Majority Element

You can find the majority element in an array by using the Boyer-Moore Voting Algorithm. This algorithm helps to find the majority element in O(n) time complexity and O(1) space complexity.

Here’s how you can implement it:

  1. Initialize Variables: Start by initializing two variables, candidate and count, to represent the current majority candidate and its count.
  2. First Pass: Iterate through the array, and for each element: a. If the count is zero, update the candidate to the current element. b. If the current element is equal to the candidate, increment the count. c. If the current element is different from the candidate, decrement the count.
  3. Second Pass (Optional): You can do a second pass to confirm that the candidate appears more than ⌊n / 2⌋ times. However, since the problem guarantees that a majority element exists, you can skip this step.
  4. Return Candidate: The variable candidate will hold the majority element.

Here’s the code:

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class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        candidate = count = 0

        for num in nums:
            if count == 0:
                candidate = num
            if num == candidate:
                count += 1
            else:
                count -= 1

        return candidate

This code efficiently finds the majority element by keeping track of the current candidate and updating it when necessary. The count variable helps in determining when a new candidate should be chosen.

counter = candidate =

[2,2,4,5,2,6,2] ^

Pick a candidate, 1 Switch the candidate to a new one when the counter is 0

  1. Define the interface Input: array of integers Output: Majority element (an integer)

  2. Find: Repeating number must be > half the length of the input (n/2) Define the term - What is majority element?

  3. Constraints: Numbers can be negative Array size is at least one, return that element

  4. Assumption You will have an answer.

  5. Examine the input and output in the given examples.

    • Manually compute the majority element.

  6. We can have more than two numbers.

Implementation.

  1. We need to have set of counts for each number

  2. Get the maximum of those numbers

  3. Check if its frequency is > n/2

Time: O(N) Space: O(N)

Sorting Approach

Time: O(n log n) Space: O(n) or O(1)

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# @param {Integer[]} nums
# @return {Integer}
def majority(element, nums)
  count = 0
    nums.each do |n|
        if n == element
            count += 1
        end
    end
    
    count
end

def majority_element(nums)
   while true
      element = nums.sample
       count = majority(element, nums)
       if count > nums.size/2
           return element
       end
   end
end

Identifying Problem Isomorphism

“Majority Element” belongs to a category of problems that focus on array traversal and frequency counting.

A simpler problem in this category is “Find All Numbers Disappeared in an Array”. In this problem, you are asked to find all the elements of [1, n] inclusive that do not appear in a given array. This problem is simpler as it only requires understanding of array indexing and does not involve frequency counting.

A more complex problem is “Top K Frequent Elements”. This requires you to find the k most frequent elements. The increased complexity arises from needing to not only count frequency but also maintain a priority queue or some other structure to track the top K frequent elements.

The reasoning behind these choices:

  • “Find All Numbers Disappeared in an Array” is simpler than “Majority Element” because it doesn’t involve frequency counting, only array indexing.

  • “Top K Frequent Elements” is more complex than “Majority Element” because, in addition to frequency counting, it requires handling a priority queue or some other structure to keep track of the top K elements.

Therefore, the order of problems from simpler to more complex, based on understanding of array traversal and frequency counting:

  1. “Find All Numbers Disappeared in an Array”
  2. “Majority Element”
  3. “Top K Frequent Elements”

This mapping is an approximation. While these problems all deal with understanding and analyzing array properties, the exact problem constraints and details can significantly influence the complexity and solution approach.

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class Solution(object):
    def majorityElement(self, nums):
        result = None
        count = 0
        for i in nums:
            if count == 0:
                result = i
            count += (1 if i == result else -1)
        return result

Problem Classification

This problem falls into the following categories:

  1. Array Manipulation: The problem involves working with an array or list of numbers.

  2. Majority Voting: The problem requires finding a majority element in the list, which is an application of the Boyer-Moore Voting Algorithm.

  3. Element Frequency: The problem involves finding the element that appears more than half the time in the list, which relates to the frequency of elements.

  4. Single Pass Algorithm: The solution requires traversing the list only once, making it an instance of single pass or linear scan problems.

Language Agnostic Coding Drills

This problem is in the domain of array manipulation and frequency counting. It requires determining the majority element in an array. The majority element is defined as the element that appears more than n/2 times, where n is the size of the array.

Let’s break down the problem solving approach into smaller drills.

  1. Understanding and working with arrays/lists: Arrays or lists are fundamental data structures in computer science used to store multiple items of the same type. You should be comfortable with creating an array, accessing elements in it, and traversing it using loops.

  2. Iterating over an array/list: Understanding how to traverse an array or list is key to solving this problem. You need to be able to go through each element in the list to check its value and count.

  3. Basic counting and comparison: This problem involves keeping a count of occurrences and comparing them with a value. The concept of counting the occurrences of a specific value in a list and keeping track of the most frequently occurring item is essential.

  4. Understanding conditional statements: If-else statements are used to compare the current count and modify the current majority element and the count.

  5. The Boyer-Moore Voting Algorithm: The final code is a implementation of the Boyer-Moore Voting Algorithm. This algorithm is designed to find the majority element in a list in linear time and constant space. This involves maintaining a count of the potential majority element and updating it as you iterate through the array.

These are the concepts you’d need to understand and the drills you’d need to go through in order to tackle this problem effectively. When combined, these drills would enable you to devise the final solution.

Targeted Drills in Python

  1. Understanding and working with arrays/lists:
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# Creating an array/list
arr = [2,2,1,1,1,2,2]

# Accessing elements
print(arr[0])  # prints 2
print(arr[-1]) # prints 2
  1. Iterating over an array/list:
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# Print each element of the array
for num in arr:
    print(num)
  1. Basic counting and comparison:
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# Count the number of times 2 appears in the list
count = 0
for num in arr:
    if num == 2:
        count += 1
print(count)  # prints 4
  1. Understanding conditional statements:
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# Print 'Hello' if count is greater than length/2, else print 'Goodbye'
if count > len(arr)/2:
    print('Hello')
else:
    print('Goodbye')
  1. Implementing the Boyer-Moore Voting Algorithm:
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result = None
count = 0
for num in arr:
    if count == 0:
        result = num
    count += (1 if num == result else -1)
print(result)  # prints 2

Each of these drills helps you understand a component of the final solution. When you put them all together, you can solve the problem with the Boyer-Moore Voting Algorithm.