EasyStringsTwo Pointers

Reverse String

1 approach, code in all languages

You are given a string represented as an array of individual characters. Rearrange the characters so that the sequence reads from back to front, effectively flipping the order of every element.

The transformation must happen directly inside the given array. You are not allowed to allocate a second array to hold the result, so the work should be done using only a constant amount of extra space.

Example 1

Input: s = ["h","e","l","l","o"]

Output: ["o","l","l","e","h"]

Each character trades place with its mirror position from the opposite end.

Example 2

Input: s = ["A","d","m","i","t"]

Output: ["t","i","m","d","A"]

The first and last characters swap, then the pair moving inward.

Constraints

1 <= s.length <= 10^5
s[i] is a printable ASCII character
You must modify the array in place with O(1) extra memory

Two Pointers Swap

Place one pointer at the start and another at the end, swap the characters they reference, and move both toward the center.

Loading animation…

Intuition

Reversing an array is the same as swapping symmetric pairs of positions: the first with the last, the second with the second-to-last, and so on. Two pointers walking inward from both ends touch every such pair exactly once. When the pointers meet or cross, every character has been placed in its mirrored slot, so a single pass of swaps completes the reversal without any auxiliary storage.

Algorithm walkthrough

1Initialize a left index at 0 and a right index at the last position of the array.
2While left is strictly less than right, exchange the characters at the two indices.
3Advance left by one and retreat right by one after each swap.
4Stop as soon as the pointers meet or pass each other; the array is now reversed in place.

Implementation

class Solution:
    def reverseString(self, s: List[str]) -> None:
        left, right = 0, len(s) - 1
        while left < right:
            s[left], s[right] = s[right], s[left]
            left += 1
            right -= 1

Time

O(n)

Space

O(1)

Each character is touched once and swaps use only a temporary variable.

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in

Back to Strings

You've got the patterns

Patterns get you through the screen. Shipping gets you hired.

FDE Coach is a cohort-based program in frontend, backend, AWS, and AI where you build real products and get referred to 200+ hiring partners. The free live workshop is the fastest way to see how we teach.

Join the free workshop Explore the program

750+ engineers trained · frontend, backend, AWS & AI