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

Left Rotate an Array by One Place using Loop - Optimal Algorithm



Problem Statement

Given an array of integers, your task is to left rotate the array by 1 position. This means that each element moves one place to the left, and the first element is moved to the end of the array.

The operation should be performed in-place using an optimal solution, without using extra space for a new array.

If the array is empty or contains only one element, it should remain unchanged after rotation.

Examples

Input ArrayOutput Array (After 1 Left Rotation)Description
[1, 2, 3, 4, 5][2, 3, 4, 5, 1]Normal case – first element moved to the end
[10][10]Single-element array remains the same
[][]Empty array remains unchanged
[5, 10, 15][10, 15, 5]All elements shifted left once, first element placed last
[7, 8][8, 7]Two-element array rotation
[0, 0, 0][0, 0, 0]All elements are the same; rotation does not change the appearance

Solution

To left rotate an array by one position means that we shift every element one step to the left, and move the very first element to the last position. Let's understand how this transformation looks and what different scenarios can occur.

Understanding with a Simple Example

Take the array [1, 2, 3, 4, 5]. After left rotating it once, every element moves left and 1 (the first element) goes to the last place. The result becomes [2, 3, 4, 5, 1].

What Happens Internally?

We can perform this efficiently in one pass:

  • Store the first element in a temporary variable (say first).
  • Shift all elements from index 1 to the end, one step to the left.
  • Finally, place first at the end of the array.

This approach ensures the rotation happens in-place without using additional space, and it runs in linear time O(n).

Special Cases to Consider

  • Empty Array: If the array has no elements, there's nothing to rotate. We return the array as is.
  • Single Element: For an array like [10], rotating it doesn’t change anything because there's only one item.
  • Identical Elements: Arrays like [0, 0, 0] look unchanged even after rotation, but internally, the positions are updated.

Why This Approach is Optimal

This technique avoids the use of any additional arrays or lists. It performs the operation in-place with constant space and only one traversal through the array. It's the most efficient way to left rotate an array by one position.

Visualization

Algorithm Steps

  1. Given an array arr of size n.
  2. Store the first element in a temporary variable first.
  3. Shift all elements of the array one position to the left (from index 1 to n-1).
  4. Assign first to the last index of the array.
  5. The array is now left-rotated by one place.

Code

Python
JavaScript
Java
C++
C
def left_rotate_by_one(arr):
    if not arr:
        return arr
    first = arr[0]
    for i in range(1, len(arr)):
        arr[i - 1] = arr[i]
    arr[-1] = first
    return arr

# Sample Input
arr = [10, 20, 30, 40, 50]
print("After Left Rotation:", left_rotate_by_one(arr))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)Every element (except the first) needs to be shifted one position to the left, which takes linear time regardless of input.
Average CaseO(n)In all scenarios, the array is traversed once to shift elements, resulting in linear time complexity.
Average CaseO(n)Even in the worst case, each element is moved exactly once, making the operation linear in the number of elements.

Space Complexity

O(1)

Explanation: Only one temporary variable is used to store the first element. No additional space is required proportional to input size.

Detailed Step by Step Example

Let's left rotate the array by one position.

{ "array": [10,20,30,40,50], "showIndices": true }

Store the first element 10 in a temporary variable.

Shift index 1

Move 20 from index 1 to index 0.

{ "array": [20,20,30,40,50], "showIndices": true, "highlightIndices": [0], "labels": { "0": "updated" } }

Shift index 2

Move 30 from index 2 to index 1.

{ "array": [20,30,30,40,50], "showIndices": true, "highlightIndices": [1], "labels": { "1": "updated" } }

Shift index 3

Move 40 from index 3 to index 2.

{ "array": [20,30,40,40,50], "showIndices": true, "highlightIndices": [2], "labels": { "2": "updated" } }

Shift index 4

Move 50 from index 4 to index 3.

{ "array": [20,30,40,50,50], "showIndices": true, "highlightIndices": [3], "labels": { "3": "updated" } }

Place the first element at the end

Set 10 to index 4.

{ "array": [20,30,40,50,10], "showIndices": true, "highlightIndices": [4], "labels": { "4": "rotated" } }

Final Result:

Array after left rotation: [20, 30, 40, 50, 10]



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