Problem Description

You have a binary array nums containing only 0s and 1s, but you cannot access it directly. Instead, you're given an ArrayReader API with two functions:

1. query(a, b, c, d): Takes four distinct indices where 0 <= a < b < c < d < array length. It returns:

   • 4 if all four elements have the same value (all 0s or all 1s)
   • 2 if three elements have one value and one element has the other value (e.g., three 0s and one 1, or three 1s and one 0)
   • 0 if there are exactly two 0s and two 1s

2. length(): Returns the size of the array

Your task is to find the index of any element that belongs to the majority value in the array. The majority value is the one that appears more frequently (either 0 or 1). If both values appear equally (tie), return -1.

Constraint: You can call query() at most 2 * n times, where n is the array length.

The solution works by establishing a reference group of 4 elements (0, 1, 2, 3) and comparing other elements against this reference. By querying different combinations and tracking which elements match the reference pattern, it determines:

• How many elements have the same value as element at index 3
• How many elements have a different value

The algorithm then compares elements at indices 0, 1, and 2 individually against the reference to determine their values relative to index 3. Variable a counts elements with the same value as index 3, while b counts elements with different values. Variable k tracks the last index found with a different value than index 3.

Finally, if a > b, the majority has the same value as index 3, so it returns 3. If b > a, the majority has a different value, so it returns k (an index with the opposite value). If a == b, there's no majority, so it returns -1.

Intuition

The key insight is that we can't directly determine the actual values (0 or 1) at any position, but we can determine relative relationships between elements - whether they're the same or different.

Think of it this way: if we pick a reference group of 4 elements and get a query result, then swap one element for another and query again, the change in result tells us about the relationship between the swapped elements.

For example, if query(0, 1, 2, 3) returns 4 (all same), and query(0, 1, 2, 4) also returns 4, we know element 4 has the same value as element 3. But if it returns 2 (three same, one different), element 4 must have a different value than element 3.

The strategy emerges from this observation:

1. Pick element 3 as our reference point - we'll classify all other elements as either "same as 3" or "different from 3"
2. For elements 4 to n-1: Compare query(0, 1, 2, i) with query(0, 1, 2, 3). If results match, element i has the same value as element 3
3. For elements 0, 1, 2: We need a different approach since they're part of our reference group. We use query(0, 1, 2, 4) as a new reference and check:
   • query(1, 2, 3, 4) vs query(0, 1, 2, 4) tells us if element 0 matches element 3
   • query(0, 2, 3, 4) vs query(0, 1, 2, 4) tells us if element 1 matches element 3
   • query(0, 1, 3, 4) vs query(0, 1, 2, 4) tells us if element 2 matches element 3

By counting how many elements are "same as 3" versus "different from 3", we determine which group is the majority. We don't need to know if element 3 is actually 0 or 1 - we just need to know which group is larger and return an index from that group.

Learn more about Math patterns.

Solution Approach

The implementation follows a systematic approach to classify all elements relative to element 3:

Step 1: Initialize and set reference

n = reader.length()
x = reader.query(0, 1, 2, 3)  # Reference query result
a, b = 1, 0  # a: count of elements same as 3, b: count of different elements
k = 0  # Track last index that's different from 3

We start with a = 1 because element 3 is in its own group.

Step 2: Compare elements 4 to n-1 with the reference

for i in range(4, n):
    if reader.query(0, 1, 2, i) == x:
        a += 1  # Element i has same value as element 3
    else:
        b += 1  # Element i has different value from element 3
        k = i   # Remember this index as different from 3

For each element i, we replace element 3 with element i in our query. If the result stays the same as x, elements 3 and i must have the same value.

Step 3: Check elements 0, 1, and 2

y = reader.query(0, 1, 2, 4)  # New reference with element 4

# Check element 0
if reader.query(1, 2, 3, 4) == y:
    a += 1  # Element 0 same as element 3
else:
    b += 1  # Element 0 different from element 3
    k = 0

# Check element 1
if reader.query(0, 2, 3, 4) == y:
    a += 1  # Element 1 same as element 3
else:
    b += 1  # Element 1 different from element 3
    k = 1

# Check element 2
if reader.query(0, 1, 3, 4) == y:
    a += 1  # Element 2 same as element 3
else:
    b += 1  # Element 2 different from element 3
    k = 2

We use element 4 to form a new reference group. By swapping out each of elements 0, 1, 2 with element 3 and comparing results:

• If swapping element 0 with 3 keeps the same result, they have the same value
• Same logic applies for elements 1 and 2

Step 4: Determine majority and return result

if a == b:
    return -1  # Tie - no majority
return 3 if a > b else k  # Return index from majority group

• If a > b: The majority has the same value as element 3, so return 3
• If b > a: The majority has a different value from element 3, so return k (any index with the opposite value)
• If a == b: No majority exists, return -1

Query Count Analysis:

• Step 2: n - 4 queries for elements 4 to n-1
• Step 3: 4 queries (one reference y and three comparisons)
• Total: n - 4 + 4 = n queries, well within the 2n limit

Example Walkthrough

Let's walk through a concrete example with array [1, 0, 1, 1, 0, 1] (length 6).

