LeetCode 1538 - Guess the Majority in a Hidden Array

LeetCode Problem 1538

Difficulty: 🟡 Medium
Topics: Array, Math, Interactive

Solution

Problem Understanding

This problem is an interactive problem. We do not receive the actual binary array directly. Instead, we can only gather information through the ArrayReader API.

The hidden array nums contains only 0 and 1. Our goal is to determine which value appears more frequently and return any index containing that majority value. If the number of 0s and 1s is equal, we must return -1.

The only way to inspect the array is through the function:

query(a, b, c, d)

This function compares four distinct indices and returns:

• 4 if all four values are identical
• 2 if exactly three values are the same and one differs
• 0 if there are two 0s and two 1s

The important observation is that the query result depends only on how many equal values exist among the four chosen indices. We never learn the actual values directly.

The array length can be as large as 10^5, and we are limited to at most 2 * n queries. This immediately rules out any solution that repeatedly compares large subsets or attempts exhaustive deduction.

The challenge is therefore not traditional majority voting, but reconstructing relative equality relationships between indices using only these 4-element queries.

Several edge cases are especially important:

• The array may contain an equal number of 0s and 1s, in which case we must return -1.
• The majority value may appear only one more time than the minority value.
• The first few elements may not all belong to the same group.
• Because we never know whether an element is actually 0 or 1, we only need to classify indices into "same as index 0" or "different from index 0".

The key insight is that we do not need the actual values. We only need to partition indices into two equivalence classes.

Approaches

Brute Force Approach

A naive approach would attempt to reconstruct the entire hidden array.

For every index, we could issue multiple queries involving known positions and infer whether the current element matches another one. Eventually, we could derive every bit value relative to a reference element.

This approach is correct because enough carefully chosen queries can determine whether two indices contain the same value. Once all indices are classified, counting frequencies becomes straightforward.

However, a brute-force implementation may require many redundant queries per index. Since the query budget is only 2 * n, any strategy using significantly more than constant queries per element will exceed the limit.

For example, comparing every pair of indices would require O(n^2) queries, which is impossible for n = 10^5.

Optimal Approach

The optimal solution uses the first four indices as a reference framework.

The core observation is:

If replacing one index in a query does not change the query result, then the replaced element must have the same value as the new element.

Suppose we compute:

base = query(0, 1, 2, 3)

Now for an index i >= 4, compare:

query(1, 2, 3, i)

The only difference between the two queries is that index 0 was replaced by i.

• If the result stays the same, then nums[i] == nums[0]
• Otherwise, nums[i] != nums[0]

This allows us to classify every index relative to index 0 using exactly one query per index.

The first four indices require special handling because we cannot query duplicate indices. For each of them, we replace that specific index with index 4 and compare against the base query.

After partitioning all indices into two groups, we simply compare their sizes:

• Larger group wins
• Equal sizes means tie

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n²)	O(1)	Repeated comparisons between indices exceed query budget
Optimal	O(n)	O(1)	Uses query invariance to classify indices efficiently

Algorithm Walkthrough

1. First, compute a baseline query using the first four indices:

base = query(0, 1, 2, 3)

This establishes a reference pattern. 2. Create two counters:

• One group contains indices equal to nums[0]
• The other group contains indices different from nums[0]

Initially:

• same_count = 1
• different_count = 0
• same_index = 0
• different_index = -1

3. Process indices 4 through n - 1.

For each index i, compute:

query(1, 2, 3, i)

This query differs from the baseline only by replacing index 0 with i.

• If the result equals base, then nums[i] == nums[0]
• Otherwise, nums[i] != nums[0]

Update the appropriate counter and store a representative index. 4. Handle indices 1, 2, and 3 separately.

For each of these indices, replace it with index 4:

• Compare query(0, 2, 3, 4) against base for index 1
• Compare query(0, 1, 3, 4) against base for index 2
• Compare query(0, 1, 2, 4) against base for index 3

If the result changes, the replaced index differs from index 0. 5. After all indices are classified, compare group sizes.

• If counts are equal, return -1
• Otherwise, return any representative index from the larger group

Why it works

Each query comparison isolates exactly one changed element. Since the other three indices remain fixed, the only reason the query result changes is because the replacement element differs in value from the removed one.

Thus, every comparison correctly determines whether an index belongs to the same value group as index 0. Since the array contains only two possible values, partitioning indices this way completely determines the majority group.

