Question

1464. Maximum Product of Two Elements in an Array

中文文档

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

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

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;
  }
};

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);
}

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)
  }
}

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);
  }
}

int maxProduct(int* nums, int numsSize) {
  int max = 0;
  int submax = 0;
  for (int i = 0; i < numsSize; i++) {
    int num = nums[i];
    if (num > max) {
      submax = max;
      max = num;
    } else if (num > submax) {
      submax = num;
    }
  }
  return (max - 1) * (submax - 1);
}

Solution 2

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

class Solution {
  public int maxProduct(int[] nums) {
    Arrays.sort(nums);
    int n = nums.length;
    return (nums[n - 1] - 1) * (nums[n - 2] - 1);
  }
}

class Solution {
public:
  int maxProduct(vector<int>& nums) {
    sort(nums.rbegin(), nums.rend());
    return (nums[0] - 1) * (nums[1] - 1);
  }
};

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

function maxProduct(nums: number[]): number {
  let max = 0;
  let submax = 0;
  for (const num of nums) {
    if (num > max) {
      submax = max;
      max = num;
    } else if (num > submax) {
      submax = num;
    }
  }
  return (max - 1) * (submax - 1);
}

Solution 3

class Solution:
  def maxProduct(self, nums: List[int]) -> int:
    a = b = 0
    for v in nums:
      if v > a:
        a, b = v, a
      elif v > b:
        b = v
    return (a - 1) * (b - 1)

class Solution {
  public int maxProduct(int[] nums) {
    int a = 0, b = 0;
    for (int v : nums) {
      if (v > a) {
        b = a;
        a = v;
      } else if (v > b) {
        b = v;
      }
    }
    return (a - 1) * (b - 1);
  }
}

class Solution {
public:
  int maxProduct(vector<int>& nums) {
    int a = 0, b = 0;
    for (int v : nums) {
      if (v > a) {
        b = a;
        a = v;
      } else if (v > b) {
        b = v;
      }
    }
    return (a - 1) * (b - 1);
  }
};

func maxProduct(nums []int) int {
  a, b := 0, 0
  for _, v := range nums {
    if v > a {
      b, a = a, v
    } else if v > b {
      b = v
    }
  }
  return (a - 1) * (b - 1)
}