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    LeetCode link: 344. Reverse String, difficulty: Easy.

    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:

    Hint 1

    The entire logic for reversing a string is based on using the opposite directional two-pointer approach!

    Intuition
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    1. This problem can be solved in one line of code using the built-in sort() method of the programming language. If this question is asked in an interview, the questioner should be testing how to do it without the built-in method.
    2. Use two pointers with opposite directions, initially one pointer points to the index 0 and the other pointer points to the index s.length - 1.
    3. Traverse the elements of the array, and the loop condition is while (left < right). In the loop body, left += 1, right -= 1.
    4. In the loop body, swap the two values.
    5. The above is the template for two pointers in opposite directions.

    Step-by-Step Solution

    1. Use two pointers with opposite directions, initially one pointer points to the index 0 and the other pointer points to the index s.length - 1.

      left = 0
right = s.length - 1

    2. Traverse the elements of the array, and the loop condition is while (left < right). In the loop body, left += 1, right -= 1.

      left = 0
right = s.length - 1

while left < right # 1
  left += 1 # 2
  right -= 1 # 3
end

    3. In the loop body, swap the two values.

      left = 0
right = s.length - 1

while left < right
  s[left], s[right] = s[right], s[left] # 1

  left += 1
  right -= 1
end

    Complexity

    Time complexity

    O(N)

    Space complexity

    O(1)

    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Java # 

    class Solution {
    public void reverseString(char[] s) {
        var left = 0;
        var right = s.length - 1;

        while (left < right) {
            var leftValue = s[left];
            s[left] = s[right];
            s[right] = leftValue;

            left++;
            right--;
        }
    }
}

    Python # 

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

        while left < right:
            s[left], s[right] = s[right], s[left]
            left += 1
            right -= 1

    C++ # 

    class Solution {
public:
    void reverseString(vector<char>& s) {
        auto left = 0;
        auto right = s.size() - 1;

        while (left < right) {
            swap(s[left], s[right]);

            left++;
            right--;
        }
    }
};

    JavaScript # 

    var reverseString = function (s) {
  let left = 0
  let right = s.length - 1

  while (left < right) {
    [s[left], s[right]] = [s[right], s[left]]

    left++
    right--
  }
};

    C# # 

    public class Solution
{
    public void ReverseString(char[] s)
    {
        int left = 0;
        int right = s.Length - 1;

        while (left < right)
        {
            (s[left], s[right]) = (s[right], s[left]);

            left++;
            right--;
        }
    }
}

    Go # 

    func reverseString(s []byte)  {
    left := 0
    right := len(s) - 1

    for left < right {
        s[left], s[right] = s[right], s[left]

        left++
        right--
    }
}

    Ruby # 

    def reverse_string(s)
  left = 0
  right = s.size - 1

  while left < right
    s[left], s[right] = s[right], s[left]

    left += 1
    right -= 1
  end
end
    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Other languages

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