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Check if a given string is sum-string
• Difficulty Level : Hard
• Last Updated : 25 Aug, 2017

Given a string of digits, determine whether it is a ‘sum-string’. A string S is called a sum-string if a rightmost substring can be written as sum of two substrings before it and same is recursively true for substrings before it.

Examples :

```“12243660” is a sum string.
Explanation : 24 + 36 = 60, 12 + 24 = 36

“1111112223” is a sum string.
Explanation: 111+112 = 223, 1+111 = 112

“2368” is not a sum string
```

In general a string S is called sum-string if it satisfies the following properties:

```sub-string(i, x) + sub-string(x+1, j)
= sub-string(j+1, l)
and
sub-string(x+1, j)+sub-string(j+1, l)
= sub-string(l+1, m)
and so on till end. ```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

From the examples, we can see that our decision depends on first two chosen numbers.
So we choose all possible first two number for given string. Then for every chosen two numbers we check whether it is sum-string or not? So the approach is very simple. We generate all possible first two numbers using two substrings s1 and s2 using two loops. then we check whether it is possible to make number s3 = (s1 + s2) or not. If we can make s3 then we recursively check for s2 + s3 so on.

 `// C++ program to check if a given string ` `// is sum-string or not ` `#include ` `using` `namespace` `std; ` ` `  `// this is function for finding sum of two ` `// numbers as string ` `string string_sum(string str1, string str2) ` `{ ` `    ``if` `(str1.size() < str2.size()) ` `       ``swap(str1, str2); ` ` `  `    ``int` `m = str1.size(); ` `    ``int` `n = str2.size(); ` `    ``string ans = ``""``; ` ` `  `    ``// sum the str2 with str1 ` `    ``int` `carry = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Sum of current digits ` `        ``int` `ds = ((str1[m - 1 - i] - ``'0'``) + ` `                  ``(str2[n - 1 - i] - ``'0'``) + ` `                  ``carry) % 10; ` ` `  `        ``carry = ((str1[m - 1 - i] - ``'0'``) + ` `                 ``(str2[n - 1 - i] - ``'0'``) + ` `                 ``carry) / 10; ` ` `  `        ``ans = ``char``(ds + ``'0'``) + ans; ` `    ``} ` ` `  `    ``for` `(``int` `i = n; i < m; i++) { ` `        ``int` `ds = (str1[m - 1 - i] - ``'0'` `+ ` `                  ``carry) % 10; ` `        ``carry = (str1[m - 1 - i] - ``'0'` `+ ` `                 ``carry) / 10; ` `        ``ans = ``char``(ds + ``'0'``) + ans; ` `    ``} ` ` `  `    ``if` `(carry) ` `        ``ans = ``char``(carry + ``'0'``) + ans; ` `    ``return` `ans; ` `} ` ` `  `// Returns true of two substrings of ginven ` `// lengths of str[beg..] can cause a positive ` `// result. ` `bool` `checkSumStrUtil(string str, ``int` `beg, ` `                     ``int` `len1, ``int` `len2) ` `{ ` ` `  `    ``// Finding two substrings of given lengths ` `    ``// and their sum ` `    ``string s1 = str.substr(beg, len1); ` `    ``string s2 = str.substr(beg + len1, len2); ` `    ``string s3 = string_sum(s1, s2); ` ` `  `    ``int` `s3_len = s3.size(); ` ` `  `    ``// if number of digits s3 is greater then ` `    ``// the available string size ` `    ``if` `(s3_len > str.size() - len1 - len2 - beg) ` `        ``return` `false``; ` ` `  `    ``// we got s3 as next number in main string ` `    ``if` `(s3 == str.substr(beg + len1 + len2, s3_len)) { ` ` `  `        ``// if we reach at the end of the string ` `        ``if` `(beg + len1 + len2 + s3_len == str.size()) ` `            ``return` `true``; ` ` `  `        ``// otherwise call recursively for n2, s3 ` `        ``return` `checkSumStrUtil(str, beg + len1, len2, ` `                                              ``s3_len); ` `    ``} ` ` `  `    ``// we do not get s3 in main string ` `    ``return` `false``; ` `} ` ` `  `// Returns true if str is sum string, else false. ` `bool` `isSumStr(string str) ` `{ ` `    ``int` `n = str.size(); ` ` `  `    ``// choosing first two numbers and checking ` `    ``// whether it is sum-string or not. ` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``for` `(``int` `j = 1; i + j < n; j++) ` `            ``if` `(checkSumStrUtil(str, 0, i, j)) ` `                ``return` `true``; ` ` `  `    ``return` `false``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``cout << isSumStr(``"1212243660"``) << endl; ` `    ``cout << isSumStr(``"123456787"``); ` `    ``return` `0; ` `} `

Output:

```1
0
```

This article is contributed by Jay Prakash Gupta. 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.