Check if a given string is sumstring (original) (raw)

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++

#include <bits/stdc++.h>

using namespace std;

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 = "" ;

`` int carry = 0;

`` for ( int i = 0; i < n; i++) {

`` 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;

}

bool checkSumStrUtil(string str, int beg, int len1,

`` int len2)

{

`` 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 (s3_len > str.size() - len1 - len2 - beg)

`` return false ;

`` if (s3 == str.substr(beg + len1 + len2, s3_len)) {

`` if (beg + len1 + len2 + s3_len == str.size())

`` return true ;

`` return checkSumStrUtil(str, beg + len1, len2,

`` s3_len);

`` }

`` return false ;

}

bool isSumStr(string str)

{

`` int n = str.size();

`` 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 ;

}

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

import java.util.*;

class GFG {

`` 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 = "" ;

`` int carry = 0 ;

`` for ( int i = 0 ; i < n; i++) {

`` 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;

`` }

`` public static boolean

`` checkSumStrUtil(String str, int beg, int len1, int len2)

`` {

`` 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 (s3_len > str.length() - len1 - len2 - beg)

`` return false ;

`` if (s3.equals(str.substring(beg + len1 + len2,

`` beg + len1 + len2

`` + s3_len))) {

`` if (beg + len1 + len2 + s3_len

`` == str.length()) {

`` return true ;

`` }

`` return checkSumStrUtil(str, beg + len1, len2,

`` s3_len);

`` }

`` return false ;

`` }

`` public static boolean isSumStr(String str)

`` {

`` int n = str.length();

`` 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 ;

`` }

`` 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" );

`` }

}

Python3

def string_sum(str1, str2):

`` if ( len (str1) < len (str2)):

`` str1, str2 = str2,str1

`` m = len (str1)

`` n = len (str2)

`` ans = ""

`` carry = 0

`` for i in range (n):

`` 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

def checkSumStrUtil( Str , beg,len1, len2):

`` s1 = Str [beg: beg + len1]

`` s2 = Str [beg + len1: beg + len1 + len2]

`` s3 = string_sum(s1, s2)

`` s3_len = len (s3)

`` if (s3_len > len ( Str ) - len1 - len2 - beg):

`` return False

`` if (s3 = = Str [beg + len1 + len2: beg + len1 + len2 + s3_len]):

`` if (beg + len1 + len2 + s3_len = = len ( Str )):

`` return True

`` return checkSumStrUtil( Str , beg + len1, len2,s3_len)

`` return False

def isSumStr( Str ):

`` n = len ( Str )

`` for i in range ( 1 ,n):

`` for j in range ( 1 ,n - i):

`` if (checkSumStrUtil( Str , 0 , i, j)):

`` return True

`` return False

print (isSumStr( "1212243660" ))

print (isSumStr( "123456787" ))

C#

using System;

class sub_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 = "" ;

`` int carry = 0;

`` for ( int i = 0; i < n; i++) {

`` 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;

`` }

`` static bool checkSumStrUtil(String str, int beg,

`` int len1, int len2)

`` {

`` 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 (s3_len > str.Length - len1 - len2 - beg)

`` return false ;

`` if (s3

`` == str.Substring(beg + len1 + len2, s3_len)) {

`` if (beg + len1 + len2 + s3_len == str.Length)

`` return true ;

`` return checkSumStrUtil(str, beg + len1, len2,

`` s3_len);

`` }

`` return false ;

`` }

`` static bool isSumStr(String str)

`` {

`` int n = str.Length;

`` 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 ;

`` }

`` public static void Main()

`` {

`` Console.WriteLine(isSumStr( "1212243660" ));

`` Console.WriteLine(isSumStr( "123456787" ));

`` }

}

Javascript

<script>

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 = ""

`` let carry = 0

`` for (let i=0;i<n;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

`` }

`` 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

}

function checkSumStrUtil(Str, beg,len1, len2){

`` 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 (s3_len > Str.length - len1 - len2 - beg)

`` return false

`` if (s3 == Str.substring(beg + len1 + len2, beg + len1 + len2 +s3_len)){

`` if (beg + len1 + len2 + s3_len == Str.length)

`` return true

`` return checkSumStrUtil(Str, beg + len1, len2,s3_len)

`` }

`` return false

}

function isSumStr(Str){

`` let n = Str.length

`` 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

}

document.write(isSumStr( "1212243660" ))

document.write(isSumStr( "123456787" ))

</script>

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.