Minimum insertions to form shortest palindrome

28 Jun, 2022
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++
Java
Python3
C#
Javascript
// 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 palindromebool 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 palindromeint 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 programint main(){    char Input[] = "JAVA";    printf("%d", findMinInsert(Input, strlen(Input)));    return 0;}
// 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 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 palindromedef 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 CodeInput = "JAVA"print(findMinInsert(Input,                    len(Input))) # This code is contributed# by Smitha
// 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
<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 palindromefunction 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 palindromefunction 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>
3
Time Complexity: O(n2)
Auxiliary Space: O(1)
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
Here we’ll preprocess the string and use the lps[] to skip the characters and store the longest palindromic prefix.
For example: s = "AABAACAABAA”, 
lps[] is [0, 1, 0, 1, 2, 0, 1, 2, 3, 4, 5]
In the above example we clearly see that the lps[] can give the length of longest proper palindromic prefix. But here we need longest palindromic prefix not proper palindromic prefix. So, to handle this we can we can append the reverse of S in S.
For Example: s = ABBC
newS = "ABBC" + "#" + "CBBA"
lps[] for newS is [0 0 0 0 0 0 0 0 1 1]
As 1 is there at the end of lps[]. so our ans is 4 - 1 = 3
Using the above logic, we can get the solution in O(N) time only.
C++
#include <bits/stdc++.h>using namespace std;// using kmp algorithmint find_lps(string str){    int n = str.length();    vector<int> lps(n, 0); // initialized with 0     int len = 0; // length of previous longest prefix suffix    // the loop calculates lps[i] for i = 1 to M-1    int i = 1;    while (i < n) {        if (str[len] == str[i]) {            lps[i++] = ++len;        }        else {            if (len != 0)                len = lps[len - 1];            else                lps[i++] = 0;        }    }    return lps[n - 1];}int main(){    string s = "AABAACAABAA";    string rev = s;    reverse(rev.begin(), rev.end());    string newS = s + "#" + rev;    int len = find_lps(newS);    cout << s.length() - len << endl;    return 0;}
0
Time Complexity: O(N),Auxiliary Space: O(N) where N is the length of newS.
Smitha Dinesh Semwal
rutvik_56
pratham76
vaibhavrabadiya117
manish_kr
shivamanandrj9
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