Even Length Substring – Equal Sum Halves: Given a string S as input which consists only of digits from 0 to 9, print the longest substring such that the sum of the digits in the first half and the second half is the same. Print -1 if such a substring does not exist.
Input Format:
The first line contains S
Output Format:
The first line contains the longest substring as per the rules defined above or -1.
Boundary Conditions:
1 <= Length of S <= 100
Example Input/Output 1:
Input:
123123
Output:
123123
Explanation:
The first half is 123 and the second half is 123. Hence the sum of the digits is equal.
Example Input/Output 2:
Input:
1538024
Output:
5380
Explanation:
The first half is 53 and the second half is 80. The sum of the digits is 8 in both the halves.
Example Input/Output 3:
Input:
12345
Output:
-1
Example Input/Output 4:
Input:
989898989
Output:
98989898
Explanation:
Here both 98989898 and 89898989 are of same length. But due to order of occurrence 98989898 is printed as the output.
s=input().strip()
p=[]
r=[s[i:j] for i in range(len(s)) for j in range(len(s)+1) if len(s[i:j])%2==0]
for i in r:
k=len(i)//2
a=i[:k];b=i[k:]
x=0;y=0
for j in a:
x+=int(j)
for h in b:
y+=int(h)
if x==y and i!="":
p.append(i)
if len(p)!=0:
print(max(p,key=len))
else:
print(-1)#include<stdio.h>
#include <stdlib.h>
int main()
{
char str[100];
scanf("%s",str);
int l=strlen(str);
int countrt=0;
for(int len=l&1? l-1: l;len>=2;len-=2){
int mid=len/2;
for(int s=0;s<=l-len;++s){
int s1=s,num1=0,num2=0;
int s2=s+mid;
for(int k=0;k<mid;++k){
num1+=(str[s1+k]-'0');
num2+=(str[s2+k]-'0');
}
if(num1==num2){
countrt=1;
for(int k=0;k<len;++k)printf("%c",str[k+s1]);
return;
}
}
}
printf("-1");
return 1;
}
