All ways to add parenthesis for evaluation - GeeksforGeeks

21 Mar, 2017
Given a string that represents an expression constituting numbers and binary operator +, – and * only. We need to parenthesize the expression in all possible way and return all evaluated values.
Input : expr = “3-2-1”
Output : {0, 2}
((3-2)-1) = 0 
(3-(2-1)) = 2

Input : expr = "5*4-3*2"
Output : {-10, 10, 14, 10, 34}
(5*(4-(3*2))) = -10
(5*((4-3)*2)) = 10
((5*4)-(3*2)) = 14
((5*(4-3))*2) = 10
(((5*4)-3)*2) = 34

We can solve this problem by parenthesizing all possible valid substring of the expression and then evaluating them, but as we can see that it will involve solving lots of repeating subproblem, to save ourselves we can follow a dynamic programming approach.We store the result for each substring in a map and if string in recursion is already solved, we return result from map instead of solving that again.Below code runs a loop in the string and if at any instant, character is operator then we divide the input from there, recursively solve each part and then combine the result in all possible ways.See the use of c_str() function, this function converts the C++ string into C char array, this function is used in below code because atoi() function expects a character array as an argument not the string. It converts character array to number.
//  C++ program to output all possible values of// an expression by parenthesizing it.#include <bits/stdc++.h>using namespace std;  //  method checks, character is operator or notbool isOperator(char op){    return (op == '+' || op == '-' || op == '*');}  //  Utility recursive method to get all possible// result of input stringvector<int> possibleResultUtil(string input,            map< string, vector<int> > memo){    //  If already calculated, then return from memo    if (memo.find(input) != memo.end())        return memo[input];      vector<int> res;    for (int i = 0; i < input.size(); i++)    {        if (isOperator(input[i]))        {            // If character is operator then split and            // calculate recursively            vector<int> resPre =              possibleResultUtil(input.substr(0, i), memo);            vector<int> resSuf =              possibleResultUtil(input.substr(i + 1), memo);              //  Combine all possible combination            for (int j = 0; j < resPre.size(); j++)            {                for (int k = 0; k < resSuf.size(); k++)                {                    if (input[i] == '+')                        res.push_back(resPre[j] + resSuf[k]);                    else if (input[i] == '-')                        res.push_back(resPre[j] - resSuf[k]);                    else if (input[i] == '*')                        res.push_back(resPre[j] * resSuf[k]);                }            }        }    }      // if input contains only number then save that     // into res vector    if (res.size() == 0)        res.push_back(atoi(input.c_str()));      // Store in memo so that input string is not     // processed repeatedly    memo[input] = res;    return res;}  //  method to return all possible output // from input expressionvector<int> possibleResult(string input){    map< string, vector<int> > memo;    return possibleResultUtil(input, memo);}  //  Driver code to test above methodsint main(){    string input = "5*4-3*2";    vector<int> res = possibleResult(input);      for (int i = 0; i < res.size(); i++)        cout << res[i] << " ";    return 0;}
Output:
-10 10 14 10 34 

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