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 Majority Element in Array Using Extended Boyer-Moore Voting Algorithm - Optimal Solution



Problem Statement

Given an integer array arr, your task is to find all elements that appear more than ⌊n/3⌋ times in the array, where n is the size of the array.

If no such element exists, return an empty list.

Examples

Input ArrayOutputDescription
[3, 2, 3][3]n = 3 → ⌊3/3⌋ = 1, 3 appears 2 times
[1][1]Single element always satisfies > n/3 condition
[1, 2][1, 2]n = 2 → ⌊2/3⌋ = 0, both 1 and 2 appear > 0 times
[1, 2, 3, 4][]No element appears more than n/3 = 1 times
[1, 1, 1, 3, 3, 2, 2, 2][1, 2]n = 8 → ⌊8/3⌋ = 2, 1 and 2 both appear 3 times
[][]Empty array has no elements, returns empty list
[2, 2][2]n = 2 → ⌊2/3⌋ = 0, 2 appears 2 times
[4, 4, 4, 5, 5, 5, 6][4, 5]n = 7 → ⌊7/3⌋ = 2, 4 and 5 appear 3 times

Solution

To solve this problem, we need to find elements that appear more than ⌊n/3⌋ times in a given array. The main challenge is to do it efficiently—ideally in linear time and using constant space.

Understanding the Constraint

When we say we want elements that occur more than n/3 times, it means we're looking for items that appear a significant number of times—but not necessarily the most frequent. Because of this threshold, there can be at most two elements that satisfy this condition. Why two? Because if three distinct numbers all appeared more than n/3 times, their total count would exceed n, which is impossible.

How Do We Approach This?

The trick is to use an extension of the Boyer-Moore Voting Algorithm. The original Boyer-Moore algorithm finds a single majority element (> n/2), but we can extend it to handle two potential candidates:

  • We go through the array once, keeping track of two candidates and their counts.
  • If we see a number matching a candidate, we increment that candidate’s count.
  • If one of the counts is zero, we replace that candidate with the current number and reset the count to 1.
  • If the current number matches neither candidate, we decrement both counts.

This gives us up to two potential candidates. But since they are just candidates, we need a second pass through the array to count how many times they actually appear.

Final Validation

Once we've completed the first pass and identified the two candidates, we run a second loop to count how many times each one appears. Only if their count is greater than n/3, do we include them in the final result.

Case-by-Case Insight

  • Single Element: If the array has only one number, that number is the majority by default.
  • All Unique Elements: If no number repeats more than ⌊n/3⌋ times, we return an empty list.
  • Multiple Valid Candidates: Sometimes two numbers will satisfy the condition—both should be returned.
  • Empty Array: There are no elements to count, so the result is an empty list.

This solution is optimal: it runs in O(n) time and uses O(1) space, as we only track two numbers and their counts during traversal.

Visualization

Algorithm Steps

  1. Given an array arr of size n.
  2. Initialize two candidates candidate1 and candidate2 and their counts to 0.
  3. Traverse the array:
  4. → If the current element equals candidate1, increment count1.
  5. → Else if it equals candidate2, increment count2.
  6. → Else if count1 is 0, set candidate1 to current element and count1 to 1.
  7. → Else if count2 is 0, set candidate2 to current element and count2 to 1.
  8. → Else decrement both count1 and count2.
  9. After the first pass, verify the candidates by counting their actual occurrences in the array.
  10. Return the candidates that appear more than n/3 times.

Code

Python
JavaScript
Java
C++
C
def majority_elements(arr):
    if not arr:
        return []
    
    candidate1 = candidate2 = None
    count1 = count2 = 0

    for num in arr:
        if num == candidate1:
            count1 += 1
        elif num == candidate2:
            count2 += 1
        elif count1 == 0:
            candidate1 = num
            count1 = 1
        elif count2 == 0:
            candidate2 = num
            count2 = 1
        else:
            count1 -= 1
            count2 -= 1

    result = []
    for candidate in [candidate1, candidate2]:
        if arr.count(candidate) > len(arr) // 3:
            result.append(candidate)
    return result

# Sample Input
arr = [3,2,3]
print("Majority Elements:", majority_elements(arr))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)The array is traversed twice: once for candidate selection and once for verification, both in linear time.
Average CaseO(n)Regardless of the input distribution, both candidate identification and validation phases take linear time.
Average CaseO(n)Even in the worst-case (e.g., all elements are different), both passes still involve scanning the entire array.

Space Complexity

O(1)

Explanation: Only a fixed number of variables are used (two candidates and their counts), independent of the input size.



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