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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

Solution 1

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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

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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;
    }
}

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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;
    }
};

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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
}

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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);
}

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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)
    }
}

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