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++ 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 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) |
# 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# 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) |
<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> |
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.