leetcode - 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 (numsi-1)*(numsj-1).

Example 1:

复制代码
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:

复制代码
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:

复制代码
Input: nums = [3,7]
Output: 12

Constraints:

复制代码
2 <= nums.length <= 500
1 <= nums[i] <= 10^3

Solution

Brute Force

Time complexity: o ( n 2 ) o(n^2) o(n2)

Space complexity: o ( 1 ) o(1) o(1)

Math Trick

The largest result must be the product of the largest element and second largest element. So go through the list and find out the largest and second largest element.

Time complexity: o ( n ) o(n) o(n)

Space complexity: o ( 1 ) o(1) o(1)

Code

Math Trick

python3 复制代码
class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        res = 0
        p1, p2 = 0, 0
        for each_num in nums:
            if each_num > p1:
                p2 = p1
                p1 = each_num
            elif each_num > p2:
                p2 = each_num
        return (p1 - 1) * (p2 - 1)
相关推荐
Navigator_Z17 分钟前
LeetCode //C - 1313. Decompress Run-Length Encoded List
c语言·算法·leetcode
dialect_1 小时前
CODEFORCES ROUND 1127 DIV2
开发语言·c++·算法
Nuanyt1 小时前
LeetCode Hot 100 链表专题(四)· 3 题(Java)
java·算法·leetcode·链表·双指针
灯澜忆梦1 小时前
LeetCode 1861. 旋转盒子|Go 双指针解法
算法
软件派2 小时前
生产动态数据智能风控:阈值判别、趋势预测、异常识别算法工程落地实战
前端·数据库·算法
ZhengEnCi10 小时前
模型训练完整流程:从数据准备到模型上线的 10 个关键步骤(全文可跑)
后端·算法
weixin_3077791311 小时前
RAG 面试陷阱——混合检索并非无条件优越:一个条件反转实验
人工智能·python·面试·职场和发展
吞下星星的少年·-·13 小时前
Div.3 CF 1131A题(排序,模拟)
算法·模拟·排序
圣保罗的大教堂14 小时前
leetcode 453. 最小操作次数使数组元素相等 中等
leetcode