Python Solution

# """
# This is the ArrayReader's API interface.
# You should not implement it, or speculate about its implementation
# """
# class ArrayReader(object):
#     def query(self, a: int, b: int, c: int, d: int) -> int:
#         pass
#
#     def length(self) -> int:
#         pass

class Solution:
    def guessMajority(self, reader: 'ArrayReader') -> int:
        n = reader.length()

        base = reader.query(0, 1, 2, 3)

        same_count = 1
        diff_count = 0

        same_index = 0
        diff_index = -1

        # Process indices 4 to n - 1
        for i in range(4, n):
            current = reader.query(1, 2, 3, i)

            if current == base:
                same_count += 1
                same_index = i
            else:
                diff_count += 1
                diff_index = i

        # Check index 1
        if reader.query(0, 2, 3, 4) == base:
            same_count += 1
            same_index = 1
        else:
            diff_count += 1
            diff_index = 1

        # Check index 2
        if reader.query(0, 1, 3, 4) == base:
            same_count += 1
            same_index = 2
        else:
            diff_count += 1
            diff_index = 2

        # Check index 3
        if reader.query(0, 1, 2, 4) == base:
            same_count += 1
            same_index = 3
        else:
            diff_count += 1
            diff_index = 3

        if same_count == diff_count:
            return -1

        if same_count > diff_count:
            return same_index

        return diff_index

The implementation starts by establishing a baseline query result for indices (0,1,2,3). Every later query is designed to differ from this baseline by exactly one index.

The loop for indices 4 and beyond compares query(1,2,3,i) with the baseline. Since only index 0 was replaced, matching results imply equal values.

Indices 1, 2, and 3 require special handling because the query API requires distinct indices. Each of them is tested by replacing that specific index with index 4.

Finally, the algorithm compares the two partition sizes and returns a representative index from the larger group.

Go Solution

/**
 * // This is the ArrayReader's API interface.
 * // You should not implement it, or speculate about its implementation
 * type ArrayReader struct {
 * }
 *
 * // Compares 4 different elements in the array
 * func (this *ArrayReader) query(a, b, c, d int) int {}
 *
 * // Returns the length of the array
 * func (this *ArrayReader) length() int {}
 */

func guessMajority(reader *ArrayReader) int {
	n := reader.length()

	base := reader.query(0, 1, 2, 3)

	sameCount := 1
	diffCount := 0

	sameIndex := 0
	diffIndex := -1

	// Process indices 4 to n - 1
	for i := 4; i < n; i++ {
		current := reader.query(1, 2, 3, i)

		if current == base {
			sameCount++
			sameIndex = i
		} else {
			diffCount++
			diffIndex = i
		}
	}

	// Check index 1
	if reader.query(0, 2, 3, 4) == base {
		sameCount++
		sameIndex = 1
	} else {
		diffCount++
		diffIndex = 1
	}

	// Check index 2
	if reader.query(0, 1, 3, 4) == base {
		sameCount++
		sameIndex = 2
	} else {
		diffCount++
		diffIndex = 2
	}

	// Check index 3
	if reader.query(0, 1, 2, 4) == base {
		sameCount++
		sameIndex = 3
	} else {
		diffCount++
		diffIndex = 3
	}

	if sameCount == diffCount {
		return -1
	}

	if sameCount > diffCount {
		return sameIndex
	}

	return diffIndex
}

The Go implementation mirrors the Python logic closely. Since Go uses explicit variable declarations and lacks Python's dynamic typing, all counters and indices are maintained as integers.

There are no overflow concerns because counts never exceed 10^5. The solution also uses constant auxiliary space since no slices or maps are required.

Worked Examples

Example 1

Input:

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

First compute:

base = query(0,1,2,3)

Indices contain:

[0,0,1,0]

Three equal and one different:

base = 2

Initial state:

Variable	Value
same_count	1
diff_count	0
same_index	0
diff_index	-1

Now process indices 4 through 7.

i	Query	Result	Relation to index 0
4	query(1,2,3,4)	0	different
5	query(1,2,3,5)	0	different
6	query(1,2,3,6)	0	different
7	query(1,2,3,7)	0	different

State becomes:

Variable	Value
same_count	1
diff_count	4

Now evaluate indices 1, 2, and 3.

Index 1

query(0,2,3,4)

Result equals base, therefore index 1 matches index 0.

Index 2

query(0,1,3,4)

Result differs from base, therefore index 2 differs from index 0.

Index 3

query(0,1,2,4)

Result equals base, therefore index 3 matches index 0.

Final counts:

Group	Count
Same as index 0	3
Different from index 0	5

The majority group is the "different" group, corresponding to value 1. Any representative from that group is valid.

Output:

5

Example 2

Input:

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

After classification:

Index	Same as index 0?
0	Yes
1	Yes
2	No
3	No
4	Yes

Counts:

Group	Count
Same	3
Different	2

Return any index from the larger group, such as 0.

