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 = “neveropen”
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++
// 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 = "neveropen"; // 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++
// 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 = "neveropen"; // 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++
// 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 = "neveropen"; // 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)
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