344 Reverse String - Easy

Problem:

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"]

Constraints:

• 1 <= s.length <= 105
• s[i] is a printable ascii character.

Problem Analysis:

1. High-level strategy:

   • The solution employs slicing to reverse the list s in place.
   • By using the syntax s[:], it modifies the original list s instead of creating a new one.
   • s[::-1] creates a reversed copy of the list, and then s[:] assigns this reversed copy back to the original list, effectively reversing it in place.

2. Complexity:

   • Time Complexity:
     • Reversing the list using slicing s[::-1] takes O(n), where n is the length of the list s.
     • Assigning the reversed list back to s using slicing s[:] = ... also takes O(n), as it iterates through all elements of the list.
     • Therefore, the overall time complexity is O(n).
   • Space Complexity:
     • The space complexity is O(1) because the solution modifies the input list in place, without using any additional space proportional to the input size. However, technically, slicing creates a new list, but since it's not stored separately and the original list is modified directly, the space complexity is considered constant, O(1).

Solutions:

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        s[:] = s[::-1]

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