Iterate over characters of a string in C++

Last Updated : 24 Nov, 2020

Given a string str of length N, the task is to traverse the string and print all the characters of the given string.

Examples:

Input: str = “GeeksforGeeks”
Output: G e e k s f o r G e e k s

Input: str = “Coder”
Output: C o d e r

Naive Approach: The simplest approach to solve this problem is to iterate a loop over the range [0, N – 1], where N denotes the length of the string, using variable i and print the value of str[i].

Below is the implementation of the above approach:

C++

// C++ program to implement 
// the above approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to traverse the string and 
// print the characters of the string 
void TraverseString(string &str, int N) 
{  
    // Traverse the string 
    for (int i = 0; i < N; i++) { 
  
        // Print current character 
        cout<< str[i]<< " "; 
    } 
      
} 
  
// Driver Code 
int main() 
{ 
    string str = "GeeksforGeeks"; 
      
    // Stores length of the string 
    int N = str.length(); 
  
    TraverseString(str, N); 
} 

Output:

G e e k s f o r G e e k s

Time Complexity: O(N)
Auxiliary Space: O(1)

Auto keyword – based Approach: The string can be traversed using auto iterator.

Below is the implementation of the above approach:

C++

// C++ program to implement 
// the above approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to traverse the string and 
// print the elements of the string 
void TraverseString(string &str, int N) 
{ 
    // Traverse the string 
    for (auto &ch : str) { 
  
        // Print current character 
        cout<< ch<< " "; 
    } 
} 
// Driver Code 
int main() 
{ 
    string str = "GeeksforGeeks"; 
  
    // Stores length of the string 
    int N = str.length(); 
  
    TraverseString(str, N); 
} 

Output:

G e e k s f o r G e e k s

Time Complexity: O(N)
Auxiliary Space: O(1)

Iterator – based Approach: The string can be traversed using iterator.

Below is the implementation of the above approach:

C++

// C++ program to implement 
// the above approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to traverse the string and 
// print the elements of the string 
void TraverseString(string &str, int N) 
{ 
      
    // Stores address of  
    // a character of str 
    string:: iterator it; 
      
    // Traverse the string 
    for (it = str.begin(); it != str.end(); 
                                   it++) { 
        // Print current character 
        cout<< *it<< " "; 
    } 
} 
  
// Driver Code 
int main() 
{ 
    string str = "GeeksforGeeks"; 
      
      
    // Stores length of the string 
    int N = str.length(); 
    TraverseString(str, N); 
} 

Output:

G e e k s f o r G e e k s

Time Complexity: O(N)
Auxiliary Space: O(1)

Minimum removal of subsequences of distinct consecutive characters required to empty a given string

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