LeetCode Python Solutions: 344. Reverse String #python #coding #leetcode

ZeroStress LeetCode Python Solutions: 344. Reverse String #python #leetcode

problem description:
Write a function that reverses a string. The input string is given as an array of characters s.
You must do this by modifying the input array in-place with O(1) extra memory.

Example 1:
Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]
Example 2:
Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

The intuition behind this solution is to reverse the given string by using two pointers. By starting from both ends of the string and gradually moving towards the middle, we swap the characters at the left and right positions until the pointers meet or cross each other.

The two-pointer approach allows us to reverse the string in-place without using any extra memory. It takes advantage of the fact that we can swap elements within the same array without needing additional space.

The intuition can be visualized as follows:

We start with the left pointer at the beginning of the string (index 0) and the right pointer at the end of the string (index len(s) - 1).

In each iteration, we swap the characters at the left and right positions. This effectively reverses the order of the characters.

After swapping, we move the left pointer one step to the right (incrementing by 1) and the right pointer one step to the left (decrementing by 1).

We continue this process until the left pointer becomes greater than or equal to the right pointer. This condition indicates that the pointers have crossed each other or met in the middle.

The intuition of this solution is based on the observation that by swapping characters from both ends towards the middle, we can reverse the order of the characters in the string efficiently.

The approach of the solution is as follows:

Initialize two pointers, left and right, pointing to the first and last characters of the input string s, respectively.

Enter a while loop that continues as long as the left pointer is less than the right pointer.

Within the loop, swap the characters at the left and right positions in the string s. This can be done using the Python shorthand notation for swapping values: s[left], s[right] = s[right], s[left].

Move the left pointer one step to the right by incrementing it (left += 1).

Move the right pointer one step to the left by decrementing it (right -= 1).

Repeat steps 3-5 until the left pointer becomes greater than or equal to the right pointer, indicating that the pointers have crossed or met in the middle.

The string s will now be reversed in-place.

The key idea behind this approach is to reverse the string by swapping characters symmetrically from both ends towards the middle. The while loop ensures that the swapping continues until the entire string is reversed.

This approach has a time complexity of O(n), where n is the length of the string. It visits each character once, performing a constant number of operations for each character. The space complexity is O(1) since the algorithm modifies the input string in-place without using any additional data structures.

Time Complexity: O(n)

The time complexity of this solution is linear, where n is the length of the string. The algorithm visits each character once, performing a constant number of operations (swapping and pointer updates) for each character. Therefore, the time complexity is directly proportional to the length of the string.

Space Complexity: O(1)

The space complexity of this solution is constant. It uses only a fixed amount of additional memory regardless of the input size. The algorithm modifies the input string in-place without using any extra data structures. Thus, the space complexity is O(1), satisfying the requirement of O(1) extra memory.

00:00 Code
01:13 Main
09:00 End

[NeedCodeLeetCode]

def reverseString(s):
left = 0
right = len(s) - 1

while left < right:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1


The approach of the solution is as follows:

Initialize two pointers, left and right, pointing to the first and last characters of the input string s, respectively.

Enter a while loop that continues as long as the left pointer is less than the right pointer.

Within the loop, swap the characters at the left and right positions in the string s. This can be done using the Python shorthand notation for swapping values: s[left], s[right] = s[right], s[left].

Move the left pointer one step to the right by incrementing it (left += 1).

Move the right pointer one step to the left by decrementing it (right -= 1).

Repeat steps 3-5 until the left pointer becomes greater than or equal to the right pointer, indicating that the pointers have crossed or met in the middle.

The string s will now be reversed in-place.

The key idea behind this approach is to reverse the string by swapping characters symmetrically from both ends towards the middle. The while loop ensures that the swapping continues until the entire string is reversed.

This approach has a time complexity of O(n), where n is the length of the string. It visits each character once, performing a constant number of operations for each character. The space complexity is O(1) since the algorithm modifies the input string in-place without using any additional data structures.