This problem statement is a part of Leetcode's learning card Array and Strings. Under the sub-heading two pointer technique.
Write a function that reverses a string. The input string is given as an array of characters char.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
You may assume all the characters consist of printable ASCII characters.
Input: ["h","e","l","l","o"] Output: ["o","l","l","e","h"]
Input: ["H","a","n","n","a","h"] Output: ["h","a","n","n","a","H"]
Use python built-in methods. There are two ways of doing the same. Suppose s in the initial list. Then s.reverse() could be done or s[::-1] can be done.
Had reaching the solution by using the built-in methods been the aim. Probably the whole exercise would not have been worth the effort.
- Iterate over the list.
- Swap the first element with the last element.
- The list would be completely reversed when we are at the mid of the list.
- When the index of the current element is equal to mid of the list. Break.
class Solution: def reverseString(self, s: List[str]) -> None: """ Do not return anything, modify s in-place instead. """ for x in range(0, len(s)): if x == int(len(s)/2): break else: temp = s[x] s[x] = s[len(s)-x-1] s[len(s)-x-1]=temp
- Time complexity - O(n), Space complexity - O(1).
A more elegant approach
- Keep two variables. One at the starting of the list at index 0. Another at the end of the list. Index len(L)-1.
- Keep swapping the elements until both the variables reach the mid of the list.
class Solution: def reverseString(self, s: List[str]) -> None: """ Do not return anything, modify s in-place instead. """ start = 0 end = len(s) - 1 while start < end: temp = s[start] s[start] = s[end] s[end] = temp start += 1 end -= 1
- The solution is still time complexity O(n) and space complexity O(1).
- Two pointer technique. Though in python there is no concept of pointer as such.
- Built-in reverse method is also implemented the same way.