Print the longest prefix of the given string which is also the suffix of the same string

Given string str, the task is to find the longest prefix which is also the suffix of the given string. The prefix and suffix should not overlap. If no such prefix exists then print -1.

Examples:

Input: str = “aabcdaabc” 
Output: aabc 
The string “aabc” is the longest 
prefix which is also suffix.

Input: str = “aaaa” 
Output: aa  

Approach: The idea is to use the pre-processing algorithm of the KMP search. In this algorithm, we build lps array which stores the following values:  

lps[i] = the longest proper prefix of pat[0..i]
which is also a suffix of pat[0..i].  

We get the length using the above approach, then print the same number of characters from the front which is our answer. 

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Returns length of the longest prefix
// which is also suffix and the two do
// not overlap. This function mainly is
// copy of computeLPSArray() in KMP Algorithm
int LengthlongestPrefixSuffix(string s)
{
    int n = s.length();

    int lps[n];

    // lps[0] is always 0
    lps[0] = 0;

    // Length of the previous
    // longest prefix suffix
    int len = 0;

    // Loop to calculate lps[i]
    // for i = 1 to n - 1
    int i = 1;
    while (i < n) {
        if (s[i] == s[len]) {
            len++;
            lps[i] = len;
            i++;
        }
        else {

            // This is tricky. Consider
            // the example. AAACAAAA
            // and i = 7. The idea is
            // similar to search step.
            if (len != 0) {
                len = lps[len - 1];

                // Also, note that we do
                // not increment i here
            }

            // If len = 0
            else {
                lps[i] = 0;
                i++;
            }
        }
    }

    int res = lps[n - 1];

    // Since we are looking for
    // non overlapping parts
    return (res > n / 2) ? n / 2 : res;
}

// Function that returns the prefix
string longestPrefixSuffix(string s)
{
    // Get the length of the longest prefix
    int len = LengthlongestPrefixSuffix(s);

    // Stores the prefix
    string prefix = "";

    // Traverse and add characters
    for (int i = 0; i < len; i++)
        prefix += s[i];

    // Returns the prefix
    return prefix;
}

// Driver code
int main()
{
    string s = "abcab";
    string ans = longestPrefixSuffix(s);
    if (ans == "")
        cout << "-1";
    else
        cout << ans;

    return 0;
}

Java

// Java implementation of the approach
class GfG
{

// Returns length of the longest prefix
// which is also suffix and the two do
// not overlap. This function mainly is
// copy of computeLPSArray() in KMP Algorithm
static int LengthlongestPrefixSuffix(String s)
{
    int n = s.length();

    int lps[] = new int[n];

    // lps[0] is always 0
    lps[0] = 0;

    // Length of the previous
    // longest prefix suffix
    int len = 0;

    // Loop to calculate lps[i]
    // for i = 1 to n - 1
    int i = 1;
    while (i < n)
    {
        if (s.charAt(i) == s.charAt(len))
        {
            len++;
            lps[i] = len;
            i++;
        }
        else
        {

            // This is tricky. Consider
            // the example. AAACAAAA
            // and i = 7. The idea is
            // similar to search step.
            if (len != 0)
            {
                len = lps[len - 1];

                // Also, note that we do
                // not increment i here
            }

            // If len = 0
            else
            {
                lps[i] = 0;
                i++;
            }
        }
    }

    int res = lps[n - 1];

    // Since we are looking for
    // non overlapping parts
    return (res > n / 2) ? n / 2 : res;
}

// Function that returns the prefix
static String longestPrefixSuffix(String s)
{
    // Get the length of the longest prefix
    int len = LengthlongestPrefixSuffix(s);

    // Stores the prefix
    String prefix = "";

    // Traverse and add characters
    for (int i = 0; i < len; i++)
        prefix += s.charAt(i);

    // Returns the prefix
    return prefix;
}

// Driver code
public static void main(String[] args)
{
    String s = "abcab";
    String ans = longestPrefixSuffix(s);
    if (ans == "")
        System.out.println("-1");
    else
        System.out.println(ans);
}
}

Python3

# Python 3 implementation of the approach

# Returns length of the longest prefix
# which is also suffix and the two do
# not overlap. This function mainly is
# copy of computeLPSArray() in KMP Algorithm
def LengthlongestPrefixSuffix(s):
    n = len(s)

    lps = [0 for i in range(n)]

    # Length of the previous
    # longest prefix suffix
    len1 = 0

    # Loop to calculate lps[i]
    # for i = 1 to n - 1
    i = 1
    while (i < n):
        if (s[i] == s[len1]):
            len1 += 1
            lps[i] = len1
            i += 1

        else:

