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

Find Leaders in an Array
Optimal Right to Left Scan Solution



Problem Statement

Given an array of integers, your task is to find all the leader elements in the array.

You need to return a list of all such leaders in left-to-right order as they appear in the array.

Examples

Input ArrayLeadersDescription
[16, 17, 4, 3, 5, 2][17, 5, 2]17 > [4,3,5,2], 5 > 2, and 2 is last so it's a leader
[1, 2, 3, 4, 5][5]Only 5 is greater than all elements to its right (none)
[5, 4, 3, 2, 1][5, 4, 3, 2, 1]Each element is greater than all elements to its right
[7][7]Single element is always a leader
[][]Empty array, so no leaders
[10, 10, 10][10]Only the last occurrence is considered a leader
[0, -1, -2, -3][0, -1, -2, -3]All elements are leaders in descending order

Solution

To solve the problem of finding leaders in an array, we need to understand what makes an element a leader. A leader is any number in the array that is strictly greater than all the numbers to its right.

We always start checking from the end of the array. This is because the last element has no elements after it, so it is always a leader by definition.

Different Cases Explained

  • Case 1: Multiple leaders exist – For example, in [16, 17, 4, 3, 5, 2], we start from the end:
    • 2 is a leader (no elements after it)
    • 5 > 2, so 5 is a leader
    • 3 < 5, so it’s not
    • 4 < 5, so skip
    • 17 > 5, so it's also a leader
    Final leaders in left-to-right order: [17, 5, 2]
  • Case 2: Increasing array – In [1, 2, 3, 4, 5], only 5 is a leader because no number is greater than it to the right.
  • Case 3: Decreasing array – In [5, 4, 3, 2, 1], each number is greater than the rest of the elements to its right, so every number is a leader.
  • Case 4: Equal elements – In [10, 10, 10], only the last 10 is a leader, because the previous ones are not greater than the next.
  • Case 5: Single element – Any array with just one element like [7] will always have that element as a leader.
  • Case 6: Empty array – No elements mean no leaders. Return an empty list.

By starting from the right and keeping track of the maximum element seen so far, we can easily identify all leader elements. Each time we find a new maximum (while scanning from right to left), we record it as a leader.

In the end, we reverse the list of leaders (since we found them from right to left) to return them in left-to-right order as required.

This approach is optimal with a time complexity of O(n) because we only scan the array once.

Visualization

Algorithm Steps

  1. Given an array arr.
  2. Initialize an empty list leaders to store leader elements.
  3. Start from the last element: initialize max_from_right = arr[n-1].
  4. Add max_from_right to the leaders list.
  5. Traverse the array from right to left:
  6. → If current element is greater than max_from_right, update max_from_right and add it to leaders.
  7. After the loop, reverse the leaders list to maintain left-to-right order.

Code

Python
JavaScript
Java
C++
C
def find_leaders(arr):
    n = len(arr)
    leaders = []
    max_from_right = arr[-1]
    leaders.append(max_from_right)
    for i in range(n - 2, -1, -1):
        if arr[i] > max_from_right:
            max_from_right = arr[i]
            leaders.append(max_from_right)
    leaders.reverse()
    return leaders

# Sample Input
arr = [16, 17, 4, 3, 5, 2]
print("Leaders:", find_leaders(arr))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)The array is scanned from right to left once, and each element is checked only once, making the operation linear.
Average CaseO(n)Regardless of input distribution, the array is always scanned once from end to start to compare with the running maximum.
Average CaseO(n)In the worst case (e.g., strictly decreasing array), every element becomes a leader, but the time taken remains linear as there is still only one pass.

Space Complexity

O(n)

Explanation: The worst-case space is O(n) if all elements are leaders and stored in the list. Additional space is used only for the leaders list and a few variables.



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