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 |
chevron_right
filter_none
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 |
chevron_right
filter_none
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") |
chevron_right
filter_none
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 |
chevron_right
filter_none
Output:
Yes
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
