Given a string s, find the length of the longest prefix, which is also a suffix. The prefix and suffix should not overlap.
Examples:
Input : S = aabcdaabc
Output : 4
Explanation: The string “aabc” is the longest prefix which is also suffix.Input : S = abcab
Output : 2Input : S = aaaa
Output : 2
Naive approach:
Since overlapping prefixes and suffixes is not allowed, we break the string from the middle and start matching left and right strings. If they are equal return size of one string, else they try for shorter lengths on both sides.
Below is a solution to the above approach:
C++
// CPP program to find length of the// longest prefix which is also suffix#include <bits/stdc++.h>using namespace std;// Function to find largest prefix// which is also a suffixint largest_prefix_suffix(const std::string& str){ int n = str.length(); // If the length of the string is less than 2, there // can't be a non-overlapping prefix and suffix, so // return 0. if (n < 2) { return 0; } // Initialize the length of the longest prefix which is // also a suffix. int len = 0; int i = 0; // Iterate through the first half of the string. while (i < n / 2) { int j1 = 0, j2 = (n - 1) - i; int isPrefixSuffix = 1; // Check if characters at corresponding positions in // the first half (j1) and the second half (j2) of // the string are equal. while (j1 <= i) { // If any pair of characters doesn't match, it's // not a prefix-suffix. if (str[j1] != str[j2]) { isPrefixSuffix = 0; } j1++; j2++; } // If it's a prefix-suffix, update the length. if (isPrefixSuffix == 1) len = i + 1; i++; } // Return the length of the longest prefix which is also // a suffix. return len;}// Driver codeint main(){ string s = "blablabla"; // Function Call to find the length of the longest // prefix which is also a suffix cout << largest_prefix_suffix(s); return 0;} |
Java
public class LongestPrefixSuffix { // Function to find the length of the longest prefix // which is also a suffix public static int largestPrefixSuffix(String str) { int n = str.length(); // If the length of the string is less than 2, there // can't be a non-overlapping prefix and suffix, so // return 0. if (n < 2) { return 0; } // Initialize the length of the longest // prefix which is also a suffix. int len = 0; int i = 0; // Iterate through the first half of the string. while (i < n / 2) { int j1 = 0, j2 = (n - 1) - i; boolean isPrefixSuffix = true; // Check if characters at corresponding // positions in the first half (j1) and the // second half (j2) of the string are equal. while (j1 <= i) { // If any pair of characters doesn't match, // it's not a prefix-suffix. if (str.charAt(j1) != str.charAt(j2)) { isPrefixSuffix = false; } j1++; j2++; } // If it's a prefix-suffix, update the length. if (isPrefixSuffix) len = i + 1; i++; } // Return the length of the longest prefix which is // also a suffix. return len; } // Driver code public static void main(String[] args) { String s = "blablabla"; // Function Call to find the length of the longest // prefix which is also a suffix System.out.println(largestPrefixSuffix(s)); }} |
Python3
# Function to find the length of the longest prefix which is also a suffixdef largest_prefix_suffix(s): n = len(s) # If the length of the string is less than 2, there can't be a non-overlapping # prefix and suffix, so return 0. if n < 2: return 0 lenn = 0 # Initialize the length of the longest prefix which is also a suffix. i = 0 # Iterate through the first half of the string. while i < n // 2: j1 = 0 j2 = (n - 1) - i is_prefix_suffix = True # Check if characters at corresponding positions in the first half (j1) # and the second half (j2) of the string are equal. while j1 <= i: if s[j1] != s[j2]: is_prefix_suffix = False # If any pair of characters doesn't match, it's not a prefix-suffix. j1 += 1 j2 += 1 if is_prefix_suffix: lenn = i + 1 # If it's a prefix-suffix, update the length. i += 1 # Return the length of the longest prefix which is also a suffix. return lenn# Driver codes = "blablabla"# Function Call to find the length of the longest prefix which is also a suffixprint(largest_prefix_suffix(s)) |
C#
