Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree

Reverse Array using Two Pointers



Problem Statement

Given an array of elements, your task is to reverse the array in-place using the two pointers technique.

The goal is to reverse the order of elements in the array without using extra space.

Examples

Input ArrayReversed OutputDescription
[1, 2, 3, 4, 5][5, 4, 3, 2, 1]Normal case with odd number of elements
[10, 20, 30, 40][40, 30, 20, 10]Normal case with even number of elements
[100][100]Single-element array remains unchanged
[][]Empty array returns empty result (edge case)
['a', 'b', 'c']['c', 'b', 'a']Non-numeric characters are also reversed
[true, false, true][true, false, true]Boolean values reversed correctly

Solution

Reversing an array is a classic problem and can be solved efficiently using the two pointers approach. The idea is simple: swap the elements from both ends of the array while gradually moving inward.

Understanding with Examples

Suppose we have the array [1, 2, 3, 4, 5]. We place one pointer at the start (index 0, value 1) and another at the end (index 4, value 5). We swap them to get [5, 2, 3, 4, 1]. Then, we move the pointers inward—next swap 2 and 4—and eventually reach the center (3), which doesn’t need to move. The array is now reversed.

What Happens in Other Cases?

  • Even-Length Arrays: Like [10, 20, 30, 40] — both sides are swapped until the middle is crossed. No element is left untouched.
  • Odd-Length Arrays: Like [1, 3, 5] — the center element (3) doesn’t move because it’s already in its correct reversed spot.
  • Single-Element Arrays: An array like [100] stays the same after reversal. Nothing changes because there’s only one element.
  • Empty Arrays: For an empty array [], there’s nothing to reverse. The result is also an empty array. No swaps happen.
  • Mixed Types: Whether it’s strings, booleans, or other values, the same logic applies. We just swap positions, not data types.

Why It’s Efficient

This approach doesn’t use any extra array or space—it modifies the array in-place. The total number of swaps is half the length of the array. And since every element is touched only once, the time complexity is O(n) with space complexity O(1).

This is a great introductory problem to understand how pointers work and how they can be used to solve array manipulation tasks efficiently.

Visualization

Algorithm Steps

  1. Initialize two pointers: one at the beginning (left) and one at the end (right) of the array.
  2. While left is less than right, swap the elements at these pointers.
  3. Increment left and decrement right to move towards the center of the array.
  4. Continue until left is no longer less than right.
  5. The array is now reversed.

Code

Python
Java
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
def reverse_array(arr):
    left, right = 0, len(arr) - 1
    while left < right:
        arr[left], arr[right] = arr[right], arr[left]
        left += 1
        right -= 1
    return arr

if __name__ == '__main__':
    arr = [1, 2, 3, 4, 5]
    print("Reversed array:", reverse_array(arr))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)Even in the best case, the algorithm must visit each element once to perform the swap. So the time complexity is linear in all cases.
Average CaseO(n)Each pair of elements is visited and swapped once, so the total operations grow linearly with the size of the array.
Average CaseO(n)All elements need to be swapped exactly once with their mirror index, requiring n/2 swaps — still considered O(n).

Space Complexity

O(1)

Explanation: Only a constant amount of extra space is used — two pointers and a temporary variable for swapping. The reversal is done in-place.

Detailed Step by Step Example

Let us take the following array and apply the Two Pointers Technique to reverse the array in-place.

{ "array": [6,3,8,2,7,4], "showIndices": true, "specialIndices": [] }

Pass 1

left = 0, right = 5

Swap elements at left index and right index.

{ "array": [6,3,8,2,7,4], "showIndices": true, "highlightIndices": [0, 5], "highlightIndicesGreen": [], "specialIndices": [], "labels": { "0": "left", "5": "right" } }

After swapping:

{ "array": [4,3,8,2,7,6], "showIndices": true, "highlightIndicesGreen": [0,5], "labels": { "0": "left", "5": "right" } }
Increment left pointer by one, and decrement right pointer by one.

Pass 2

left = 1, right = 4

Swap elements at left index and right index.

{ "array": [4,3,8,2,7,6], "showIndices": true, "highlightIndices": [1, 4], "highlightIndicesGreen": [0,5], "specialIndices": [], "labels": { "1": "left", "4": "right" } }

After swapping:

{ "array": [4,7,8,2,3,6], "showIndices": true, "highlightIndicesGreen": [0,1,4,5], "labels": { "1": "left", "4": "right" } }
Increment left pointer by one, and decrement right pointer by one.

Pass 3

left = 2, right = 3

Swap elements at left index and right index.

{ "array": [4,7,8,2,3,6], "showIndices": true, "highlightIndices": [2, 3], "highlightIndicesGreen": [0,1,4,5], "specialIndices": [], "labels": { "2": "left", "3": "right" } }

After swapping:

{ "array": [4,7,2,8,3,6], "showIndices": true, "highlightIndicesGreen": [0,1,2,3,4,5], "labels": { "2": "left", "3": "right" } }
Increment left pointer by one, and decrement right pointer by one.

left = 3, right = 2

left pointer crosses right pointer, end of loop.

Array is fully reversed.

{ "array": [4,7,2,8,3,6], "showIndices": true, "highlightIndicesGreen": [0,1,2,3,4,5], "specialIndices": [] }


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