Guess the Majority in a Hidden Array

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Arrays

You have access to an interactive array (i.e., you can query values at specified indices).

You are given an instance of the class ArrayReader which has the following API:

int query(int a, int b, int c, int d): where 0 <= a < b < c < d < ArrayReader.length(). The function returns:

1 if the majority element among the values [ArrayReader.get(a), ArrayReader.get(b), ArrayReader.get(c), ArrayReader.get(d)] is equal to ArrayReader.get(a).
0 otherwise.

int length(): Returns the length of the array.

You are allowed to make at most 2 * n calls to ArrayReader.query().

Return the index of the majority element in the array. If there is no majority element, return -1.

Example 1:

Input: arr = [1,1,2,1,2]
Output: 0
Explanation:
We have the following interaction:
ArrayReader.length() -> 5
ArrayReader.query(0, 1, 2, 3) -> 1
ArrayReader.query(1, 2, 3, 4) -> 0
The most frequent element is 1, which appears 3 times.

Example 2:

Input: arr = [1,1,2,2,1,1]
Output: 0

Constraints:

n == ArrayReader.length()
1 <= n <= 10⁵
Each element of the array is in the range [0, 10⁹].

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the size limit of the hidden array?
What is the range of possible values in the hidden array?
If there is no majority element, what should my function return?
Can I assume the existence of the `ArrayReader` class as described, or do I need to account for its potential absence or errors?
How many times can I call the `query` method in the `ArrayReader`? Is there a call limit?

Brute Force Solution

Approach

The brute force approach to finding the majority element in a hidden array means we directly compare every element against every other element. We will be systematically checking each possibility and counting how many times each element appears. The element that appears more than half the time is the majority.

Here's how the algorithm would work step-by-step:

Consider the first element in the hidden array.
Compare this element to every other element in the array, one by one.
Count how many times it's the same as the element you're comparing it to.
Check if the count is greater than half the total number of elements in the array. If it is, you've found the majority element!
If the count is not greater than half, move on to the next element in the array.
Repeat the comparison and counting process with this new element.
Keep doing this for every element in the array until you find an element that appears more than half the time or you run out of elements to check.

Code Implementation

def guess_the_majority_brute_force(hidden_array_length, guess, all):    # We can't access the hidden array directly. This emulates the API interaction.    def get_element(index): 
        return all[index]

    for first_element_index in range(hidden_array_length):
        current_element = get_element(first_element_index)
        element_count = 0

        # Iterate through the entire array to compare against the current element.
        for second_element_index in range(hidden_array_length):
            other_element = get_element(second_element_index)
            if current_element == other_element:
                element_count += 1

        # Check if the current element's count exceeds the majority threshold.
        if element_count > hidden_array_length // 2:
            return current_element

    return -1

Big(O) Analysis

Time Complexity

O(n²) – The described brute force approach involves comparing each element in the hidden array to every other element. For each of the n elements, a comparison is made with approximately n other elements in the worst case. Therefore, the total number of comparisons grows proportionally to n multiplied by n, which results in approximately n * n operations. Thus, the time complexity is O(n²).

Space Complexity

O(1) – The brute force approach described only uses a single counter variable to track the number of times an element matches another. No auxiliary data structures like arrays, hash maps, or significant recursion are involved. The space used by the counter variable is constant regardless of the size of the hidden array, denoted as N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The problem involves figuring out which number appears more often in a hidden list. We can't directly see the list, but we can ask questions to compare pairs of numbers to figure out the majority element efficiently without checking every single position.

Here's how the algorithm would work step-by-step:

Start by picking the first number in the hidden list as our current guess for the majority number.
Compare our current guess with the next number in the list. If they are the same, we still believe our guess is correct. If they are different, we think our guess might be wrong.
If we suspected our guess was wrong in the last step, we have to compare the next value with the value we started with (our guess) to confirm which one to use next. Reset the count to one, starting our guess for majority with the new value.
Keep doing this until we have gone through the entire hidden list. This approach smartly cancels out different numbers, and leaves us with the remaining majority number.
The number we are left with is a potential majority element. To be certain, we now count how many times it shows up by comparing it to every number in the hidden list.
If the final count is more than half of the numbers in the list, then it is indeed the majority. If not, then there is no majority element.

Code Implementation

class GuessTheMajority:

    def majority(self, array_length):
        pass

    def guess(self, index_a, index_b):
        pass

def find_majority(guesser, array_length):

    majority_index_so_far = 0
    count = 1

    for i in range(1, array_length):
        # Check if current element is the majority candidate
        if guesser.guess(majority_index_so_far, i) == 0:
            count += 1
        else:
            count -= 1

            # If the count is 0, update the majority index
            if count == 0:
                majority_index_so_far = i
                count = 1

    # Potential majority element is found; verify it
    actual_majority_count = 0

    for i in range(0, array_length):
        if guesser.guess(majority_index_so_far, i) == 0:
            actual_majority_count += 1

    # Check if the candidate is truly the majority
    if actual_majority_count > array_length // 2:
        return majority_index_so_far
    else:
        return -1

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the hidden array once to find a candidate for the majority element. Then, it iterates through the array a second time to count the occurrences of this candidate. Both iterations are proportional to the input size n, where n is the number of elements in the hidden array. Therefore, the total time complexity is O(n + n), which simplifies to O(n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It stores a few variables like the current guess for the majority element, and a counter to track the occurrences of the potential majority. The number of these variables does not depend on the size of the input list (N). Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty array	Return -1 immediately since there's no majority element.
Array with one element	Return the index 0 since that single element is the majority.
Array with all elements identical	The algorithm should correctly identify the first element as the majority, as all queries will return 0 or 1 for equality.
Array with a single element appearing more than n/2 times, and all other elements appearing only once	Moore's Voting Algorithm will isolate the frequent element correctly and verify its majority status through query.
Large array size exceeding memory limits	Moore's Voting Algorithm only needs constant space so it does not have memory issues with a large number of queries.
Hidden API throws exceptions or returns unexpected values	Add try-except block around the calls to the compare function and return -1 if an exception occurs to indicate invalid input.
No majority element exists	After finding a candidate with Moore's Voting Algorithm, verify with queries that the candidate is indeed the majority by counting its occurrences using compare.
Integer overflow if using counts for comparison	Moore's Voting Algorithm avoids explicit counting so integer overflow is not a direct concern, but verify with compare function result.

Empty array

How to Handle:

Return -1 immediately since there's no majority element.

Array with one element

How to Handle:

Return the index 0 since that single element is the majority.

Array with all elements identical

How to Handle:

The algorithm should correctly identify the first element as the majority, as all queries will return 0 or 1 for equality.

Array with a single element appearing more than n/2 times, and all other elements appearing only once

How to Handle:

Moore's Voting Algorithm will isolate the frequent element correctly and verify its majority status through query.

Large array size exceeding memory limits

How to Handle:

Moore's Voting Algorithm only needs constant space so it does not have memory issues with a large number of queries.

Hidden API throws exceptions or returns unexpected values

How to Handle:

Add try-except block around the calls to the compare function and return -1 if an exception occurs to indicate invalid input.

No majority element exists

How to Handle:

After finding a candidate with Moore's Voting Algorithm, verify with queries that the candidate is indeed the majority by counting its occurrences using compare.

Integer overflow if using counts for comparison

How to Handle:

Moore's Voting Algorithm avoids explicit counting so integer overflow is not a direct concern, but verify with compare function result.