Given a string str and an integer K, the task is to check whether the given string is K-periodic. A string is k-periodic if the string is a repetition of the sub-string str[0 … k-1] i.e. the string “ababab” is 2-periodic. Print Yes if the given string is k-periodic, else print No.
Examples:
Input: str = “geeksgeeks”, k = 5
Output: Yes
Given string can be generated by repeating the prefix of length k i.e. “geeks”Input: str = “geeksforgeeks”, k = 3
Output: No
Approach: Starting with the sub-string str[k, 2k-1], str[2k, 3k-1] and so on, check whether all of these sub-strings are equal to the prefix of the string of length k i.e. str[0, k-1]. If the condition is true for all such sub-strings, then print Yes else print No.
Below is the implementation of the above approach:
// CPP implementation of the approach #include<bits/stdc++.h> using namespace std;
// Function that return true if sub-string
// of length k starting at index i is also
// a prefix of the string
bool isPrefix(string str, int len, int i, int k)
{
// k length sub-string cannot start at index i
if (i + k > len)
return false ;
for ( int j = 0; j < k; j++)
{
// Character mismatch between the prefix
// and the sub-string starting at index i
if (str[i] != str[j])
return false ;
i++;
}
return true ;
}
// Function that returns true if str is K-periodic
bool isKPeriodic(string str, int len, int k)
{
// Check whether all the sub-strings
// str[0, k-1], str[k, 2k-1] ... are equal
// to the k length prefix of the string
for ( int i = k; i < len; i += k)
if (!isPrefix(str, len, i, k))
return false ;
return true ;
}
// Driver code
int main()
{
string str = "geeksgeeks" ;
int len = str.length();
int k = 5;
if (isKPeriodic(str, len, k))
cout << ( "Yes" );
else
cout << ( "No" );
}
// This code is contributed by // Surendra_Gangwar |
// Java implementation of the approach class GFG {
// Function that return true if sub-string
// of length k starting at index i is also
// a prefix of the string
static boolean isPrefix(String str, int len, int i, int k)
{
// k length sub-string cannot start at index i
if (i + k > len)
return false ;
for ( int j = 0 ; j < k; j++) {
// Character mismatch between the prefix
// and the sub-string starting at index i
if (str.charAt(i) != str.charAt(j))
return false ;
i++;
}
return true ;
}
// Function that returns true if str is K-periodic
static boolean isKPeriodic(String str, int len, int k)
{
// Check whether all the sub-strings
// str[0, k-1], str[k, 2k-1] ... are equal
// to the k length prefix of the string
for ( int i = k; i < len; i += k)
if (!isPrefix(str, len, i, k))
return false ;
return true ;
}
// Driver code
public static void main(String[] args)
{
String str = "geeksgeeks" ;
int len = str.length();
int k = 5 ;
if (isKPeriodic(str, len, k))
System.out.print( "Yes" );
else
System.out.print( "No" );
}
} |
# Python3 implementation of the approach # Function that returns true if sub-string # of length k starting at index i # is also a prefix of the string def isPrefix(string, length, i, k):
# k length sub-string cannot
# start at index i
if i + k > length:
return False
for j in range ( 0 , k):
# Character mismatch between the prefix
# and the sub-string starting at index i
if string[i] ! = string[j]:
return False
i + = 1
return True
# Function that returns true if # str is K-periodic def isKPeriodic(string, length, k):
# Check whether all the sub-strings
# str[0, k-1], str[k, 2k-1] ... are equal
# to the k length prefix of the string
for i in range (k, length, k):
if isPrefix(string, length, i, k) = = False :
return False
return True
# Driver code if __name__ = = "__main__" :
string = "geeksgeeks"
length = len (string)
k = 5
if isKPeriodic(string, length, k) = = True :
print ( "Yes" )
else :
print ( "No" )
# This code is contributed # by Rituraj Jain |
// C# implementation of the approach using System;
class GFG
{ // Function that return true if sub-string
// of length k starting at index i is also
// a prefix of the string
static bool isPrefix(String str, int len, int i, int k)
{
// k length sub-string cannot start at index i
if (i + k > len)
return false ;
for ( int j = 0; j < k; j++)
{
// Character mismatch between the prefix
// and the sub-string starting at index i
if (str[i] != str[j])
return false ;
i++;
}
return true ;
}
// Function that returns true if str is K-periodic
static bool isKPeriodic(String str, int len, int k)
{
// Check whether all the sub-strings
// str[0, k-1], str[k, 2k-1] ... are equal
// to the k length prefix of the string
for ( int i = k; i < len; i += k)
if (!isPrefix(str, len, i, k))
return false ;
return true ;
}
// Driver code
public static void Main()
{
String str = "geeksgeeks" ;
int len = str.Length;
int k = 5;
if (isKPeriodic(str, len, k))
Console.Write( "Yes" );
else
Console.Write( "No" );
}
} /* This code contributed by PrinciRaj1992 */ |
<?php // PHP implementation of the approach // Function that return true if sub- // of length $k starting at index $i // is also a prefix of the string function isPrefix( $str , $len , $i , $k )
{ // $k length sub- cannot start at index $i
if ( $i + $k > $len )
return false;
for ( $j = 0; $j < $k ; $j ++)
{
// Character mismatch between the prefix
// and the sub- starting at index $i
if ( $str [ $i ] != $str [ $j ])
return false;
$i ++;
}
return true;
} // Function that returns true if $str is K-periodic function isKPeriodic( $str , $len , $k )
{ // Check whether all the sub-strings
// $str[0, $k-1], $str[$k, 2k-1] ... are equal
// to the $k length prefix of the
for ( $i = $k ; $i < $len ; $i += $k )
if (!isPrefix( $str , $len , $i , $k ))
return false;
return true;
} // Driver code $str = "geeksgeeks" ;
$len = strlen ( $str );
$k = 5;
if (isKPeriodic( $str , $len , $k ))
echo ( "Yes" );
else echo ( "No" );
// This code is contributed by ihritik ?> |
<script> // js implementation of the approach // Function that return true if sub-string // of length k starting at index i is also // a prefix of the string function isPrefix(str, len, i, k){
// k length sub-string cannot start at index i
if (i + k > len)
return false ;
for (let j = 0; j < k; j++)
{
// Character mismatch between the prefix
// and the sub-string starting at index i
if (str[i] != str[j])
return false ;
i++;
}
return true ;
} // Function that returns true if str is K-periodic function isKPeriodic(str, len, k)
{ // Check whether all the sub-strings
// str[0, k-1], str[k, 2k-1] ... are equal
// to the k length prefix of the string
for (let i = k; i < len; i += k)
if (!isPrefix(str, len, i, k))
return false ;
return true ;
} // Driver code let str = "geeksgeeks" ;
let len = str.length; let k = 5; if (isKPeriodic(str, len, k))
document.write( "Yes" );
else document.write( "No" );
</script> |
Yes
Time Complexity: O(K * log(len))
Auxiliary Space: O(1)