Example 3

Input:

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

Classification:

Index	Same as index 0?
0	Yes
1	No
2	Yes
3	No
4	Yes
5	No
6	Yes
7	No

Counts:

Group	Count
Same	4
Different	4

Since the counts are equal, return:

-1

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each index is processed with a constant number of queries
Space	O(1)	Only a few counters and indices are stored

The algorithm performs one baseline query, one query for every index from 4 to n-1, and three additional queries for indices 1, 2, and 3.

That totals exactly:

1 + (n - 4) + 3 = n

queries, which is comfortably below the 2 * n limit.

No auxiliary data structures proportional to input size are used.

Test Cases

# These tests simulate the hidden API locally.

class ArrayReader:
    def __init__(self, nums):
        self.nums = nums

    def query(self, a, b, c, d):
        values = [
            self.nums[a],
            self.nums[b],
            self.nums[c],
            self.nums[d]
        ]

        zeros = values.count(0)
        ones = values.count(1)

        if zeros == 4 or ones == 4:
            return 4

        if zeros == 2 and ones == 2:
            return 0

        return 2

    def length(self):
        return len(self.nums)

class Solution:
    def guessMajority(self, reader):
        n = reader.length()

        base = reader.query(0, 1, 2, 3)

        same_count = 1
        diff_count = 0

        same_index = 0
        diff_index = -1

        for i in range(4, n):
            if reader.query(1, 2, 3, i) == base:
                same_count += 1
                same_index = i
            else:
                diff_count += 1
                diff_index = i

        if reader.query(0, 2, 3, 4) == base:
            same_count += 1
            same_index = 1
        else:
            diff_count += 1
            diff_index = 1

        if reader.query(0, 1, 3, 4) == base:
            same_count += 1
            same_index = 2
        else:
            diff_count += 1
            diff_index = 2

        if reader.query(0, 1, 2, 4) == base:
            same_count += 1
            same_index = 3
        else:
            diff_count += 1
            diff_index = 3

        if same_count == diff_count:
            return -1

        if same_count > diff_count:
            return same_index

        return diff_index

sol = Solution()

# Example 1
nums = [0,0,1,0,1,1,1,1]
ans = sol.guessMajority(ArrayReader(nums))
assert nums[ans] == 1  # majority is 1

# Example 2
nums = [0,0,1,1,0]
ans = sol.guessMajority(ArrayReader(nums))
assert nums[ans] == 0  # majority is 0

# Example 3
nums = [1,0,1,0,1,0,1,0]
assert sol.guessMajority(ArrayReader(nums)) == -1  # tie case

# Minimum size array
nums = [1,1,1,0,1]
ans = sol.guessMajority(ArrayReader(nums))
assert nums[ans] == 1  # smallest valid n

# All elements equal
nums = [0,0,0,0,0,0]
ans = sol.guessMajority(ArrayReader(nums))
assert nums[ans] == 0  # all identical

# Majority by one element
nums = [1,0,1,0,1]
ans = sol.guessMajority(ArrayReader(nums))
assert nums[ans] == 1  # narrow majority

# Alternating but odd length
nums = [0,1,0,1,0,1,0]
ans = sol.guessMajority(ArrayReader(nums))
assert nums[ans] == 0  # alternating pattern

# Large majority
nums = [1] * 90 + [0] * 10
ans = sol.guessMajority(ArrayReader(nums))
assert nums[ans] == 1  # dominant majority

Test	Why
[0,0,1,0,1,1,1,1]	Validates standard majority detection
[0,0,1,1,0]	Tests small odd-sized input
[1,0,1,0,1,0,1,0]	Verifies tie handling
[1,1,1,0,1]	Tests minimum valid size
[0,0,0,0,0,0]	Ensures all-equal arrays work
[1,0,1,0,1]	Tests majority by one element
[0,1,0,1,0,1,0]	Tests alternating pattern
[1]*90 + [0]*10	Tests strongly skewed distribution

Edge Cases

Tie Between Zeros and Ones

A very important edge case occurs when the array contains exactly the same number of 0s and 1s.

A buggy implementation might always return one of the groups without checking equality carefully. This solution explicitly compares same_count and diff_count at the end and returns -1 when they match.

All Elements Identical

If every element in the array is identical, every comparison query will match the baseline query.

This can expose bugs in implementations that incorrectly initialize counts or fail to update representative indices properly. Here, all indices correctly accumulate into the "same" group, and any valid index is returned.

Majority by Only One Element

Arrays such as:

[1,0,1,0,1]

are easy to mishandle because a single classification mistake changes the final answer.

The solution avoids ambiguity because every comparison isolates exactly one replaced element. Each index is classified independently relative to index 0, guaranteeing accurate counts even when the margin is minimal.