Given a string ‘str’, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.

Examples:

Input: str = "abcabcabc" Output: true The given string is 3 times repetition of "abc" Input: str = "abadabad" Output: true The given string is 2 times repetition of "abad" Input: str = "aabaabaabaab" Output: true The given string is 4 times repetition of "aab" Input: str = "abcdabc" Output: false

Source: Google Interview Question

## We strongly recommend that you click here and practice it, before moving on to the solution.

There can be many solutions to this problem. The challenging part is to solve the problem in O(n) time. Below is a O(n) algorithm.

Let the given string be ‘str’ and length of given string be ‘n’.

1) Find length of the longest proper prefix of ‘str’ which is also a suffix. Let the length of the longest proper prefix suffix be ‘len’. This can be computed in O(n) time using pre-processing step of KMP string matching algorithm.

2) If value of ‘n – len’ divides n (or ‘n % (n-len)’ is 0), then return true, else return false.

In case of ‘true’ , the substring ‘str[0..n-len-1]’ is the substring that repeats n%(n-len) times.

Let us take few examples.

Input: str = “ABCDABCD”, n = 8 (Number of characters in ‘str’)

The value of len is 4 (“ABCD” is the longest substring which is both prefix and suffix)

Since (n-len) divides n, the answer is true.

Input: str = “ABCDABC”, n = 7 (Number of characters in ‘str’)

The value of len is 3 (“ABC” is the longest substring which is both prefix and suffix)

Since (n-len) doesn’t divides n, the answer is false.

Input: str = “ABCABCABCABCABC”, n = 15 (Number of characters in ‘str’)

The value of len is 12 (“ABCABCABCABC” is the longest substring which is both prefix and suffix)

Since (n-len) divides n, the answer is true.

**How does this work?**

length of longest proper prefix-suffix (or len) is always between 0 to n-1. If len is n-1, then all characters in string are same. For example len is 3 for “AAAA”. If len is n-2 and n is even, then two characters in string repeat n/2 times. For example “ABABABAB”, length of lps is 6. The reason is if the first n-2 characters are same as last n-2 character, the starting from the first pair, every pair of characters is identical to the next pair. The following diagram demonstrates same for substring of length 4.

Following is the implementation of above algorithm:

## C++

// A C++ program to check if a string is 'n' times // repetition of one of its substrings #include<iostream> #include<cstring> using namespace std; // A utility function to fill lps[] or compute prefix funcrion // used in KMP string matching algorithm. Refer // http://www.geeksforgeeks.org/archives/11902 for details void computeLPSArray(char str[], int M, int lps[]) { int len = 0; //lenght of the previous longest prefix suffix int i; lps[0] = 0; //lps[0] is always 0 i = 1; // the loop calculates lps[i] for i = 1 to M-1 while (i < M) { if (str[i] == str[len]) { len++; lps[i] = len; i++; } else // (pat[i] != pat[len]) { if (len != 0) { // This is tricky. Consider the example AAACAAAA // and i = 7. len = lps[len-1]; // Also, note that we do not increment i here } else // if (len == 0) { lps[i] = 0; i++; } } } } // Returns true if str is repetition of one of its substrings // else return false. bool isRepeat(char str[]) { // Find length of string and create an array to // store lps values used in KMP int n = strlen(str); int lps[n]; // Preprocess the pattern (calculate lps[] array) computeLPSArray(str, n, lps); // Find length of longest suffix which is also // prefix of str. int len = lps[n-1]; // If there exist a suffix which is also prefix AND // Length of the remaining substring divides total // length, then str[0..n-len-1] is the substring that // repeats n/(n-len) times (Readers can print substring // and value of n/(n-len) for more clarity. return (len > 0 && n%(n-len) == 0)? true: false; } // Driver program to test above function int main() { char txt[][100] = {"ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC"}; int n = sizeof(txt)/sizeof(txt[0]); for (int i=0; i<n; i++) isRepeat(txt[i])? cout << "True\n" : cout << "False\n"; return 0; }

## Java

// Java program to check if a string is 'n' // times repetition of one of its substrings import java.io.*; class GFG { // A utility function to fill lps[] or compute // prefix funcrion used in KMP string matching // algorithm. Refer // http://www.geeksforgeeks.org/archives/11902 // for details static void computeLPSArray(String str, int M, int lps[]) { // lenght of the previous // longest prefix suffix int len = 0; int i; lps[0] = 0; // lps[0] is always 0 i = 1; // the loop calculates lps[i] // for i = 1 to M-1 while (i < M) { if (str.charAt(i) == str.charAt(len)) { len++; lps[i] = len; i++; } else // (pat[i] != pat[len]) { if (len != 0) { // This is tricky. Consider the // example AAACAAAA and i = 7. len = lps[len-1]; // Also, note that we do // not increment i here } else // if (len == 0) { lps[i] = 0; i++; } } } } // Returns true if str is repetition of // one of its substrings else return false. static boolean isRepeat(String str) { // Find length of string and create // an array to store lps values used in KMP int n = str.length(); int lps[] = new int[n]; // Preprocess the pattern (calculate lps[] array) computeLPSArray(str, n, lps); // Find length of longest suffix // which is also prefix of str. int len = lps[n-1]; // If there exist a suffix which is also // prefix AND Length of the remaining substring // divides total length, then str[0..n-len-1] // is the substring that repeats n/(n-len) // times (Readers can print substring and // value of n/(n-len) for more clarity. return (len > 0 && n%(n-len) == 0)? true: false; } // Driver program to test above function public static void main(String[] args) { String txt[] = {"ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC"}; int n = txt.length; for (int i = 0; i < n; i++) { if(isRepeat(txt[i]) == true) System.out.println("True"); else System.out.println("False"); } } } // This code is contributed by Prerna Saini

## Python

# A Python program to check if a string is 'n' times # repetition of one of its substrings # A utility function to fill lps[] or compute prefix funcrion # used in KMP string matching algorithm. Refer # http://www.geeksforgeeks.org/archives/11902 for details def computeLPSArray(string, M, lps): length = 0 # length of the previous longest prefix suffix i = 1 lps[0] = 0 # lps[0] is always 0 # the loop calculates lps[i] for i = 1 to M-1 while i < M: if string[i] == string[length]: length += 1 lps[i] = length i += 1 else: if length != 0: # This is tricky. Consider the example AAACAAAA # and i = 7. length = lps[length-1] # Also, note that we do not increment i here else: lps[i] = 0 i += 1 # Returns true if string is repetition of one of its substrings # else return false. def isRepeat(string): # Find length of string and create an array to # store lps values used in KMP n = len(string) lps = [0] * n # Preprocess the pattern (calculate lps[] array) computeLPSArray(string, n, lps) # Find length of longest suffix which is also # prefix of str. length = lps[n-1] # If there exist a suffix which is also prefix AND # Length of the remaining substring divides total # length, then str[0..n-len-1] is the substring that # repeats n/(n-len) times (Readers can print substring # and value of n/(n-len) for more clarity. if len > 0 and n%(n-length) == 0: return True else: False # Driver program txt = ["ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC"] n = len(txt) for i in xrange(n): if isRepeat(txt[i]): print "True" else: print "False" # This code is contributed by BHAVYA JAIN

Output:

True True True False True True False

Time Complexity: Time complexity of the above solution is O(n) as it uses KMP preprocessing algorithm which is linear time algorithm.

This article is contributed by **Harshit Agrawal**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.