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Guess the Majority in a Hidden Array

Updated on 02 June, 2025
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Problem Statement

In this problem, we are provided with an integer array nums which consists exclusively of the integers 0 and 1. However, we cannot directly interact with this array. Instead, we are given access through a provided API called ArrayReader that allows us to query the array indirectly. Here is what each method of the API accomplishes:

  • int query(int a, int b, int c, int d): This method accepts four indices (a, b, c, d with 0 <= a < b < c < d < ArrayReader.length()). It evaluates the elements at these indices in the hidden array and returns:
    • 4: If all four elements are identical.
    • 2: If three elements are of one value and the other is of another (i.e., three 0s and one 1 or three 1s and one 0).
    • 0: If there are two 0s and two 1s.
  • int length(): Provides the total number of elements in the hidden array nums.

Our goal utilizing these API functions is to determine the index of the most frequent element (0 or 1) in the array nums. If neither number is more frequent than the other, the function should return -1. We are restricted to making 2 * n query calls, where n is the length of the array.

Examples

Example 1

Input:

nums = [0,0,1,0,1,1,1,1]

Output:

5

Explanation:

The following calls to the API
reader.length() // returns 8 because there are 8 elements in the hidden array.
reader.query(0,1,2,3) // returns 2 this is a query that compares the elements nums[0], nums[1], nums[2], nums[3]
// Three elements have a value equal to 0 and one element has value equal to 1 or viceversa.
reader.query(4,5,6,7) // returns 4 because nums[4], nums[5], nums[6], nums[7] have the same value.
we can infer that the most frequent value is found in the last 4 elements.
Index 2, 4, 6, 7 is also a correct answer.

Example 2

Input:

nums = [0,0,1,1,0]

Output:

0

Example 3

Input:

nums = [1,0,1,0,1,0,1,0]

Output:

-1

Constraints

  • 5 <= nums.length <= 105
  • 0 <= nums[i] <= 1

Approach and Intuition

Given the example inputs and the constraints mentioned, here's a logical way to break down the approach:

Example 1:

  • In this scenario, the array nums = [0,0,1,0,1,1,1,1].
  • Our initial steps would look like:
    1. Query the first half of the array reader.query(0, 1, 2, 3). If this query returns 2, it means there is a split in the majority, with three positions taken by one number (either 0 or 1), and one by the other.
    2. Next, query the second half of the array reader.query(4, 5, 6, 7). If this returns a 4, it means all the elements in that segment are either all 0s or all 1s.
    3. From this information, we can infer the most frequent element appears more within the queried segments.

Example 2:

  • The array nums = [0,0,1,1,0] would be split and queried similarly:
    1. This short example could be managed by one or two queries across different sections to determine the majority frequently appearing number. Here, since the length is odd and the distribution after one or two queries can directly determine the majority (since one section might not need querying if already known).

Example 3:

  • A more evenly distributed array like nums = [1,0,1,0,1,0,1,0] might return equal distributions from multiple queries:
    1. The returned values, all being 0, more systematically from various segments helps establish there's no frequency prevalence.
    2. In this evenly split condition, leading to a returned value of -1 due to the tie is the optimal outcome.

From this, we see that the approach generally involves segmenting the array using the query function, deducing the occurrence patterns from the returned values, and systematically determining the prevalent frequency to find the index of the most frequently occurring number or recognize a balanced distribution.

Solutions

  • C++
  • Java
  • Python
cpp 
class Solution {
    int cntMatches = 1, cntNonMatches = 0, unmatchedIndex = -1;

public:
    int detectMajority(ArrayReader& arrReader) {
        int size = arrReader.length(), initialQuery = arrReader.query(0, 1, 2, 3),
            secondQuery = arrReader.query(1, 2, 3, 4);
        function<void(bool, int)> process = [this](bool isEqual, int index) {
            if (isEqual) {
                cntMatches++;
            } else {
                cntNonMatches++;
                unmatchedIndex = index;
            }
        };

        process(secondQuery == initialQuery, 4);
        for (int i = 5; i < size; i++) {
            process(arrReader.query(1, 2, 3, i) == initialQuery, i);
        }
        process(arrReader.query(0, 2, 3, 4) == secondQuery, 1);
        process(arrReader.query(0, 1, 3, 4) == secondQuery, 2);
        process(arrReader.query(0, 1, 2, 4) == secondQuery, 3);
        return cntMatches > cntNonMatches ? 0 : cntNonMatches > cntMatches ? unmatchedIndex : -1;
    }
};

In this solution for the "Guess the Majority in a Hidden Array" problem using C++, an approach is taken to determine the majority element without directly accessing array elements, as the array is hidden.

Here’s what unfolded in the implementation:

  • First, the implementation uses an ArrayReader object, which is a class to interact with the hidden data. Two initial queries are made with the query() method to compare indices in the array to get clues about majority elements.

  • The queries made are strategic; first to determine initial patterns by checking set groups of indices. This gives a sense of whether there are more matching or non-matching elements.

  • A lambda function process is defined within the class to update the count of elements matching the initial pattern (cntMatches) and elements that don't match (cntNonMatches, noting the index with unmatchedIndex if a discrepancy occurs).

