344. Reverse String

INFO

Problem: Reverse String

Pattern: Two pointers / in-place mutation

Reverse String asks us to reverse an array of characters in-place. Since the input is already an array, we do not need to create a new reversed array.

Approach

Use one pointer at the start and one pointer at the end:

1. Swap the characters at i and j.
2. Move i forward.
3. Move j backward.
4. Stop when the pointers meet or cross.

Every swap places two characters into their final positions. The middle character in an odd-length array does not need to move.

Swift Notes

The function receives s as inout [Character], which means changes are applied directly to the caller's array. Swift's swapAt method performs the in-place swap clearly and safely.

This is a useful pattern to remember because it also appears in array reversal, palindrome checks, and many two-pointer interview problems.

Complexity

• Time complexity: O(n).
• Space complexity: O(1).

Edge Cases

• Empty and single-character arrays are already reversed.
• Even-length arrays require all positions to be swapped in pairs.
• Odd-length arrays leave the center character untouched.

class Solution {
    func reverseString(_ s: inout [Character]) {
        var i = 0
        var j = s.count - 1
        while i < j {
            s.swapAt(i, j)
            i = i + 1
            j = j - 1
        }
    }
}

var solution = Solution()

var testcase1: [Character] = ["h","e","l","l","o"]
solution.reverseString(&testcase1)
print(testcase1) // ["o","l","l","e","h"]

var testcase2: [Character] = ["H","a","n","n","a","h"]
solution.reverseString(&testcase2)
print(testcase2) // ["h","a","n","n","a","H"]