using System;class Program { // Function to find the length of the longest prefix // which is also a suffix static int LargestPrefixSuffix(string str) { int n = str.Length; // If the length of the string is less than 2, there // can't be a non-overlapping prefix and suffix, so // return 0. if (n < 2) { return 0; } int len = 0; // Initialize the length of the longest // prefix which is also a suffix. int i = 0; // Iterate through the first half of the string. while (i < n / 2) { int j1 = 0; int j2 = (n - 1) - i; bool isPrefixSuffix = true; // Check if characters at corresponding // positions in the first half (j1) and the // second half (j2) of the string are equal. while (j1 <= i) { if (str[j1] != str[j2]) { isPrefixSuffix = false; // If any pair of // characters doesn't // match, it's not a // prefix-suffix. } j1++; j2++; } if (isPrefixSuffix) len = i + 1; // If it's a prefix-suffix, // update the length. i++; } // Return the length of the longest prefix which is // also a suffix. return len; } // Driver code static void Main() { string s = "blablabla"; // Function Call to find the length of the longest // prefix which is also a suffix Console.WriteLine(LargestPrefixSuffix(s)); }} |
Javascript
// Function to find the length of the longest prefix which is also a suffixfunction largestPrefixSuffix(str) { const n = str.length; // If the length of the string is less than 2, there can't be a non-overlapping // prefix and suffix, so return 0. if (n < 2) { return 0; } let len = 0; // Initialize the length of the longest prefix which is also a suffix. let i = 0; // Iterate through the first half of the string. while (i < Math.floor(n / 2)) { let j1 = 0; let j2 = (n - 1) - i; let isPrefixSuffix = true; // Check if characters at corresponding positions in the first half (j1) // and the second half (j2) of the string are equal. while (j1 <= i) { if (str.charAt(j1) !== str.charAt(j2)) { isPrefixSuffix = false; // If any pair of characters doesn't match, it's not a prefix-suffix. } j1++; j2++; } if (isPrefixSuffix) { len = i + 1; // If it's a prefix-suffix, update the length. } i++; } // Return the length of the longest prefix which is also a suffix. return len;}// Driver codeconst s = "blablabla";// Function Call to find the length of the longest prefix which is also a suffixconsole.log(largestPrefixSuffix(s)); |
Output
3
Time Complexity: O(n^2)
Auxiliary Space: O(1)
Longest prefix which is also suffix using KMP algorithm:
The idea is to use the preprocessing algorithm KMP search. In the preprocessing 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].
Below is the implementation:
C++
// Efficient CPP program to find length of // the longest prefix which is also suffix#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 computeLPSArray() of in below postint longestPrefixSuffix(string s){ int n = s.length(); int lps[n]; lps[0] = 0; // lps[0] is always 0 // length of the previous // longest prefix suffix int len = 0; // the loop calculates lps[i] // for i = 1 to n-1 int i = (n+1)/2; while (i < n) { if (s[i] == s[len]) { len++; lps[i] = len; i++; } else // (pat[i] != pat[len]) { // 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 } else // if (len == 0) { lps[i] = 0; i++; } } } int res = lps[n-1]; // Since we are looking for // non overlapping parts. return res;}// Driver program to test above functionint main(){ string s = "bbabbabb"; cout << longestPrefixSuffix(s); return 0;}// Corrected by Nilanshu Yadav |
C
#include <stdio.h>#include <string.h>int longestPrefixSuffix(const char* s){ int n = strlen(s); int lps[n]; lps[0] = 0; // lps[0] is always 0 int len = 0; int i = (n + 1) / 2; while (i < n) { if (s[i] == s[len]) { len++; lps[i] = len; i++; } else { if (len != 0) { len = lps[len - 1]; } else { lps[i] = 0; i++; } } } int res = lps[n - 1]; return res;}int main(){ const char* s = "bbabbabb"; printf("%d\n", longestPrefixSuffix(s)); return 0;} |
Java
