Find all possible outcomes of a given expression in C++

C++Server Side ProgrammingProgramming

Suppose we have an arithmetic expression without parentheses. Our task is to find all possible outcomes of that expression. Suppose the expression is like 1+2*3-4, this can be interpreted like below −

  • 1+(2*(3-4)) = 1 + (2* -1) = -1
  • (1+2)*(3-4) = 3 * -1 = -3
  • 1+((2*3)-4) = 1 + (6 - 4) = 3
  • ((1+2)*3)-4 = (3 * 3) - 4 = 5
  • 1+(2*3)-4 = 1 + 6 – 4 = 3

To solve this problem, we have to follow these steps −

  • Initially set res as empty

  • For every operator x, do the following −

    • Recursively evaluate all possible values on left of x, let the list of values be L

    • Recursively evaluate all possible values on right of x, let the list of values be R

    • Loop through all values in L:

      • Loop through all values in R−

        • Apply current operator x on current element in L and R, and add the evaluated values to res

  • return the res as output

Example

#include<iostream>
#include<vector>
using namespace std;
int solve(int a, char op, int b) {
   if (op=='+')
      return a+b;
   if (op=='-')
      return a-b;
   if (op == '*')
      return a*b;
}
vector<int> getAllResults(string expr, int low, int high) {
   vector<int> res;
   if (low == high) {
      res.push_back(expr[low] - '0');
      return res;
   }
   if (low == (high-2)) {
      int num = solve(expr[low]-'0', expr[low+1], expr[low+2]-'0');
      res.push_back(num);
      return res;
   }
   for (int i=low+1; i<=high; i+=2) {
      vector<int> L = evaluateAll(expr, low, i-1);
      vector R = evaluateAll(expr, i+1, high);
      for (int s1=0; s1<L.size(); s1++) {
         for (int s2=0; s2<R.size(); s2++) {
            int val = solve(L[s1], expr[i], R[s2]);
            res.push_back(val);
         }
      }
   }
   return res;
}
int main() {
   string expr = "1+2*3-4";
   vector<int> ans = getAllResults(expr, 0, expr.length()-1);
   for (int i=0; i< ans.size(); i++)
      cout << ans[i] << endl;
}

Output

2 1 4 3 5 6 7 8
raja
Published on 18-Dec-2019 06:53:17
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