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Count all possible Paths between two Vertices

Check if a given string is sum-string

Last Updated : 22 Aug, 2022

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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.

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 <bits/stdc++.h>
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

   <script>
 
// JavaScript code to implement the approach
 
// this is function for finding sum of two
// numbers as string
function string_sum(str1, str2){
 
    if (str1.length < str2.length){
        let temp = str1
        str1 = str2
        str2 = temp
 
    }
    let m = str1.length
    let n = str2.length
    let ans = ""
 
    // sum the str2 with str1
    let carry = 0
    for(let i=0;i<n;i++){
 
        // Sum of current digits
        let ds = ((str1.charCodeAt(m - 1 - i) - '0'.charCodeAt(0)) +
                (str2.charCodeAt(n - 1 - i) - '0'.charCodeAt(0)) +
                carry) % 10
 
        carry = Math.floor(((str1.charCodeAt(m - 1 - i) - '0'.charCodeAt(0)) +
                (str2.charCodeAt(n - 1 - i) - '0'.charCodeAt(0)) +
                carry) / 10)
 
        ans = ds.toString() + ans
    }
 
    for(let i=n;i<m;i++){
         
        let ds = ((str1.charCodeAt(m - 1 - i) - '0'.charCodeAt(0)) +
                (str2.charCodeAt(n - 1 - i) - '0'.charCodeAt(0)) +
                carry) % 10
 
        carry = Math.floor(((str1.charCodeAt(m - 1 - i) - '0'.charCodeAt(0)) +
                (str2.charCodeAt(n - 1 - i) - '0'.charCodeAt(0)) +
                carry) / 10)
 
        ans = ds.toString() + ans
    }
 
    if (carry)
        ans = carry.toString() + ans
    return ans
}
 
// Returns true if two substrings of given
// lengths of str[beg..] can cause a positive
// result.
function checkSumStrUtil(Str, beg,len1, len2){
    // Finding two substrings of given lengths
    // and their sum
    let s1 = Str.substring(beg,beg+len1)
    let s2 = Str.substring(beg + len1, beg + len1 +len2)
    let s3 = string_sum(s1, s2)
 
    let 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, 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.
function isSumStr(Str){
 
    let n = Str.length
 
    // choosing first two numbers and checking
    // whether it is sum-string or not.
    for(let i=1;i<n;i++){
        for(let j=1;j<n-i;j++){
            if (checkSumStrUtil(Str, 0, i, j))
                return true
        }
    }
 
    return false
}
 
 
// Driver code
document.write(isSumStr("1212243660"))
document.write(isSumStr("123456787"))
 
// This code is contributed by shinjanpatra
 
</script>
 
  

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.

Count all possible Paths between two Vertices

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