// Efficient Java program to find length of // the longest prefix which is also suffixclass GFG { // Returns length of the longest prefix // which is also suffix and the two do // not overlap. This function mainly is // copy computeLPSArray() of in below post // for-patterns-set-2-kmp-algorithm/ static int longestPrefixSuffix(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; // the loop calculates lps[i] // for i = 1 to n-1 int i = (n+1)/2; while (i < n) { if (s.charAt(i) == s.charAt(len)) { len++; lps[i] = len; i++; } // (pat[i] != pat[len]) 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; } // Driver program public static void main (String[] args) { String s = "bbabbabb"; System.out.println(longestPrefixSuffix(s)); }}// This code is contributed by Anant Agarwal.// Corrected by Nilanshu Yadav |
Python3
# Efficient Python 3 program# to find length of # the longest prefix # which is also suffix# Returns length of the longest prefix# which is also suffix and the two do# not overlap. This function mainly is# copy computeLPSArray() of in below postdef longestPrefixSuffix(s) : n = len(s) lps = [0] * n # lps[0] is always 0 # length of the previous # longest prefix suffix l = 0 # the loop calculates lps[i] # for i = 1 to n-1 i = (n+1)//2; while (i < n) : if (s[i] == s[l]) : l = l + 1 lps[i] = l i = i + 1 else : # (pat[i] != pat[len]) # This is tricky. Consider # the example. AAACAAAA # and i = 7. The idea is # similar to search step. if (l != 0) : l = lps[l-1] # Also, note that we do # not increment i here else : # if (len == 0) lps[i] = 0 i = i + 1 res = lps[n-1] # Since we are looking for # non overlapping parts. return res; # Driver program to test above functions = "bbabbabb"print(longestPrefixSuffix(s))# This code is contributed# by Nikita Tiwari.#Corrected by Nilanshu Yadav |
C#
// Efficient C# program to find length of // the longest prefix which is also suffixusing System;class GFG { // Returns length of the longest prefix // which is also suffix and the two do // not overlap. This function mainly is // copy computeLPSArray() of in below post // for-patterns-set-2-kmp-algorithm/ static int longestPrefixSuffix(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; // the loop calculates lps[i] // for i = 1 to n-1 int i = 1; while (i < n) { if (s[i] == s[len]) { len++; lps[i] = len; i++; } // (pat[i] != pat[len]) 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; } // Driver program public static void Main () { string s = "abcab"; Console.WriteLine(longestPrefixSuffix(s)); }}// This code is contributed by vt_m. |
Javascript
<script>// Efficient javascript program to find length of // the longest prefix which is also suffix{// Returns length of the longest prefix// which is also suffix and the two do// not overlap. This function mainly is// copy computeLPSArray() of in below post// for-patterns-set-2-kmp-algorithm/function longestPrefixSuffix(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; // the loop calculates 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++; } // (pat[i] != pat[len]) 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;}// Driver program var s = "abcab";document.write(longestPrefixSuffix(s));// This code is contributed by 29AjayKumar </script> |
PHP
<?php// Efficient PHP program to find length of // the longest prefix which is also suffix// Returns length of the longest prefix// which is also suffix and the two do// not overlap. This function mainly is// copy computeLPSArray() of in below postfunction longestPrefixSuffix($s){ $n = strlen($s); $lps[$n] = NULL; // lps[0] is always 0 $lps[0] = 0; // length of the previous // longest prefix suffix $len = 0; // the loop calculates lps[i] // for i = 1 to n-1 $i = 1; while ($i < $n) { if ($s[$i] == $s[$len]) { $len++; $lps[$i] = $len; $i++; } // (pat[i] != pat[len]) else { // This is tricky. Consider // the example. AAACAAAA // and i = 7. The idea is // similar to search step. if ($len != 0) { $len = $lps[$len-1]; // Also, note that we do // not increment i here } // if (len == 0) else { $lps[$i] = 0; $i++; } } } $res = $lps[$n-1]; // Since we are looking for // non overlapping parts. return ($res > $n/2)? $n/2 : $res;} // Driver Code $s = "abcab"; echo longestPrefixSuffix($s);// This code is contributed by nitin mittal?> |
Output
2
Time Complexity: O(n)
Auxiliary Space: O(n)
Please refer computeLPSArray() of KMP search for an explanation.
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