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
有点。5 小时前
C++03阶段练习(练习题)
数据结构·算法·图论
周末也要写八哥6 小时前
经典算法实例:游戏中弱角色的数量(二)
算法
是Yu欸7 小时前
鸿蒙PC移植:2048 从网页小游戏到 AI 桌面应用
大数据·人工智能·算法·数据挖掘·openharmony·codex
鹿角片ljp7 小时前
KV Cache 解析
java·算法
liliangcsdn7 小时前
IVOL与偏度因子的对比测量分析
算法
threerocks9 小时前
Jev 入门第一课
算法
西柚研究生1234569 小时前
论文分析17:YOLOv11_UAVNet:无人机航拍图像专用目标检测算法
人工智能·python·深度学习·算法·目标检测
hetao173383710 小时前
2026-09-17 hetao1733837 的刷题记录
c++·算法
午彦琳11 小时前
2026.9.17
数据结构·算法·leetcode
木井巳11 小时前
【记忆化搜索】不同路径
java·算法·leetcode·深度优先·剪枝·推荐算法