Prerequisites

Before attempting this problem, you should be comfortable with:

  • Interactive/Query-Based Problems - Understanding how to extract information through limited API calls rather than direct array access
  • Counting and Grouping - Tracking counts of elements that share properties without knowing their actual values
  • Logical Deduction - Using comparison results to infer relationships between elements

1. n Queries

Intuition

We cannot access array elements directly, only through queries that return the count of identical elements among four indices. The key observation is that comparing two queries that differ by exactly one index tells us whether that differing element matches the rest.

If query(0, 1, 2, 3) equals query(1, 2, 3, 4), then indices 0 and 4 must have the same value (since replacing one with the other kept the count unchanged). Using this principle, we can determine which elements match index 0 and which differ, without knowing the actual values.

Algorithm

  1. Use index 0 as the reference. Track counts for elements equal to index 0 and those that differ.
  2. For indices 4 through n-1: compare query(1, 2, 3, i) with query(0, 1, 2, 3). If equal, index i matches index 0.
  3. For indices 1, 2, and 3: compare queries that swap index 0 with the target index. For example, compare query(0, 2, 3, 4) with query(1, 2, 3, 4) to determine if index 1 matches index 0.
  4. Return index 0 if the equal count is larger, return the last differing index if that group is larger, or return -1 if counts are equal.
class Solution:
    def guessMajority(self, reader: 'ArrayReader') -> int:
        n = reader.length()
        cnt_equal = 1
        cnt_differ = 0
        index_differ = -1

        def f(equal, i):
            nonlocal cnt_equal, cnt_differ, index_differ
            if equal:
                cnt_equal += 1
            else:
                cnt_differ += 1
                index_differ = i

        query0123 = reader.query(0, 1, 2, 3)
        query1234 = reader.query(1, 2, 3, 4)
        f(reader.query(1, 2, 3, 4) == query0123, 4)
        
        for i in range(5, n):
            f(reader.query(1, 2, 3, i) == query0123, i)

        f(reader.query(0, 2, 3, 4) == query1234, 1)
        f(reader.query(0, 1, 3, 4) == query1234, 2)
        f(reader.query(0, 1, 2, 4) == query1234, 3)

        return (0 if cnt_equal > cnt_differ else index_differ
                if cnt_differ > cnt_equal else -1)

Time & Space Complexity

  • Time complexity: O(n)O(n)

  • Space complexity: O(1)O(1)

Where nn is the number of queries.


Common Pitfalls

Misunderstanding What Equal Query Results Mean

When two queries that differ by exactly one index return the same count, the two differing indices must have the same value. Conversely, if the counts differ, those indices have different values. Confusing this relationship leads to incorrectly classifying elements as equal or different from the reference index.

Forgetting to Handle Indices 1, 2, and 3 Separately

The standard approach compares query(1, 2, 3, i) with query(0, 1, 2, 3) to determine if index i matches index 0. However, this only works for indices 4 and above. Indices 1, 2, and 3 require different query comparisons using query(1, 2, 3, 4) as the reference since they overlap with the base query indices.

Returning Wrong Index When Groups Are Equal

When the count of elements equal to index 0 equals the count of different elements, there is no strict majority, so the function should return -1. A common mistake is returning 0 or indexDiffer in this case. The problem requires a strict majority, meaning one group must have more than half the elements.