Print all possible expressions that evaluate to a target

Given a string that contains only digits from 0 to 9, and an integer value, target. Find out how many expressions are possible which evaluate to target using binary operator +, – and * in given string of digits.

Input : "123",  Target : 6
Output : {“1+2+3”, “1*2*3”}

Input : “125”, Target : 7
Output : {“1*2+5”, “12-5”}

This problem can be solved by putting all possible binary operator in mid between to digits and evaluating them and then check they evaluate to target or not.

  • While writing the recursive code, we need to keep these variable as argument of recursive method – result vector, input string, current expression string, target value, position till which input is processed, current evaluated value and last value in evaluation.
  • Last value is kept in recursion because of multiplication operation, while doing multiplication we need last value for correct evaluation.

See below example for better understanding –

Input is 125, suppose we have reached till 1+2 now,
Input = “125”, current expression = “1+2”, 
position = 2, current val = 3, last = 2

Now when we go for multiplication, we need last 
value for evaluation as follows:

current val = current val - last + last * current val

First we subtract last and then add last * current 
val for evaluation, new last is last * current val.
current val = 3 – 2 + 2*5 = 11
last = 2*5 = 10 

Another thing to note in below code is, we have ignored all numbers which start from 0 by imposing a condition as first condition inside the loop so that we will not process number like 03, 05 etc.

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.

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find all possible expression which
// evaluate to target
#include <bits/stdc++.h>
using namespace std;
  
// Utility recursive method to generate all possible
// expressions
void getExprUtil(vector<string>& res, string curExp,
                 string input, int target, int pos,
                 int curVal, int last)
{
    // true if whole input is processed with some
    // operators
    if (pos == input.length())
    {
        // if current value is equal to target
        //then only add to final solution
        // if question is : all possible o/p then just
        //push_back without condition
        if (curVal == target)
            res.push_back(curExp);
        return;
    }
  
    // loop to put operator at all positions
    for (int i = pos; i < input.length(); i++)
    {
        // ignoring case which start with 0 as they
        // are useless for evaluation
        if (i != pos && input[pos] == '0')
            break;
  
        // take part of input from pos to i
        string part = input.substr(pos, i + 1 - pos);
  
        // take numeric value of part
        int cur = atoi(part.c_str());
  
        // if pos is 0 then just send numeric value
        // for next recurion
        if (pos == 0)
            getExprUtil(res, curExp + part, input,
                     target, i + 1, cur, cur);
  
  
        // try all given binary operator for evaluation
        else
        {
            getExprUtil(res, curExp + "+" + part, input,
                     target, i + 1, curVal + cur, cur);
            getExprUtil(res, curExp + "-" + part, input,
                     target, i + 1, curVal - cur, -cur);
            getExprUtil(res, curExp + "*" + part, input,
                     target, i + 1, curVal - last + last * cur,
                     last * cur);
        }
    }
}
  
// Below method returns all possible expression
// evaluating to target
vector<string> getExprs(string input, int target)
{
    vector<string> res;
    getExprUtil(res, "", input, target, 0, 0, 0);
    return res;
}
  
// method to print result
void printResult(vector<string> res)
{
    for (int i = 0; i < res.size(); i++)
        cout << res[i] << " ";
    cout << endl;
}
  
// Driver code to test above methods
int main()
{
    string input = "123";
    int target = 6;
    vector<string> res = getExprs(input, target);
    printResult(res);
  
    input = "125";
    target = 7;
    res = getExprs(input, target);
    printResult(res);
    return 0;
}

chevron_right


Output:

1+2+3 1*2*3 
1*2+5 12-5

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up


Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.