            # This is tricky. Consider
            # the example. AAACAAAA
            # and i = 7. The idea is
            # similar to search step.
            if (len1 != 0):
                len1 = lps[len1 - 1]

                # Also, note that we do
                # not increment i here

            # If len = 0
            else:
                lps[i] = 0
                i += 1

    res = lps[n - 1]

    # Since we are looking for
    # non overlapping parts
    if (res > int(n / 2)):
        return int(n / 2)
    else:
        return res

# Function that returns the prefix
def longestPrefixSuffix(s):

    # Get the length of the longest prefix
    len1 = LengthlongestPrefixSuffix(s)

    # Stores the prefix
    prefix = ""

    # Traverse and add characters
    for i in range(len1):
        prefix += s[i]

    # Returns the prefix
    return prefix

# Driver code
if __name__ == '__main__':
    s = "abcab"
    ans = longestPrefixSuffix(s)
    if (ans == ""):
        print("-1")
    else:
        print(ans)

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;

class GfG
{

    // Returns length of the longest prefix
    // which is also suffix and the two do
    // not overlap. This function mainly is
    // copy of computeLPSArray() in KMP Algorithm
    static int LengthlongestPrefixSuffix(string s)
    {
        int n = s.Length;

        int []lps = new int[n];

        // lps[0] is always 0
        lps[0] = 0;

        // Length of the previous
        // longest prefix suffix
        int len = 0;

        // Loop to calculate lps[i]
        // for i = 1 to n - 1
        int i = 1;
        while (i < n)
        {
            if (s[i] == s[len])
            {
                len++;
                lps[i] = len;
                i++;
            }
            else
            {

                // This is tricky. Consider
                // the example. AAACAAAA
                // and i = 7. The idea is
                // similar to search step.
                if (len != 0)
                {
                    len = lps[len - 1];

                    // Also, note that we do
                    // not increment i here
                }

                // If len = 0
                else
                {
                    lps[i] = 0;
                    i++;
                }
            }
        }

        int res = lps[n - 1];

        // Since we are looking for
        // non overlapping parts
        return (res > n / 2) ? n / 2 : res;
    }

    // Function that returns the prefix
    static String longestPrefixSuffix(string s)
    {
        // Get the length of the longest prefix
        int len = LengthlongestPrefixSuffix(s);

        // Stores the prefix
        string prefix = "";

        // Traverse and add characters
        for (int i = 0; i < len; i++)
            prefix += s[i];

        // Returns the prefix
        return prefix;
    }

    // Driver code
    public static void Main()
    {
        string s = "abcab";
        string ans = longestPrefixSuffix(s);
        if (ans == "")
            Console.WriteLine("-1");
        else
            Console.WriteLine(ans);
    }
}

// This code is contributed by Ryuga

Javascript

<script>

// JavaScript implementation of the approach


// Returns length of the longest prefix
// which is also suffix and the two do
// not overlap. This function mainly is
// copy of computeLPSArray() in KMP Algorithm
function LengthlongestPrefixSuffix(s)
{
    var n = s.length;

    var lps = Array.from({length: n}, (_, i) => 0);

    // lps[0] is always 0
    lps[0] = 0;

    // Length of the previous
    // longest prefix suffix
    var len = 0;

    // Loop to calculate lps[i]
    // for i = 1 to n - 1
    var i = 1;
    while (i < n)
    {
        if (s.charAt(i) == s.charAt(len))
        {
            len++;
            lps[i] = len;
            i++;
        }
        else
        {

            // This is tricky. Consider
            // the example. AAACAAAA
            // and i = 7. The idea is
            // similar to search step.
            if (len != 0)
            {
                len = lps[len - 1];

                // Also, note that we do
                // not increment i here
            }

            // If len = 0
            else
            {
                lps[i] = 0;
                i++;
            }
        }
    }

    var res = lps[n - 1];

    // Since we are looking for
    // non overlapping parts
    return (res > n / 2) ? n / 2 : res;
}

// Function that returns the prefix
function longestPrefixSuffix(s)
{
    // Get the length of the longest prefix
    var len = LengthlongestPrefixSuffix(s);

    // Stores the prefix
    var prefix = "";

    // Traverse and add characters
    for (var i = 0; i < len; i++)
        prefix += s.charAt(i);

    // Returns the prefix
    return prefix;
}

// Driver code
var s = "abcab";
var ans = longestPrefixSuffix(s);
if (ans == "")
    document.write("-1");
else
    document.write(ans);

// This code contributed by shikhasingrajput

</script>

Output

ab

Time Complexity: O(N), as we are using a loop to traverse N times to build los array. Where N is the length of the string.
Auxiliary Space: O(N), as we are using extra space for the lps array. Where N is the length of the string.