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# Different Ways to Add Parentheses LeetCode Solution

## Problem – Different Ways to Add Parentheses LeetCode Solution

Given a string `expression` of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order.

The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does not exceed `104`.

Example 1:

``````Input: expression = "2-1-1"
Output: [0,2]
Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2
``````

Example 2:

``````Input: expression = "2*3-4*5"
Output: [-34,-14,-10,-10,10]
Explanation:
(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10
``````

Constraints:

• `1 <= expression.length <= 20`
• `expression` consists of digits and the operator `'+'``'-'`, and `'*'`.
• All the integer values in the input expression are in the range `[0, 99]`.

## Different Ways to Add Parentheses LeetCode Solution in Java

``````public class Solution {
public List<Integer> diffWaysToCompute(String input) {
for (int i=0; i<input.length(); i++) {
if (input.charAt(i) == '-' ||
input.charAt(i) == '*' ||
input.charAt(i) == '+' ) {
String part1 = input.substring(0, i);
String part2 = input.substring(i+1);
List<Integer> part1Ret = diffWaysToCompute(part1);
List<Integer> part2Ret = diffWaysToCompute(part2);
for (Integer p1 :   part1Ret) {
for (Integer p2 :   part2Ret) {
int c = 0;
switch (input.charAt(i)) {
case '+': c = p1+p2;
break;
case '-': c = p1-p2;
break;
case '*': c = p1*p2;
break;
}
}
}
}
}
if (ret.size() == 0) {
}
return ret;
}
}
``````

## Different Ways to Add Parentheses LeetCode Solution in C++

``````class Solution {
public:
vector<int> diffWaysToCompute(string input) {
unordered_map<string, vector<int>> dpMap;
return computeWithDP(input, dpMap);
}

vector<int> computeWithDP(string input, unordered_map<string, vector<int>> &dpMap) {
vector<int> result;
int size = input.size();
for (int i = 0; i < size; i++) {
char cur = input[i];
if (cur == '+' || cur == '-' || cur == '*') {
// Split input string into two parts and solve them recursively
vector<int> result1, result2;
string substr = input.substr(0, i);
// check if dpMap has the result for substr
if (dpMap.find(substr) != dpMap.end())
result1 = dpMap[substr];
else
result1 = computeWithDP(substr, dpMap);

substr = input.substr(i + 1);
if (dpMap.find(substr) != dpMap.end())
result2 = dpMap[substr];
else
result2 = computeWithDP(substr, dpMap);

for (auto n1 : result1) {
for (auto n2 : result2) {
if (cur == '+')
result.push_back(n1 + n2);
else if (cur == '-')
result.push_back(n1 - n2);
else
result.push_back(n1 * n2);
}
}
}
}
// if the input string contains only number
if (result.empty())
result.push_back(atoi(input.c_str()));
// save to dpMap
dpMap[input] = result;
return result;
}
};
``````

## Different Ways to Add Parentheses LeetCode Solution in Python

``````class Solution(object):
def diffWaysToCompute(self, input):
m = {}
return self.dfs(input, m)

def dfs(self, input, m):
if input in m:
return m[input]
if input.isdigit():
m[input] = int(input)
return [int(input)]
ret = []
for i, c in enumerate(input):
if c in "+-*":
l = self.diffWaysToCompute(input[:i])
r = self.diffWaysToCompute(input[i+1:])
ret.extend(eval(str(x)+c+str(y)) for x in l for y in r)
m[input] = ret
return ret
``````
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