/**
aacecaaa
01234567
1. find the first pos j that left the string non-palindrome (j=7)
2. get suffix s.substr(j)
3. get prefix = reverse of the suffix
4. recursion: mid = shortestPalindrome(s.substr(0,j))
5. return prefix+mid+suffix
step 1 and step 4 are essential
abbb
*/
string shortestPalindrome(string s) {
int n = s.length();
if(n<2) return s;
int j = 0;
for(int i=n-1; i>=0; i--){
if(s[i]==s[j]){
j++;
}
}
if(j==n) {
// s is already a palindrome, so the shortest one is itself
return s;
}
string suffix = s.substr(j);
string prefix = suffix;
reverse(prefix.begin(), prefix.end());
string mid = shortestPalindrome(s.substr(0, j));
return prefix+mid+suffix;
}
int main()
{
char a[]="aaaaa";
shortestPalindrome(a);
return 0;
}
