原文：

给定仅包含小写字符的字符串str。 任务是按照出现的顺序打印偶数频率的字符。

注意：重复出现的偶数频率按出现的次数打印。

示例：

输入：str = "geeksforgeeks"

输出：geeksgeeks

字符 频率
g 2
e 4
k 2
s 2
f 1
o 1
r 1

仅有g，e，k和s是频率偶数的字符。

输入：str = "aeroplane"

输出：aeae

方法：创建一个频率数组，以存储给定字符串str的每个字符的频率。 再次遍历字符串str，并检查该字符的频率是否为偶数。 如果是，则打印字符。

下面是上述方法的实现：

c 

// c   implementation of the approach 
#include  
using namespace std; 
#define size 26 
// function to print the even frequency characters 
// in the order of their occurrence 
void printchar(string str, int n) 
{ 
    // to store the frequency of each of 
    // the character of the string 
    int freq[size]; 
    // initialize all elements of freq[] to 0 
    memset(freq, 0, sizeof(freq)); 
    // update the frequency of each character 
    for (int i = 0; i < n; i  ) 
        freq[str[i] - 'a']  ; 
    // traverse str character by character 
    for (int i = 0; i < n; i  ) { 
        // if frequency of current character is even 
        if (freq[str[i] - 'a'] % 2 == 0) { 
            cout << str[i]; 
        } 
    } 
} 
// driver code 
int main() 
{ 
    string str = "geeksforgeeks"; 
    int n = str.length(); 
    printchar(str, n); 
    return 0; 
}

java

// java implementation of the approach 
import java.util.*; 
class gfg  
{ 
static int size = 26; 
// function to print the even frequency characters 
// in the order of their occurrence 
static void printchar(string str, int n) 
{ 
    // to store the frequency of each of 
    // the character of the string 
    int []freq = new int[size]; 
    // update the frequency of each character 
    for (int i = 0; i < n; i  ) 
        freq[str.charat(i) - 'a']  ; 
    // traverse str character by character 
    for (int i = 0; i < n; i  )  
    { 
        // if frequency of current character is even 
        if (freq[str.charat(i) - 'a'] % 2 == 0) 
        { 
            system.out.print(str.charat(i)); 
        } 
    } 
} 
// driver code 
public static void main(string[] args) 
{ 
    string str = "geeksforgeeks"; 
    int n = str.length(); 
    printchar(str, n); 
} 
}  
// this code is contributed by 29ajaykumar

python3

# python3 implementation of the approach 
size = 26
# function to print the even frequency characters 
# in the order of their occurrence 
def printchar(string, n): 
    # to store the frequency of each of 
    # the character of the stringing 
    # initialize all elements of freq[] to 0 
    freq = [0] * size 
    # update the frequency of each character 
    for i in range(0, n): 
        freq[ord(string[i]) - ord('a')]  = 1
    # traverse string character by character 
    for i in range(0, n):  
        # if frequency of current character is even 
        if (freq[ord(string[i]) - 
                 ord('a')] % 2 == 0): 
            print(string[i], end = "") 
# driver code 
if __name__ == '__main__': 
    string = "geeksforgeeks"
    n = len(string) 
    printchar(string, n) 
# this code is contributed by ashutosh450

c

// c# implementation of the approach 
using system; 
using system.collections.generic; 
class gfg  
{ 
static int size = 26; 
// function to print the even frequency characters 
// in the order of their occurrence 
static void printchar(string str, int n) 
{ 
    // to store the frequency of each of 
    // the character of the string 
    int []freq = new int[size]; 
    // update the frequency of each character 
    for (int i = 0; i < n; i  ) 
        freq[str[i] - 'a']  ; 
    // traverse str character by character 
    for (int i = 0; i < n; i  )  
    { 
        // if frequency of current character is even 
        if (freq[str[i] - 'a'] % 2 == 0) 
        { 
            console.write(str[i]); 
        } 
    } 
} 
// driver code 
public static void main(string[] args) 
{ 
    string str = "geeksforgeeks"; 
    int n = str.length; 
    printchar(str, n); 
} 
} 
// this code is contributed by princi singh

输出：

geeksgeeks

时间复杂度：o(n)。

辅助空间：o(1)。

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