Step 1: Initialize and set reference

• n = 6
• x = query(0, 1, 2, 3) = query([1, 0, 1, 1]) = 2 (three 1s, one 0)
• a = 1 (count of elements same as index 3, which has value 1)
• b = 0 (count of elements different from index 3)
• k = 0 (will track last index with different value)

Step 2: Compare elements 4 and 5 with reference

For index 4:

• query(0, 1, 2, 4) = query([1, 0, 1, 0]) = 0 (two 1s, two 0s)
• Since 0 ≠ x(2), element 4 has a different value than element 3
• b = 1, k = 4

For index 5:

• query(0, 1, 2, 5) = query([1, 0, 1, 1]) = 2 (three 1s, one 0)
• Since 2 == x(2), element 5 has the same value as element 3
• a = 2

Current counts: a = 2 (indices 3 and 5 have value 1), b = 1 (index 4 has value 0)

Step 3: Check elements 0, 1, and 2

Set new reference: y = query(0, 1, 2, 4) = query([1, 0, 1, 0]) = 0

Check element 0:

• query(1, 2, 3, 4) = query([0, 1, 1, 0]) = 0 (two 1s, two 0s)
• Since 0 == y(0), when we swap element 0 with 3, the result stays the same
• This means element 0 and element 3 have the same value
• a = 3

Check element 1:

• query(0, 2, 3, 4) = query([1, 1, 1, 0]) = 2 (three 1s, one 0)
• Since 2 ≠ y(0), when we swap element 1 with 3, the result changes
• This means element 1 and element 3 have different values
• b = 2, k = 1

Check element 2:

• query(0, 1, 3, 4) = query([1, 0, 1, 0]) = 0 (two 1s, two 0s)
• Since 0 == y(0), when we swap element 2 with 3, the result stays the same
• This means element 2 and element 3 have the same value
• a = 4

Step 4: Determine majority

Final counts:

• a = 4 (indices 0, 2, 3, 5 all have value 1)
• b = 2 (indices 1, 4 have value 0)

Since a(4) > b(2), the majority has the same value as element 3. Return: 3

This correctly identifies index 3 as belonging to the majority value (1 appears 4 times vs 0 appearing 2 times).

Query count: We made 1 + 2 + 4 = 7 queries total, which is less than 2n = 12.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm makes a constant number of queries for each element in the array:

• Initial query: reader.query(0, 1, 2, 3) - 1 query
• Loop from index 4 to n-1: Each iteration makes exactly 1 query via reader.query(0, 1, 2, i), resulting in n - 4 queries
• Three additional queries after the loop:
  • reader.query(0, 1, 2, 4) - 1 query
  • reader.query(1, 2, 3, 4) - 1 query
  • reader.query(0, 2, 3, 4) - 1 query
  • reader.query(0, 1, 3, 4) - 1 query

Total number of queries: 1 + (n - 4) + 4 = n + 1 = O(n)

Since each query operation is assumed to be O(1), the overall time complexity is O(n).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• Variables n, x, y store integer values
• Variables a, b, k store integer counters/indices
• Loop variable i for iteration

All these variables require constant space regardless of the input size, resulting in O(1) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Initialization of Counter Variables

A common mistake is initializing same_as_index3_count to 0 instead of 1. Since we're using index 3 as our reference point and counting how many elements share the same value as index 3, we must start with same_as_index3_count = 1 because index 3 itself belongs to its own group.

Incorrect:

same_as_index3_count = 0  # Wrong! Forgets to count index 3 itself
different_from_index3_count = 0

Correct:

same_as_index3_count = 1  # Correct! Index 3 is counted in its own group
different_from_index3_count = 0

2. Misunderstanding Query Result Interpretation

Another pitfall is misinterpreting what matching query results mean. When reader.query(0, 1, 2, i) == baseline_query, it means element i has the same value as element 3, not that they're different.

Incorrect Logic:

for i in range(4, n):
    if reader.query(0, 1, 2, i) == baseline_query:
        different_from_index3_count += 1  # Wrong interpretation!
    else:
        same_as_index3_count += 1  # Wrong interpretation!

Correct Logic:

for i in range(4, n):
    if reader.query(0, 1, 2, i) == baseline_query:
        same_as_index3_count += 1  # Same query result = same value
    else:
        different_from_index3_count += 1  # Different query result = different value

3. Forgetting to Update last_different_index for Elements 0, 1, 2

When checking elements 0, 1, and 2, it's crucial to update last_different_index if any of them are different from index 3. Forgetting this could result in returning an invalid index when the majority group is different from index 3.

Incorrect:

# Check index 0
if reader.query(1, 2, 3, 4) == reference_query_with_4:
    same_as_index3_count += 1
else:
    different_from_index3_count += 1
    # Forgot to update last_different_index!

Correct:

# Check index 0
if reader.query(1, 2, 3, 4) == reference_query_with_4:
    same_as_index3_count += 1
else:
    different_from_index3_count += 1
    last_different_index = 0  # Important: track this index

4. Using Wrong Indices in Verification Queries

When checking elements 0, 1, and 2, the queries must correctly swap out the element being tested with index 3. A common error is using the wrong combination of indices.

Incorrect (checking index 0):

# Wrong: should exclude index 0 and include index 3
if reader.query(0, 2, 3, 4) == reference_query_with_4:
    # This actually tests index 1, not index 0!

Correct (checking index 0):

# Correct: exclude index 0, include index 3
if reader.query(1, 2, 3, 4) == reference_query_with_4:
    # Properly tests if index 0 has same value as index 3

Solution to Avoid These Pitfalls:

1. Document your reference point clearly - Always comment that index 3 is your reference and you're comparing all other elements against it
2. Use descriptive variable names - Instead of a and b, use names like same_as_index3_count and different_from_index3_count
3. Add assertions or checks - Verify that same_as_index3_count + different_from_index3_count == n at the end as a sanity check
4. Test with small examples - Manually trace through arrays like [0,0,1,1] or [1,1,1,0,0] to verify your logic