Minimum insertions to form shortest palindrome

Difficulty Level : Medium
Last Updated : 28 May, 2021

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:

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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

Javascript

<script>
// javascript program to find minimum number of insertions
// on left side to form a palindrome.
 
// Returns true if a string str[st..end] is palindrome
function isPalin(str,st,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
function findMinInsert(str,n)
{
    // Find the largest prefix of given string
    // that is palindrome.
    for (let 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);
    }
}
 
    let Input = "JAVA";
    document.write(findMinInsert(Input,Input.length));
     
 // This code is contributed by vaibhavrabadiya07.  
</script>

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

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