Find if a given string can be represented from a substring by iterating the substring “n” times

4.4

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.

isRepeat

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.

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



4.4 Average Difficulty : 4.4/5.0
Based on 50 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.