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## 1. The two-pointer algorithm

**Idea:**
Swap elements from the start and the end of the array, gradually moving toward the center.

```javascript
function reverseArray(arr) {
  let left = 0
  let right = arr.length - 1

while (left < right) {
    // Swap arr[left] and arr[right]
    const temp = arr[left]
    arr[left] = arr[right]
    arr[right] = temp

left++
    right--
  }

return arr
}

// Example:
console.log(reverseArray([1, 2, 3, 4, 5])) // [5, 4, 3, 2, 1]
```

### Complexity

- **Time:** `O(n)`
- **Memory:** `O(1)`, no new array is created.

---

## 2. Reversing without changing the original array

If the task does not allow changing the original array, you can create a new one:

```javascript
function reversedCopy(arr) {
  const result = []
  for (let i = arr.length - 1; i >= 0; i--) {
    result.push(arr[i])
  }
  return result
}

console.log(reversedCopy([10, 20, 30])) // [30, 20, 10]
```

### Complexity

- **Time:** `O(n)`
- **Memory:** `O(n)`, a new array.

---

## 3. Via recursion (less common, but often asked about)

```javascript
function reverseRecursive(arr, left = 0, right = arr.length - 1) {
  if (left >= right) return arr

const temp = arr[left]
  arr[left] = arr[right]
  arr[right] = temp

return reverseRecursive(arr, left + 1, right - 1)
}

console.log(reverseRecursive([1, 2, 3, 4])) // [4, 3, 2, 1]
```

Recursion is less memory-efficient (the call stack), but it neatly demonstrates the "edges to center" logic.

---

## Summary

| Method | Changes the array | Time | Memory | Notes |
|---|---|---|---|---|
| **Two pointers** | Yes | O(n) | O(1) | The optimal approach |
| **Loop with a new array** | No | O(n) | O(n) | Does not modify the original |
| **Recursion** | Yes | O(n) | O(n) | Less efficient, but illustrative |

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