Circle Majority

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Unit 7 Session 2 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 20 mins
• 🛠️ Topics: Divide and Conquer, Recursion, Majority Element Problem

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• Q: What happens if no single element appears more than ⌊n/2⌋ times?
  • A: By problem constraints, it’s given that a majority element always exists in the array, so there’s always an answer.

HAPPY CASE
Input: [3, 2, 3]
Output: 3
Explanation: 3 appears more than ⌊3/2⌋ = 1 times.

EDGE CASE
Input: [2, 2, 1, 1, 1, 2, 2]
Output: 2
Explanation: 2 appears more than ⌊7/2⌋ = 3 times.

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

This problem is a classic case for the divide and conquer approach:

• Splitting the array and finding the majority element in each half recursively.
• Combining results from two halves to determine the overall majority element.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Recursively divide the array into halves, find the majority element in each half, and then determine the majority for the entire segment by counting.

1) Divide the array into two halves, breaking the problem into smaller sub-problems.
2) Recursively find the majority element in each half.
3) Once you have potential majority elements 
  - Use a helper function to count the occurrences of these two candidates in the original segment.
4) Determine which candidate is the true majority by comparing their counts.
  - Return the candidate that has the higher count

⚠️ Common Mistakes

• Failing to handle cases where two halves have different majority elements.
• Incorrectly counting occurrences in the combined step.

4: I-mplement

Implement the code to solve the algorithm.

def count_frequency(nums, num, start, end):
    " Counts how many times 'num' appears in 'nums' between 'start' and 'end' inclusive. "
    count = 0
    for i in range(start, end + 1):
        if nums[i] == num:
            count += 1
    return count

def find_majority(nums, left, right):
    " Recursively finds the majority element in nums between left and right indices. "
    # Base case: single element, the majority element is trivially the element itself
    if left == right:
        return nums[left]

    # Recursively find the majority element in the left and right halves
    mid = (left + right) // 2
    left_majority = find_majority(nums, left, mid)
    right_majority = find_majority(nums, mid + 1, right)

    # If the two halves agree on the majority, return it
    if left_majority == right_majority:
        return left_majority

    # Count each majority candidate in the entire current range
    left_count = count_frequency(nums, left_majority, left, right)
    right_count = count_frequency(nums, right_majority, left, right)

    # Return the element with the higher count
    if left_count > right_count:
        return left_majority
    else:
        return right_majority

def majority_element(nums):
    " Uses divide and conquer to find the majority element in nums. "
    return find_majority(nums, 0, len(nums) - 1)

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Test the function with inputs [3, 2, 3] to ensure it correctly identifies 3 as the majority.
• Check with input [2, 2, 1, 1, 1, 2, 2] to confirm that it correctly identifies 2 as the majority.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(n log n) due to the recursive division of the array and the counting process within each divided segment.
• Space Complexity: O(log n) due to the recursion stack space required for the divide and conquer process.