# 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 : 6Output :{“1+2+3”, “1*2*3”}Input :“125”, Target : 7Output :{“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.

`// 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*

*filter_none*

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.

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