Reverse String

easy string two pointers

Problem

Reverse a string in place. The input is given as an array of characters; modify it directly using O(1) extra memory.

Inputs = ["h","e","l","l","o"]

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

Walk two pointers from the ends and swap until they meet.

def reverse_string(s):
    l, r = 0, len(s) - 1
    while l < r:
        s[l], s[r] = s[r], s[l]
        l += 1
        r -= 1

function reverseString(s) {
  let l = 0, r = s.length - 1;
  while (l < r) {
    [s[l], s[r]] = [s[r], s[l]];
    l++;
    r--;
  }
}

class Solution {
    public void reverseString(char[] s) {
        int l = 0, r = s.length - 1;
        while (l < r) {
            char t = s[l]; s[l] = s[r]; s[r] = t;
            l++; r--;
        }
    }
}

void reverseString(vector<char>& s) {
    int l = 0, r = s.size() - 1;
    while (l < r) {
        swap(s[l], s[r]);
        l++; r--;
    }
}

Explanation

To reverse the array in place without any extra storage, we use the classic two-pointer technique: one pointer at the front and one at the back, swapping their values and walking toward the middle.

Pointer l starts at index 0 and r at the last index. We swap s[l] with s[r], then do l += 1 and r -= 1. Each swap puts two characters into their final mirrored positions at once.

The loop runs while l < r. Once they meet (or cross), every pair has been swapped and the array is fully reversed. A middle character in an odd-length array simply stays where it is.

This uses only O(1) extra memory and a single pass, since each character is touched just once.

Example: ["h","e","l","l","o"]. Swap index 0 and 4 → o...h; swap index 1 and 3 → l and l (no visible change); the middle l stays. Result: ["o","l","l","e","h"].

Time: O(n) Space: O(1)