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’.
- Find the 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.
- 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 <bits/stdc++.h>using namespace std;// A utility function to fill lps[] or compute prefix// function used in KMP string matching algorithm. Refer// detailsvoid computeLPSArray(char str[], int M, int lps[]){ // length 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[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 functionint 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;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// A C++ program to check if a string is 'n' times// repetition of one of its substrings#include <stdbool.h>#include <stdio.h>#include <string.h>// A utility function to fill lps[] or compute prefix// function used in KMP string matching algorithm. Refer// detailsvoid computeLPSArray(char str[], int M, int lps[]){ // length 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[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 functionint 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]) ? printf("True\n") : printf("False\n"); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program to check if a string is 'n'// times repetition of one of its substringsimport java.io.*;class GFG { // A utility function to fill lps[] or compute // prefix function used in KMP string matching // algorithm. Refer // for details static void computeLPSArray(String str, int M, int lps[]) { // length 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 Aditya Kumar (adityakumar129) |
Python3
# 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 function# used in KMP string matching algorithm. Refer# https://www.geeksforgeeks.org/archives/11902 for detailsdef 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 length > 0 and n%(n-length) == 0: return True else: False# Driver programtxt = ["ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC"]n = len(txt)for i in range(n): if isRepeat(txt[i]): print (True) else: print (False)# This code is contributed by BHAVYA JAIN |
C#
// C# program to check if a string is 'n'// times repetition of one of its substringsusing System;class GFG { // A utility function to fill lps[] or // compute prefix function used in KMP // string matching algorithm. Refer // for details static void computeLPSArray(String str, int M, int []lps) { // length 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[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. static bool 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[] txt = {"ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC"}; int n = txt.Length; for (int i = 0; i < n; i++) { if(isRepeat(txt[i]) == true) Console.WriteLine("True"); else Console.WriteLine("False"); } }}// This code is contributed by Sam007. |
Javascript
<script> // Javascript program to check if a string is 'n' // times repetition of one of its substrings // A utility function to fill lps[] or // compute prefix function used in KMP // string matching algorithm. Refer // for details function computeLPSArray(str, M, lps) { // length of the previous // longest prefix suffix let len = 0; let 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. function isRepeat(str) { // Find length of string and create // an array to store lps values used // in KMP let n = str.length; let lps = new Array(n); lps.fill(0); // Preprocess the pattern (calculate // lps[] array) computeLPSArray(str, n, lps); // Find length of longest suffix // which is also prefix of str. let 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; } let txt = ["ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC"]; let n = txt.length; for (let i = 0; i < n; i++) { if(isRepeat(txt[i]) == true) document.write("True" + "</br>"); else document.write("False" + "</br>"); }</script> |
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.
Auxiliary Space: O(n) because it is using extra space for array “lps”
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.
Another approach:
Above problem can be solved with out using KMP algorithm and extra space.
This approach uses two pointers to check the minimum period of a String as a first step. Period of a String is the length of prefix substring which can be repeated x(x=length/period) times to construct the given string.
for eg: Input string “abcabcabcabc” is having a period 3. which means we can construct the given string by repeating first 3 characters 4 (length/3=4) number of times.
Approach:
- Initially set first pointer – i at 0th index of given string and second pointer – j at 1st index.
- check the characters at both index. if both matches, take period as (j-i) and increment i and j.
- if doesn’t match, Once again check if current character matches with first character at 0th index. if matches, update period as j(j-0=j) and set i to next character.
- f both characters not matches, set i to 0 and update period to -1.
- At the end check if the calculated period exactly divides the length of the string. if not, update period to -1. This check eliminates the presence of strings ending with suffix less than the period. for e.g period for the string “GEEKSFORGEEKS” is calculated as 8 but the suffix string(GEEKS) is having 5 characters only and is incomplete.
