3 Palindrome Codevita 9 Solution

3 Palindrome Codevita 9 Solution

Given an input string word, split the string into exactly 3 palindromic substrings.

Working from left to right, choose the smallest split for the first substring that still allows the remaining word to be split into 2 palindromes.

Similarly, choose the smallest second palindromic substring that leaves a third palindromic substring.

If there is no way to split the word into exactly three palindromic substrings, print “”Impossible”” (without quotes). Every character of the string needs to be consumed.

Cases not allowed –

After finding 3 palindromes using above instructions, if any character of the original string remains unconsumed.

No character may be shared in forming 3 palindromes.

Constraints

1 <= the length of input sting <= 1000

Input

First line contains the input string consisting of characters between [a-z].

Output

Print 3 substrings one on each line.

Time Limit

1

Examples
Example 1

Input

nayannamantenet

Output

nayan

naman

tenet

Explanation

The original string can be split into 3 palindromes as mentioned in the output.

However, if the input was nayanamantenet, then the answer would be “”Impossible””.

Example 2

Input

aaaaa

Output

a

a

aaa

Explanation

The other ways to split the given string into 3 palindromes are as follows –

[a, aaa, a], [aaa, a, a], [aa, aa, a], etc.

Since we want to minimize the length of the first palindromic substring using left to right processing, the correct way to split is [a, a, aaa].

Example 3

Input

aaaabaaaa

Output

a

aaabaaa

a

Explanation

The other ways to split the given string into 3 palindromes are as follows –

[aaaa, b, aaaa], [aa, aabaa, aa], etc.

Since we want to minimize the length of the first palindromic substring using left to right processing, the correct way to split is [a, aaabaaa, a].”

SOLUTION

Program: 3 Palindrome Codevita 9 Solution in C++

#include<bits/stdc++.h>
typedef long long int lld;
#define mod 1000000007
using namespace std;
bool pali(string s)
{
    if(s.length()==1) return true;
    string s1=s;reverse(s1.begin(),s1.end());
    return (s1==s);
}
int main()
{
    speed;
    string s,s1,s2,s3;
    cin>>s;
    int l=s.length();
    for(int i=1;i<l-1;i++)
    {
        s1=s.substr(0,i);
        if(pali(s1))
            for(int j=1;j<l-i;j++)
        {
            s2=s.substr(i,j);s3=s.substr(i+j,l-i-j);
            if(pali(s2)&&pali(s3))
            {
                cout<<s1<<endl<<s2<<endl<<s3;return 0;
            }
        }
    }
    cout<<"Impossible";
    return 0;
}

import sys
def if_palindrome(s):
    if len(s)==1:
        return True
    s1=s[::-1]
    return s1==s
s=input()
l=len(s)
for i in range(1,l-1):
    s1=s[:i]
    if if_palindrome(s1):
        for j in range(1,l-i):
            s2=s[i:i+j]
            s3=s[i+j:]
            if if_palindrome(s2) and if_palindrome(s3):
                print(s1)
                print(s2)
                print(s3)
                sys.exit()
print("Impossible")