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
0 0 04 分钟前
CCF-CSP 37-3 模板展开(templating)【C++】
c++·算法
im_AMBER5 分钟前
Leetcode 71 买卖股票的最佳时机 | 增量元素之间的最大差值
笔记·学习·算法·leetcode
良木生香5 分钟前
【数据结构-初阶】详解线性表(1)---顺序表
数据结构
bulingg8 分钟前
聚类方法(kmeans,DBSCAN,层次聚类,GMM,EM算法)
算法·kmeans·聚类
lally.10 分钟前
Kaggle Binary Classification with a Bank Dataset逻辑回归实现(准确率0.94539)
人工智能·算法·机器学习·逻辑回归
埃伊蟹黄面10 分钟前
二分查找算法
c++·算法·leetcode
野蛮人6号13 分钟前
力扣热题100道之78子集
算法·leetcode·职场和发展
悦来客栈的老板25 分钟前
AST反混淆实战|reese84_jsvmp反编译前的优化处理
java·前端·javascript·数据库·算法
dragoooon3429 分钟前
[优选算法专题十一.字符串 ——NO.60~63 最长公共前缀、5最长回文子串、 二进制求和 、字符串相乘]
算法·leetcode·动态规划
lengxuenong30 分钟前
第四届挑战赛二轮题解
c++·算法