Implementation:
C++
/*package whatever //do not write package name here */#include <bits/stdc++.h>using namespace std;int findPeriod(string A){ int length = A.size(); // Initially consider there is no period for given // String int period = -1; /*set two pointers one(i) at index 0 and other(j) at index 1. increment j till first character is obtained in the string*/ int i = 0; for (int j = 1; j < length; j++) { int currChar = A[j]; int comparator = A[i]; /*If current character matches with first *character *update period as the difference of j and i*/ if (currChar == comparator) { period = (j - i); i++; } /* if any mismatch occurs in between set i to * zero also check if current character again * matches * with starting character. If matches, update * period*/ else { if (currChar == A[0]) { i = 1; period = j; } else { i = 0; period = -1; } } } /*check if the period is exactly dividing * string if not reset the period to -1 * this eliminates partial substrings like * e.g string -"GEEKSFORGEEKS" */ period = (length % period != 0) ? -1 : period; return period;}int main(){ vector<string> testStrings = { "ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC" }; int n = testStrings.size(); for (int i = 0; i < n; i++) { if (findPeriod(testStrings[i]) == -1) cout << "false\n"; else cout << "True\n"; } return 0;} // This code is contributed by rakeshsahni |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { public static int findPeriod(String A) { int length = A.length(); // Initially consider there is no period for given // String int period = -1; /*set two pointers one(i) at index 0 and other(j) at index 1. increment j till first character is obtained in the string*/ int i = 0; for (int j = 1; j < length; j++) { int currChar = A.charAt(j); int comparator = A.charAt(i); /*If current character matches with first *character *update period as the difference of j and i*/ if (currChar == comparator) { period = (j - i); i++; } /* if any mismatch occurs in between set i to * zero also check if current character again * matches * with starting character. If matches, update * period*/ else { if (currChar == A.charAt(0)) { i = 1; period = j; } else { i = 0; period = -1; } } } /*check if the period is exactly dividing * string if not reset the period to -1 * this eliminates partial substrings like * e.g string -"GEEKSFORGEEKS" */ period = (length % period != 0) ? -1 : period; return period; } public static void main(String[] args) { String[] testStrings = { "ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC" }; int n = testStrings.length; for (int i = 0; i < n; i++) { if (findPeriod(testStrings[i]) == -1) System.out.println("false"); else System.out.println("True"); } }} |
Python3
def findPeriod(A): length = len(A) # Initially consider there is no period for given # String period = -1 # set two pointers one(i) at # index 0 and other(j) at index 1. increment j till # first character is obtained in the string i = 0 for j in range(1,length): currChar = A[j] comparator = A[i] # If current character matches with first # character # update period as the difference of j and i if (currChar == comparator): period = (j - i) i += 1 # if any mismatch occurs in between set i to # zero also check if current character again # matches # with starting character. If matches, update # period else: if (currChar == A[0]): i = 1 period = j else: i = 0 period = -1 # check if the period is exactly dividing # string if not reset the period to -1 # this eliminates partial substrings like # e.g string -"GEEKSFORGEEKS" period = -1 if (length % period != 0) else period return periodtestStrings = [ "ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC" ]n = len(testStrings)for i in range(n): if (findPeriod(testStrings[i]) == -1): print("false") else: print("True")# This code is contributed by shinjanpatra |
C#
using System;public class GFG{ public static int findPeriod(String A) { int length = A.Length; // Initially consider there is no period for given // String int period = -1; /*set two pointers one(i) at index 0 and other(j) at index 1. increment j till first character is obtained in the string*/ int i = 0; for (int j = 1; j < length; j++) { int currChar = A[j]; int comparator = A[i]; /*If current character matches with first *character *update period as the difference of j and i*/ if (currChar == comparator) { period = (j - i); i++; } /* if any mismatch occurs in between set i to * zero also check if current character again * matches * with starting character. If matches, update * period*/ else { if (currChar == A[0]) { i = 1; period = j; } else { i = 0; period = -1; } } } /*check if the period is exactly dividing * string if not reset the period to -1 * this eliminates partial substrings like * e.g string -"GEEKSFORGEEKS" */ period = (length % period != 0) ? -1 : period; return period; } static public void Main (){ // Code String[] txt = {"ABCABC", "ABABAB", "ABCDABCD", "GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC", "ABCDABC"}; int n = txt.Length; for (int i = 0; i < n; i++) { if(findPeriod(txt[i]) == -1) Console.WriteLine("False"); else Console.WriteLine("True"); } }}// This code is contributed by lokeshpotta20. |
Javascript
<script>function findPeriod(A){ let length = A.length; // Initially consider there is no period for given // String let period = -1; /*set two pointers one(i) at index 0 and other(j) at index 1. increment j till first character is obtained in the string*/ let i = 0; for (let j = 1; j < length; j++) { let currChar = A[j]; let comparator = A[i]; /*If current character matches with first *character *update period as the difference of j and i*/ if (currChar == comparator) { period = (j - i); i++; } /* if any mismatch occurs in between set i to * zero also check if current character again * matches * with starting character. If matches, update * period*/ else { if (currChar == A[0]) { i = 1; period = j; } else { i = 0; period = -1; } } } /*check if the period is exactly dividing * string if not reset the period to -1 * this eliminates partial substrings like * e.g string -"GEEKSFORGEEKS" */ period = (length % period != 0) ? -1 : period; return period;}// driver codelet testStrings = [ "ABCABC", "ABABAB", "ABCDABCD","GEEKSFORGEEKS", "GEEKGEEK", "AAAACAAAAC","ABCDABC" ];let n = testStrings.length;for (let i = 0; i < n; i++) { if (findPeriod(testStrings[i]) == -1) document.write("false","</br>"); else document.write("True","</br>");}// This code is contributed by shinjanpatra</script> |
Output
True True True false True True false
Time Complexity : O(n)
Auxiliary Space : O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
