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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

• Java
• C++
• Python
• Go
• TypeScript
• Php
• RenderScript

• 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)
      }
  }
  
  

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