Open In App
Related Articles

Check if a given string is sum-string

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

From the examples, we can see that our decision depends on the first two chosen numbers. So we choose all possible first two numbers for a 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 the number s3 = (s1 + s2) or not. If we can make s3 then we recursively check for s2 + s3 and so on.

Implementation:

C++

 `// 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 if two substrings of given``// 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 than``    ``// 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()``{``    ``bool` `result;` `    ``result = isSumStr(``"1212243660"``);``    ``cout << (result == 1 ? ``"True\n"` `: ``"False\n"``);``      ` `    ``result = isSumStr(``"123456787"``);``    ``cout << (result == 1 ? ``"True\n"` `: ``"False\n"``);``    ``return` `0;``}`

Java

 `// Java program to check if a given string``// is sum-string or not``import` `java.util.*;` `class` `GFG {``    ``// this is function for finding sum of two``    ``// numbers as string``    ``public` `static` `String string_sum(String str1,``                                    ``String str2)``    ``{``        ``if` `(str1.length() < str2.length()) {``            ``String temp = str1;``            ``str1 = str2;``            ``str2 = temp;``        ``}``        ``int` `m = str1.length();``        ``int` `n = str2.length();``        ``String ans = ``""``;` `        ``// sum the str2 with str1``        ``int` `carry = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// Sum of current digits``            ``int` `ds``                ``= ((str1.charAt(m - ``1` `- i) - ``'0'``)``                   ``+ (str2.charAt(n - ``1` `- i) - ``'0'``) + carry)``                  ``% ``10``;` `            ``carry``                ``= ((str1.charAt(m - ``1` `- i) - ``'0'``)``                   ``+ (str2.charAt(n - ``1` `- i) - ``'0'``) + carry)``                  ``/ ``10``;` `            ``ans = Character.toString((``char``)(ds + ``'0'``))``                  ``+ ans;``        ``}` `        ``for` `(``int` `i = n; i < m; i++) {``            ``int` `ds = (str1.charAt(m - ``1` `- i) - ``'0'` `+ carry)``                     ``% ``10``;``            ``carry = (str1.charAt(m - ``1` `- i) - ``'0'` `+ carry)``                    ``/ ``10``;``            ``ans = Character.toString((``char``)(ds + ``'0'``))``                  ``+ ans;``        ``}` `        ``if` `(carry != ``0``) {``            ``ans = Character.toString((``char``)(carry + ``'0'``))``                  ``+ ans;``        ``}``        ``return` `ans;``    ``}``    ``// Returns true if two substrings of given``    ``// lengths of str[beg..] can cause a positive``    ``// result.``    ``public` `static` `boolean``    ``checkSumStrUtil(String str, ``int` `beg, ``int` `len1, ``int` `len2)``    ``{` `        ``// Finding two substrings of given lengths``        ``// and their sum``        ``String s1 = str.substring(beg, beg + len1);``        ``String s2``            ``= str.substring(beg + len1, beg + len1 + len2);``        ``String s3 = string_sum(s1, s2);` `        ``int` `s3_len = s3.length();` `        ``// if number of digits s3 is greater than``        ``// the available string size``        ``if` `(s3_len > str.length() - len1 - len2 - beg)``            ``return` `false``;` `        ``// we got s3 as next number in main string``        ``if` `(s3.equals(str.substring(beg + len1 + len2,``                                    ``beg + len1 + len2``                                        ``+ s3_len))) {``            ``// if we reach at the end of the string``            ``if` `(beg + len1 + len2 + s3_len``                ``== str.length()) {``                ``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.``    ``public` `static` `boolean` `isSumStr(String str)``    ``{``        ``int` `n = str.length();` `        ``// 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``    ``public` `static` `void` `main(String[] args)``    ``{``        ``boolean` `result;``        ``result = isSumStr(``"1212243660"``);``        ``System.out.println(result == ``true` `? ``"True"``                                          ``: ``"False"``);` `        ``result = isSumStr(``"123456787"``);``        ``System.out.println(result == ``true` `? ``"True"``                                          ``: ``"False"``);``    ``}``}` `// This code is contributed by Tapesh (tapeshdua420)`

