1464. Maximum Product of Two Elements in an Array
Easy
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 <= 5001 <= nums[i] <= 10^3
class Solution:
def maxProduct(self, nums: List[int]) -> int:
nums = [-num for num in nums]
heapq.heapify(nums)
#print(nums)
return(heapq.heappop(nums) +1) * (heapq.heappop(nums)+1)
#return (nums[0]+1)*(nums[1]+1)
# n1 = nums[0]
# n2 = nums[1]
# for num in nums[2:]:
# if num > n1:
# if n1>n2:
# n2=n1
# n1 = num
# elif num > n2:
# n2 = num
# if n2>n1:
# n1=n2
# #print(n1, n2)
# return (n1-1)*(n2-1)