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++
#include <bits/stdc++.h>
using namespace std;
void TraverseString(string &str, int N)
{
for (int i = 0; i < N; i++) {
cout<< str[i]<< " ";
}
}
int main()
{
string str = "GeeksforGeeks";
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++
#include <bits/stdc++.h>
using namespace std;
void TraverseString(string &str, int N)
{
for (auto &ch : str) {
cout<< ch<< " ";
}
}
int main()
{
string str = "GeeksforGeeks";
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++
#include <bits/stdc++.h>
using namespace std;
void TraverseString(string &str, int N)
{
string:: iterator it;
for (it = str.begin(); it != str.end();
it++) {
cout<< *it<< " ";
}
}
int main()
{
string str = "GeeksforGeeks";
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)
Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!