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 sInput: 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++ 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);
} |
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++ 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);
} |
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++ 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);
} |
G e e k s f o r G e e k s
Time Complexity: O(N)
Auxiliary Space: O(1)