Python3

 `# Python code for the above approach` `# this is function for finding sum of two``# numbers as string``def` `string_sum(str1, str2):` `    ``if` `(``len``(str1) < ``len``(str2)):``        ``str1, str2 ``=` `str2,str1` `    ``m ``=` `len``(str1)``    ``n ``=` `len``(str2)``    ``ans ``=` `""` `    ``# sum the str2 with str1``    ``carry ``=` `0``    ``for` `i ``in` `range``(n):` `        ``# Sum of current digits``        ``ds ``=` `((``ord``(str1[m ``-` `1` `-` `i]) ``-` `ord``(``'0'``)) ``+``                ``(``ord``(str2[n ``-` `1` `-` `i]) ``-` `ord``(``'0'``)) ``+``                ``carry) ``%` `10` `        ``carry ``=` `((``ord``(str1[m ``-` `1` `-` `i]) ``-` `ord``(``'0'``)) ``+``                ``(``ord``(str2[n ``-` `1` `-` `i]) ``-` `ord``(``'0'``)) ``+``                ``carry) ``/``/` `10` `        ``ans ``=` `str``(ds) ``+` `ans` `    ``for` `i ``in` `range``(n,m):``        ``ds ``=` `(``ord``(str1[m ``-` `1` `-` `i]) ``-` `ord``(``'0'``) ``+``                ``carry) ``%` `10``        ``carry ``=` `(``ord``(str1[m ``-` `1` `-` `i]) ``-` `ord``(``'0'``) ``+``                ``carry) ``/``/` `10``        ``ans ``=` `str``(ds) ``+` `ans` `    ``if` `(carry):``        ``ans ``=` `str``(carry) ``+` `ans``    ``return` `ans` `# Returns True if two substrings of given``# lengths of str[beg..] can cause a positive``# result.``def` `checkSumStrUtil(``Str``, beg,len1, len2):` `    ``# Finding two substrings of given lengths``    ``# and their sum``    ``s1 ``=` `Str``[beg: beg``+``len1]``    ``s2 ``=` `Str``[beg ``+` `len1: beg ``+` `len1 ``+``len2]``    ``s3 ``=` `string_sum(s1, s2)` `    ``s3_len ``=` `len``(s3)` `    ``# if number of digits s3 is greater than``    ``# the available string size``    ``if` `(s3_len > ``len``(``Str``) ``-` `len1 ``-` `len2 ``-` `beg):``        ``return` `False` `    ``# we got s3 as next number in main string``    ``if` `(s3 ``=``=` `Str``[beg ``+` `len1 ``+` `len2: beg ``+` `len1 ``+` `len2 ``+``s3_len]):` `        ``# if we reach at the end of the string``        ``if` `(beg ``+` `len1 ``+` `len2 ``+` `s3_len ``=``=` `len``(``Str``)):``            ``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.``def` `isSumStr(``Str``):` `    ``n ``=` `len``(``Str``)` `    ``# choosing first two numbers and checking``    ``# whether it is sum-string or not.``    ``for` `i ``in` `range``(``1``,n):``        ``for` `j ``in` `range``(``1``,n``-``i):``            ``if` `(checkSumStrUtil(``Str``, ``0``, i, j)):``                ``return` `True` `    ``return` `False`  `# Driver code``print``(isSumStr(``"1212243660"``))``print``(isSumStr(``"123456787"``))` `# This code is contributed by shinjanpatra`

C#

 `// C# program to check if a given string``// is sum-string or not` `using` `System;` `class` `sub_string {``    ``// this is function for finding sum of two``    ``// numbers as string``    ``static` `String string_sum(String str1, String str2)``    ``{``        ``if` `(str1.Length < str2.Length) {``            ``String temp = str1;``            ``str1 = str2;``            ``str2 = temp;``        ``}` `        ``int` `m = str1.Length;``        ``int` `n = str2.Length;``        ``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 > 0)``            ``ans = (``char``)(carry + ``'0'``) + ans;``        ``return` `ans;``    ``}` `    ``// Returns true if two substrings of given``    ``// lengths of str[beg..] can cause a positive``    ``// result.``    ``static` `bool` `checkSumStrUtil(String str, ``int` `beg,``                                ``int` `len1, ``int` `len2)``    ``{` `        ``// Finding two substrings of given lengths``        ``// and their sum``        ``String s1 = str.Substring(beg, len1);``        ``String s2 = str.Substring(beg + len1, len2);``        ``String s3 = string_sum(s1, s2);` `        ``int` `s3_len = s3.Length;` `        ``// if number of digits s3 is greater than``        ``// the available string size``        ``if` `(s3_len > str.Length - len1 - len2 - beg)``            ``return` `false``;` `        ``// we got s3 as next number in main string``        ``if` `(s3``            ``== str.Substring(beg + len1 + len2, s3_len)) {` `            ``// if we reach at the end of the string``            ``if` `(beg + len1 + len2 + s3_len == str.Length)``                ``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.``    ``static` `bool` `isSumStr(String str)``    ``{``        ``int` `n = str.Length;` `        ``// 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``    ``public` `static` `void` `Main()``    ``{``        ``Console.WriteLine(isSumStr(``"1212243660"``));``        ``Console.WriteLine(isSumStr(``"123456787"``));``    ``}``}` `// This code is contributed by Abhijeet Kumar(abhijeet19403)`

Javascript

 ``

Output

```True
False```

Time Complexity: O(n*n*n), where n is the length of the string.
Auxiliary Space: O(n), where n is the length of the string.

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