leetcode - 2232. Minimize Result by Adding Parentheses to Expression

Description

You are given a 0-indexed string expression of the form "+" where and represent positive integers.

Add a pair of parentheses to expression such that after the addition of parentheses, expression is a valid mathematical expression and evaluates to the smallest possible value. The left parenthesis must be added to the left of '+' and the right parenthesis must be added to the right of '+'.

Return expression after adding a pair of parentheses such that expression evaluates to the smallest possible value. If there are multiple answers that yield the same result, return any of them.

The input has been generated such that the original value of expression, and the value of expression after adding any pair of parentheses that meets the requirements fits within a signed 32-bit integer.

Example 1:

复制代码
Input: expression = "247+38"
Output: "2(47+38)"
Explanation: The expression evaluates to 2 * (47 + 38) = 2 * 85 = 170.
Note that "2(4)7+38" is invalid because the right parenthesis must be to the right of the '+'.
It can be shown that 170 is the smallest possible value.

Example 2:

复制代码
Input: expression = "12+34"
Output: "1(2+3)4"
Explanation: The expression evaluates to 1 * (2 + 3) * 4 = 1 * 5 * 4 = 20.

Example 3:

复制代码
Input: expression = "999+999"
Output: "(999+999)"
Explanation: The expression evaluates to 999 + 999 = 1998.

Constraints:

复制代码
3 <= expression.length <= 10
expression consists of digits from '1' to '9' and '+'.
expression starts and ends with digits.
expression contains exactly one '+'.
The original value of expression, and the value of expression after adding any pair of parentheses that meets the requirements fits within a signed 32-bit integer.

Solution

Try every possible position to put the parentheses.

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

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

Code

python3 复制代码
class Solution:
    def minimizeResult(self, expression: str) -> str:
        def get_value_from_expression(expression: str) -> int:
            left, right = expression.find('('), expression.find(')')
            res = eval(expression[left + 1: right])
            if left == 0 and right == len(expression) - 1:
                pass
            elif left == 0:
                res *= int(expression[right + 1:])
            elif right == len(expression) - 1:
                res *= int(expression[:left])
            else:
                res *= int(expression[:left]) * int(expression[right + 1:])
            return res

        left_index = []
        right_index = []
        add_index = expression.find('+')
        for i in range(len(expression)):
            if i < add_index:
                left_index.append(i)
            elif i > add_index:
                right_index.append(i + 1)
        min_val = float('inf')
        res = ''
        for l in left_index:
            for r in right_index:
                cur_expression = f'{expression[:l]}({expression[l:r]}){expression[r:]}'
                cur_value = get_value_from_expression(cur_expression)
                if cur_value < min_val:
                    min_val = cur_value
                    res = cur_expression
        return res
相关推荐
大怪v14 小时前
前端:人工智能?我也会啊!来个花活,😎😎😎“自动驾驶”整起!
前端·javascript·算法
惯导马工16 小时前
【论文导读】ORB-SLAM3:An Accurate Open-Source Library for Visual, Visual-Inertial and
深度学习·算法
骑自行车的码农17 小时前
【React用到的一些算法】游标和栈
算法·react.js
博笙困了18 小时前
AcWing学习——双指针算法
c++·算法
moonlifesudo18 小时前
322:零钱兑换(三种方法)
算法
NAGNIP1 天前
大模型框架性能优化策略:延迟、吞吐量与成本权衡
算法
美团技术团队2 天前
LongCat-Flash:如何使用 SGLang 部署美团 Agentic 模型
人工智能·算法
Fanxt_Ja2 天前
【LeetCode】算法详解#15 ---环形链表II
数据结构·算法·leetcode·链表
侃侃_天下2 天前
最终的信号类
开发语言·c++·算法
茉莉玫瑰花茶2 天前
算法 --- 字符串
算法