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++ 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 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 |
# 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# 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 |
<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
// C++ program for the above approach #include <iostream> #include <unordered_map> using namespace std;
bool checkString(string s) {
// Counting the frequency of all character
unordered_map< char , int > frequency;
for ( int i = 0; i < s.length(); i++) {
char ch = s.at(i);
frequency[ch]++;
}
// Traversing frequency
for ( auto value : frequency) {
// Checking if any element has odd count
if (value.second % 2 == 1) {
return false ;
}
}
return true ;
} // Driver code int main() {
string s = "geeksforgeeksfor" ;
if (checkString(s)) {
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
return 0;
} // This code is contributed by adityashatmfh |
import java.util.Map;
import java.util.HashMap;
public class Main {
static boolean checkString(String s) {
// Counting the frequency of all character
Map<Character, Integer> frequency = new HashMap<>();
for ( int i = 0 ; i < s.length(); i++) {
char ch = s.charAt(i);
frequency.put(ch, frequency.getOrDefault(ch, 0 ) + 1 );
}
// Traversing frequency
for ( int value : frequency.values()) {
// Checking if any element has odd count
if (value % 2 == 1 ) {
return false ;
}
}
return true ;
}
// Driver code
public static void main(String[] args) {
String s = "geeksforgeeksfor" ;
if (checkString(s)) {
System.out.println( "Yes" );
} else {
System.out.println( "No" );
}
}
} |
# 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 |
// C# program for the above approach using System;
using System.Collections.Generic;
class MainClass {
static bool CheckString( string s) {
// Counting the frequency of all characters
Dictionary< char , int > frequency = new Dictionary< char , int >();
foreach ( char ch in s) {
if (frequency.ContainsKey(ch)) {
frequency[ch]++;
} else {
frequency.Add(ch, 1);
}
}
// Traversing frequency
foreach ( int value in frequency.Values) {
// Checking if any element has odd count
if (value % 2 == 1) {
return false ;
}
}
return true ;
}
// Driver code
public static void Main( string [] args) {
string s = "geeksforgeeksfor" ;
if (CheckString(s)) {
Console.WriteLine( "Yes" );
} else {
Console.WriteLine( "No" );
}
}
} // This code is contributed by codebraxnzt |
// JavaScript implementation of the above approach function checkString(s)
{ // Counting the frequency of all character let frequency = new Map();
for (let i = 0; i < s.length; i++) {
if (frequency.has(s[i])) {
frequency.set(s[i], frequency.get(s[i]) + 1);
} else {
frequency.set(s[i], 1);
}
} // Traversing frequency for (let value of frequency.values())
{ // Checking if any element has odd count if (value % 2 == 1) {
return false ;
}
} return true ;
} // Driver code let s = "geeksforgeeksfor" ;
if (checkString(s)) {
console.log( "Yes" );
} else {
console.log( "No" );
} // This code is contributed by codebraxnzt |
Yes