) -> int:
ans = 0
for i, a in enumerate(nums):
for b in nums[i + 1 :]:
ans = max(ans, (a - 1) * (b - 1))
return ans


• func maxProduct(nums []int) int {
ans := 0
for i, a := range nums {
for _, b := range nums[i+1:] {
t := (a - 1) * (b - 1)
if ans < t {
ans = t
}
}
}
return ans
}

• function maxProduct(nums: number[]): number {
const n = nums.length;
for (let i = 0; i < 2; i++) {
let maxIdx = i;
for (let j = i + 1; j < n; j++) {
if (nums[j] > nums[maxIdx]) {
maxIdx = j;
}
}
[nums[i], nums[maxIdx]] = [nums[maxIdx], nums[i]];
}
return (nums[0] - 1) * (nums[1] - 1);
}


• class Solution {
/**
* @param Integer[] $nums * @return Integer */ function maxProduct($nums) {
$max = 0;$submax = 0;
for ($i = 0;$i < count($nums);$i++) {
if ($nums[$i] > $max) {$submax = $max;$max = $nums[$i];
} elseif ($nums[$i] > $submax) {$submax = $nums[$i];
}
}
return ($max - 1) * ($submax - 1);
}
}


• impl Solution {
pub fn max_product(nums: Vec<i32>) -> i32 {
let mut max = 0;
let mut submax = 0;
for &num in nums.iter() {
if num > max {
submax = max;
max = num;
} else if num > submax {
submax = num;
}
}
(max - 1) * (submax - 1)
}
}