# Longest prefix which is also suffix

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 : 2

Input : S = aaaa
Output : 2

Recommended Practice

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 ` `using` `namespace` `std;`   `// Function to find largest prefix` `// which is also a suffix` `int` `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 code` `int` `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 suffix` `def` `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 code` `s ``=` `"blablabla"` `# Function Call to find the length of the longest prefix which is also a suffix` `print``(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 suffix` `function` `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 code` `const s = ``"blablabla"``;`   `// Function Call to find the length of the longest prefix which is also a suffix` `console.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` `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 post` `// https://www.geeksforgeeks.org/searching-for-patterns-set-2-kmp-algorithm/` `int` `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 function` `int` `main()` `{` `    ``string s = ``"bbabbabb"``;` `    ``cout << longestPrefixSuffix(s);` `    ``return` `0;` `}`   `// Corrected by Nilanshu Yadav`

## C

 `#include ` `#include `   `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 suffix`   `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` `    ``// https://www.geeksforgeeks.org/searching-` `    ``// 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 post` `# https://www.geeksforgeeks.org/searching-for-patterns-set-2-kmp-algorithm/` `def` `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 function` `s ``=` `"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 suffix` `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 computeLPSArray() of in below post` `    ``// https://www.geeksforgeeks.org/searching-` `    ``// 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

 ``

## PHP

 ` ``\$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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