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Check if a string is made up of K alternating characters

  • Last Updated : 27 May, 2021

Given a string str and an integer K, the task is to check if it is made up of K alternating characters.
Examples: 
 

Input: str = “acdeac”, K = 4 
Output: Yes
Input: str = “abcdcab”, K = 2 
Output: No 
 

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Approach: In order for the string to be made up of K alternating characters, it must satisfy the following conditions:
 

  1. All the characters at (i + mK) indices must be same, where i is the current index and mK represents the mth multiple of K. It means that after every K indices, the character must get repeated.
  2. The adjacent characters must not be the same. This is because if the string is of type “AAAAA”, where a single character is repeated any number of time, the above condition will get matched, but the answer must be ‘No’ in this case.
    Below code is the implementation of the above approach:
     

C++




// C++ implementation of the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a string
// is made up of k alternating characters
bool isKAlternating(string s, int k)
{
    if (s.length() < k)
        return false;
 
    int checker = 0;
 
    // Check if all the characters at
    // indices 0 to K-1 are different
    for (int i = 0; i < k; i++) {
 
        int bitAtIndex = s[i] - 'a';
 
        // If that bit is already set in
        // checker, return false
        if ((checker & (1 << bitAtIndex)) > 0) {
            return false;
        }
 
        // Otherwise update and continue by
        // setting that bit in the checker
        checker = checker | (1 << bitAtIndex);
    }
 
    for (int i = k; i < s.length(); i++)
        if (s[i - k] != s[i])
            return false;
 
    return true;
}
 
// Driver code
int main()
{
    string str = "acdeac";
    int K = 4;
 
    if (isKAlternating(str, K))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
 
    return 0;
}

Java




// Java implementation of the approach
class GFG{
  
// Function to check if a String
// is made up of k alternating characters
static boolean isKAlternating(String s, int k)
{
    if (s.length() < k)
        return false;
  
    int checker = 0;
  
    // Check if all the characters at
    // indices 0 to K-1 are different
    for (int i = 0; i < k; i++) {
  
        int bitAtIndex = s.charAt(i) - 'a';
  
        // If that bit is already set in
        // checker, return false
        if ((checker & (1 << bitAtIndex)) > 0) {
            return false;
        }
  
        // Otherwise update and continue by
        // setting that bit in the checker
        checker = checker | (1 << bitAtIndex);
    }
  
    for (int i = k; i < s.length(); i++)
        if (s.charAt(i - k) != s.charAt(i) )
            return false;
  
    return true;
}
  
// Driver code
public static void main(String[] args)
{
    String str = "acdeac";
    int K = 4;
  
    if (isKAlternating(str, K))
        System.out.print("Yes" +"\n");
    else
        System.out.print("No" +"\n");
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python 3 implementation of the approach
  
# Function to check if a string
# is made up of k alternating characters
def isKAlternating( s, k):
    if (len(s) < k):
        return False
  
    checker = 0
  
    # Check if all the characters at
    # indices 0 to K-1 are different
    for i in range( k):
  
        bitAtIndex = ord(s[i]) - ord('a')
  
        # If that bit is already set in
        # checker, return false
        if ((checker & (1 << bitAtIndex)) > 0):
            return False
  
        # Otherwise update and continue by
        # setting that bit in the checker
        checker = checker | (1 << bitAtIndex)
  
    for i in range(k,len(s)):
        if (s[i - k] != s[i]):
            return False
  
    return True
  
# Driver code
if __name__ =="__main__":
 
    st = "acdeac"
    K = 4
  
    if (isKAlternating(st, K)):
        print ("Yes")
    else:
        print ("No")
  
# This code is contributed by chitranayal  

C#




// C# implementation of the approach
using System;
 
class GFG{
 
// Function to check if a String
// is made up of k alternating characters
static bool isKAlternating(String s, int k)
{
    if (s.Length < k)
        return false;
 
    int checker = 0;
 
    // Check if all the characters at
    // indices 0 to K-1 are different
    for (int i = 0; i < k; i++) {
 
        int bitAtIndex = s[i] - 'a';
 
        // If that bit is already set in
        // checker, return false
        if ((checker & (1 << bitAtIndex)) > 0) {
            return false;
        }
 
        // Otherwise update and continue by
        // setting that bit in the checker
        checker = checker | (1 << bitAtIndex);
    }
 
    for (int i = k; i < s.Length; i++)
        if (s[i - k] != s[i] )
            return false;
 
    return true;
}
 
// Driver code
public static void Main()
{
    String str = "acdeac";
    int K = 4;
 
    if (isKAlternating(str, K))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This article contributed by AbhiThakur

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to check if a string
// is made up of k alternating characters
function isKAlternating(s, k)
{
    if (s.length < k)
        return false;
 
    var checker = 0;
 
    // Check if all the characters at
    // indices 0 to K-1 are different
    for (var i = 0; i < k; i++) {
 
        var bitAtIndex = s[i].charCodeAt(0) - 'a'.charCodeAt(0);
 
        // If that bit is already set in
        // checker, return false
        if ((checker & (1 << bitAtIndex)) > 0) {
            return false;
        }
 
        // Otherwise update and continue by
        // setting that bit in the checker
        checker = checker | (1 << bitAtIndex);
    }
 
    for (var i = k; i < s.length; i++)
        if (s[i - k] != s[i])
            return false;
 
    return true;
}
 
// Driver code
var str = "acdeac";
var K = 4;
if (isKAlternating(str, K))
    document.write( "Yes" );
else
    document.write( "No" );
 
// This code is contributed by importantly.
</script>
  1.  
Output: 
Yes

 

  1. Time Complexity: O(N)
     



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