Reverse a string
You are given a string 'STR'. The string contains [a-z] [A-Z] [0-9] [special characters]. You have to find the reverse of the string.

For example:

If the given string is: STR = "abcde". You have to print the string "edcba".

follow up:

Try to solve the problem in O(1) space complexity.

Input Format:

The first line of input contains a single integer 'T', representing the number of test cases or queries to be run. Then the 'T' test cases follow. The first and only line of each test case contains a string 'STR'.

Output Format:

For each test case, print a single line containing a single string denoting the reverse of the given string 'STR'. The output of each test case will be printed in a separate line.

Note:

You are not required to print the expected output, it has already been taken care of. Just implement the function.

Constraints:

1 ≤ T ≤ 10 1 ≤ |STR| ≤ 10 ^ 5 Where |STR| is the length of the string STR. Time limit: 1 sec.

Question

Reverse a string

You are given a string 'STR'. The string contains [a-z] [A-Z] [0-9] [special characters]. You have to find the reverse of the string.

For example:

 If the given string is: STR = "abcde". You have to print the string "edcba".

follow up:

Try to solve the problem in O(1) space complexity.

Input Format:

The first line of input contains a single integer 'T', representing the number of test cases or queries to be run. 

Then the 'T' test cases follow.

The first and only line of each test case contains a string 'STR'.

Output Format:

For each test case, print a single line containing a single string denoting the reverse of the given string 'STR'.

The output of each test case will be printed in a separate line.

Note:

You are not required to print the expected output, it has already been taken care of. Just implement the function.

Constraints:

1 ≤ T ≤ 10
1 ≤ |STR| ≤ 10 ^ 5 

Where |STR| is the length of the string STR.

Time limit: 1 sec.

Accepted Answer

This can be done by iterative swapping using two pointers. The first pointer points to the beginning of the string, whereas the second pointer points to the end. Both pointers keep swapping their elements and go towards each other. Essentially, the algorithm simulates the rotation of a string with respect to its midpoint.
Time Complexity : O(n)

Accepted Answer

Optimal Solution

Traverse the string and swap the first character with the last character, the second character with the second last character and so on.
Basically, you need to swap the i-th character with the (N-i-1)-th character where N is the length of the string and 0-based indexing is considered.

Space Complexity: O(1)Explanation:

O(1).

In the worst case, only constant extra space is required.

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the length of the string.

Since we are iterating through the string only once, the overall time complexity is O(N).