  • A loop iterates through the array starting from index 5 (since the first few queries cover indices 0 to 4). Within the loop, the same query strategy determines if the majority logic holds true for subsequent elements.

  • The solution determines which indices to compare based on previous majority clues (initialQuery and secondQuery). Leveraging this, the program decides if matches and non-matches keep the initial query conditions intact.

  • At the end, the detectMajority function returns:

    • Index 0 if cntMatches is greater indicating consistent matches across queries.
    • unmatchedIndex if cntNonMatches is greater suggesting the majority are the non-matching cases.
    • -1 if neither are conclusive in majority.

The implementation essentially balances finding patterns and checking their consistency, enabling the deduction of the majority without revealing actual elements of the hidden array. This utilizes the provided methods on the ArrayReader to conduct minimal and necessary checks, ensuring efficiency considering the constraints.

java 
class Solution {
    int countSame = 1, countDifferent = 0, differentIndex = -1;

    private void checkEquality(boolean condition, int index) {
        if (condition) {
            countSame++;
        } else {
            countDifferent++;
            differentIndex = index;
        }
    }

    public int findMajorityIndex(ArrayReader reader) {
        int size = reader.length(), firstFour = reader.query(0, 1, 2, 3), lastFour = reader.query(1, 2, 3, 4);
        checkEquality(lastFour == firstFour, 4);
        for (int j = 5; j < size; j++) {
            checkEquality(reader.query(1, 2, 3, j) == firstFour, j);
        }
        checkEquality(reader.query(0, 2, 3, 4) == lastFour, 1);
        checkEquality(reader.query(0, 1, 3, 4) == lastFour, 2);
        checkEquality(reader.query(0, 1, 2, 4) == lastFour, 3);
        return countSame > countDifferent ? 0 : countDifferent > countSame ? differentIndex : -1;
    }
}

The given Java solution aims to solve the problem of identifying the index of the majority element in a hidden array. An interface ArrayReader is utilised, which includes methods length() to fetch the size of the array and query() to determine equality among selected indices. The approach hinges on comparing the results from various subsets of indices to deduce the majority.

Here is a breakdown of the program logic:

  • Variables such as countSame, countDifferent, and differentIndex are initialized to keep track of the number of times indices have the same response as the initial check (firstFour), times they differ, and the index of the first occurrence of a differing response, respectively.
  • The method checkEquality() updates these variables based on the condition passed—whether the result of a new query matches the previously established pattern.
  • In findMajorityIndex(), the size of the array is gathered, and initial queries (firstFour for the first four indices and lastFour for indices 1 to 4) are made to establish a baseline comparison.
  • Iterative checks from the sixth element onward adjust the counts based on their conformity to the baseline.
  • Additional checks for indices 1 through 3 against the lastFour results update counts and potential differing index.
  • Finally, it evaluates which among countSame or countDifferent` prevails; if there is a majority, the respective index or condition is returned, else -1 indicates no clear majority.

This solution assesses majority through strategic querying and comparisons, optimizing the interrogation of the hidden array to deduce the majority element efficiently.

python 
class Solution:
    def findMajorityElement(self, reader: 'ArrayReader') -> int:
        length = reader.length()
        countSame = 1
        countDifferent = 0
        differentIndex = -1

        def updateCount(isSame, index):
            nonlocal countSame, countDifferent, differentIndex
            if isSame:
                countSame += 1
            else:
                countDifferent += 1
                differentIndex = index

        firstQuery = reader.query(0, 1, 2, 3)
        secondQuery = reader.query(1, 2, 3, 4)
        updateCount(reader.query(1, 2, 3, 4) == firstQuery, 4)
        for j in range(5, length):
            updateCount(reader.query(1, 2, 3, j) == firstQuery, j)
        updateCount(reader.query(0, 2, 3, 4) == secondQuery, 1)
        updateCount(reader.query(0, 1, 3, 4) == secondQuery, 2)
        updateCount(reader.query(0, 1, 2, 4) == secondQuery, 3)
        return (0 if countSame > countDifferent else differentIndex
                if countDifferent > countSame else -1)

This Python solution addresses the problem of determining the majority element in a hidden array by leveraging queries provided through an ArrayReader interface. The objective is to identify an index that represents the majority element based on comparison queries across elements of the array.

  • Initiate the algorithm by fetching the total length of the array from the reader.
  • Two main counters, countSame and countDifferent, are used to track how many times the results of queries match or differ from initially established baseline queries.
  • A function updateCount is defined locally to update these counts based on the results of subsequent queries.
  • Two initial queries compare groups of four elements consecutively. These are pivotal in setting a reference for later comparisons.
  • The results of these base queries are used to extend comparison to elements beyond the initial sets.
  • Iterate over each subsequent element, comparing it using a query against a consistent subset from the base query and updating the counts.
  • Three additional queries adjust the baseline by checking one element at a time against the rest, refining the results.
  • Decide the output based on the majority identified: return 0 if the majority result matches the base case, the specific different index if a new majority is found, or -1 if no conclusive majority exists.

By following this robust querying process, you efficiently use limited available operations to deduce the majority element, an essential strategy when directly accessing elements is restricted.

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