Given a string S, determine the least number of characters that should be added on to the left side of S so that the complete string becomes a palindrome.
Examples:
Input: S = "LOL" Output: 0 LOL is already a palindrome Input: S = "JAVA" Output: 3 We need to add 3 characters to form AVAJAVA.
The idea is to find the longest palindromic prefix of given string. The count of characters after the prefix is our answer. The longest palindromic prefix can be found by looping from last char to first char. For example, in “JAVA”, the longest palindromic prefix is “J”, so we need to add remaining 3 at the beginning characters to form palindrome.
C++
// C++ program to find minimum number of insertions // on left side to form a palindrome. #include <bits/stdc++.h> using namespace std; // Returns true if a string str[st..end] is palindrome bool isPalin( char str[], int st, int end) { while (st < end) { if (str[st] != str[end]) return false ; st++; end--; } return true ; } // Returns count of insertions on left side to make // str[] a palindrome int findMinInsert( char str[], int n) { // Find the largest prefix of given string // that is palindrome. for ( int i=n-1; i>=0; i--) { // Characters after the palindromic prefix // must be added at the beginning also to make // the complete string palindrome if (isPalin(str, 0, i)) return (n-i-1); } } // Driver program int main() { char Input[] = "JAVA" ; printf ( "%d" , findMinInsert(Input, strlen (Input))); return 0; } |
Java
// Java program to find minimum number // of insertions on left side to form // a palindrome. import java.util.*; class GFG{ // Returns true if a string // str[st..end] is palindrome static boolean isPalin( char []str, int st, int end) { while (st < end) { if (str[st] != str[end]) return false ; st++; end--; } return true ; } // Returns count of insertions on // left side to make str[] a palindrome static int findMinInsert( char []str, int n) { // Find the largest prefix of given // string that is palindrome. for ( int i = n - 1 ; i >= 0 ; i--) { // Characters after the palindromic // prefix must be added at the // beginning also to make the // complete string palindrome if (isPalin(str, 0 , i)) return (n - i - 1 ); } return 0 ; } // Driver Code public static void main(String []args) { char []Input = "JAVA" .toCharArray(); System.out.println(findMinInsert(Input, Input.length)); } } // This code is contributed by pratham76 |
Python3
# Python3 program to find # minimum number of insertions # on left side to form a palindrome. # Returns true if a string # str[st..end] is palindrome def isPalin( str , st, end): while (st < end): if ( str [st] ! = str [end]): return False st + = 1 end - - 1 return True # Returns count of insertions # on left side to make # str[] a palindrome def findMinInsert( str , n): # Find the largest # prefix of given string # that is palindrome. for i in range (n - 1 , - 1 , - 1 ): # Characters after the # palindromic prefix must # be added at the beginning # also to make the complete # string palindrome if (isPalin( str , 0 , i)): return (n - i - 1 ) # Driver Code Input = "JAVA" print (findMinInsert( Input , len ( Input ))) # This code is contributed # by Smitha |
C#
// C# program to find minimum number // of insertions on left side to form // a palindrome. using System; using System.Text; class GFG{ // Returns true if a string // str[st..end] is palindrome static bool isPalin( char []str, int st, int end) { while (st < end) { if (str[st] != str[end]) return false ; st++; end--; } return true ; } // Returns count of insertions on // left side to make str[] a palindrome static int findMinInsert( char []str, int n) { // Find the largest prefix of given string // that is palindrome. for ( int i = n - 1; i >= 0; i--) { // Characters after the palindromic // prefix must be added at the // beginning also to make the // complete string palindrome if (isPalin(str, 0, i)) return (n - i - 1); } return 0; } // Driver Code public static void Main( string []args) { char []Input = "JAVA" .ToCharArray(); Console.Write(findMinInsert(Input, Input.Length)); } } // This code is contributed by rutvik_56 |
Output:
3
Time Complexity: O(n2)
Thanks to Utkarsh Trivedi for suggesting this solution.
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