Program to check if all characters have even frequency
Given a string S consisting only of lowercase letters check if the string has all characters appearing even times.
Examples:
Input : abaccaba
Output : Yes
Explanation: ‘a’ occurs four times, ‘b’ occurs twice, ‘c’ occurs twice and the other letters occur zero times.Input: hthth
Output : No
Approach: We will go through the string and count the occurrence of all characters after that we will check if the occurrences are even or not if there is any odd frequency character then we immediately print No.
C++
// C++ implementation of the above approach #include <iostream> using namespace std; bool check(string s) { // creating a frequency array int freq[26] = {0}; // Finding length of s int n = s.length(); for ( int i = 0; i < s.length(); i++) // counting frequency of all characters freq[s[i] - 97]++; // checking if any odd frequency // is there or not for ( int i = 0; i < 26; i++) if (freq[i] % 2 == 1) return false ; return true ; } // Driver Code int main() { string s = "abaccaba" ; check(s) ? cout << "Yes" << endl : cout << "No" << endl; return 0; } // This code is contributed by // sanjeev2552 |
Java
// Java implementation of the above approach class GFG { static boolean check(String s) { // creating a frequency array int [] freq = new int [ 26 ]; // Finding length of s int n = s.length(); // counting frequency of all characters for ( int i = 0 ; i < s.length(); i++) { freq[(s.charAt(i)) - 97 ] += 1 ; } // checking if any odd frequency // is there or not for ( int i = 0 ; i < freq.length; i++) { if (freq[i] % 2 == 1 ) { return false ; } } return true ; } // Driver Code public static void main(String[] args) { String s = "abaccaba" ; if (check(s)) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } } } // This code is contributed by Rajput-Ji |
Python3
# Python implementation of the above approach def check(s): # creating a frequency array freq = [ 0 ] * 26 # Finding length of s n = len (s) for i in range (n): # counting frequency of all characters freq[ ord (s[i]) - 97 ] + = 1 for i in range ( 26 ): # checking if any odd frequency # is there or not if (freq[i] % 2 = = 1 ): return False return True # Driver code s = "abaccaba" if (check(s)): print ( "Yes" ) else : print ( "No" ) |
C#
// C# implementation of the approach using System; class GFG { static Boolean check(String s) { // creating a frequency array int [] freq = new int [26]; // Finding length of s int n = s.Length; // counting frequency of all characters for ( int i = 0; i < s.Length; i++) { freq[(s[i]) - 97] += 1; } // checking if any odd frequency // is there or not for ( int i = 0; i < freq.Length; i++) { if (freq[i] % 2 == 1) { return false ; } } return true ; } // Driver Code public static void Main(String[] args) { String s = "abaccaba" ; if (check(s)) { Console.WriteLine( "Yes" ); } else { Console.WriteLine( "No" ); } } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript implementation of the above approach function check(s) { // creating a frequency array var freq = Array(26).fill(0); // Finding length of s var n = s.length; for ( var i = 0; i < s.length; i++) // counting frequency of all characters freq[s[i] - 97]++; // checking if any odd frequency // is there or not for ( var i = 0; i < 26; i++) if (freq[i] % 2 == 1) return false ; return true ; } // Driver Code var s = "abaccaba" ; check(s) ? document.write( "Yes" ) : document.write( "No" ); </script> |
Yes
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(26) ⇒ O(1), no extra space is required, so it is a constant.
Method #2:Using built-in python functions.
Approach: We will scan the string and count the occurrence of all characters using the built-in Counter() function after that we traverse the counter list and check if the occurrences are even or not if there is any odd frequency character then we immediately print No.
Note: This method is applicable for all types of characters
Python3
# Python implementation for # the above approach # importing Counter function from collections import Counter # Function to check if all # elements occur even times def checkString(s): # Counting the frequency of all # character using Counter function frequency = Counter(s) # Traversing frequency for i in frequency: # Checking if any element # has odd count if (frequency[i] % 2 = = 1 ): return False return True # Driver code s = "geeksforgeeksfor" if (checkString(s)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by vikkycirus |
Yes
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