Minimum insertions to form shortest palindrome

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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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; 
} 

Python 3

# 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 

Output:

Time Complexity: O(n²)

Thanks to Utkarsh Trivedi for suggesting this solution.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Python 3

Recommended Posts: