Guess the Majority in a Hidden Array - Practice Coding Problems

Guess the Majority in a Hidden Array - Problem

Imagine you're a detective trying to solve a mystery about a hidden binary array containing only 0s and 1s. You can't see the array directly, but you have a special tool - an ArrayReader API that can examine groups of 4 elements at once!

Your mission: Find any index of the most frequent value in the array. If there's a tie between 0s and 1s, return -1.

The ArrayReader API provides:

query(a, b, c, d): Returns information about 4 elements at positions a < b < c < d:
- 4: All 4 elements are the same (all 0s or all 1s)
- 2: Three elements have one value, one element has the other
- 0: Two elements are 0, two elements are 1
length(): Returns the array size

Challenge: You can only call query() at most 2×n times where n is the array length!

Input & Output

example_1.py — Basic Case

$ Input: Hidden array: [0,0,1,0,0,1,1,1], n = 8

› Output: 5 (or any index containing 1)

💡 Note: The array has 3 zeros and 5 ones. Since 1 is the majority element (5 > 3), we can return any index containing 1, such as index 5.

example_2.py — Tie Case

$ Input: Hidden array: [0,0,1,1], n = 4

› Output: -1

💡 Note: The array has 2 zeros and 2 ones. Since there's a tie (2 = 2), we return -1 as specified.

example_3.py — Edge Case

$ Input: Hidden array: [1,1,1,1,1], n = 5

› Output: 0 (or any index from 0-4)

💡 Note: All elements are 1, so 1 is clearly the majority. We can return any valid index.

Visualization

Tap to expand

Understanding the Visualization

Establish Reference

Query(0,1,2,3) gives us a baseline pattern to compare against

Strategic Deduction

Use additional queries to logically determine what nums[0] contains

Systematic Comparison

Replace nums[0] with each other element to detect matches/differences

Count and Conclude

Tally the evidence to determine which value is in the majority

Key Takeaway

🎯 Key Insight: By establishing element 0 as our reference point and using strategic replacement queries, we can efficiently determine the majority element while staying within the 2n query limit!

Time & Space Complexity

Time Complexity

⏱️

O(n)

We make at most 2n queries, each taking constant time

✓ Linear Growth

Space Complexity

O(1)

Only need to track counts, not store entire array

✓ Linear Space

Constraints

5 ≤ nums.length ≤ 10⁵
nums[i] is either 0 or 1
You can call query() at most 2 × nums.length times
0 ≤ a < b < c < d < ArrayReader.length() for all query calls

Asked in

G Google 15 f Meta 12 a Amazon 8 ⊞ Microsoft 6

The optimal solution uses a smart reference strategy: first determine the value of element 0 through strategic queries, then efficiently compare all other elements by replacing element 0 in queries to detect if values match or differ. This approach uses at most 2n queries and runs in O(n) time with O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Query Strategy	O(n)	O(n)	Query all possible combinations and try to reconstruct the array
Smart Reference Strategy (Optimal)	O(n)	O(1)	Determine one element's value, then efficiently compare others against it

Brute Force Query Strategy — Algorithm Steps

Query consecutive groups of 4 elements: (0,1,2,3), (1,2,3,4), etc.
Try to infer individual element values from overlapping queries
Count the number of 0s and 1s based on inferred values
Return index of majority element or -1 if tie

Visualization

Tap to expand

Step-by-Step Walkthrough

Query Groups

Query consecutive 4-element groups

Overlap Analysis

Analyze overlapping elements between queries

Value Inference

Try to infer individual element values

Majority Count

Count inferred 0s and 1s to find majority

Code -

solution.c — C

// Simplified brute force approach
int guessMajority(ArrayReader* reader) {
    int n = length(reader);
    if (n < 4) return -1;
    
    int queriesUsed = 0;
    int inferred[1000] = {0}; // Assuming max array size
    int inferredCount = 0;
    
    // Query consecutive groups
    for (int i = 0; i <= n - 4; i++) {
        if (queriesUsed >= 2 * n) break;
        
        int result = query(reader, i, i + 1, i + 2, i + 3);
        queriesUsed++;
        
        // Simplified inference logic
        if (result == 4) {
            for (int j = i; j < i + 4; j++) {
                inferred[j] = 1;
                if (j >= inferredCount) inferredCount = j + 1;
            }
        }
    }
    
    // Count majority
    if (inferredCount == 0) return -1;
    
    int ones = 0;
    for (int i = 0; i < inferredCount; i++) {
        if (inferred[i] == 1) ones++;
    }
    
    int zeros = inferredCount - ones;
    
    if (ones > zeros) {
        for (int i = 0; i < inferredCount; i++) {
            if (inferred[i] == 1) return i;
        }
    } else if (zeros > ones) {
        for (int i = 0; i < inferredCount; i++) {
            if (inferred[i] == 0) return i;
        }
    }
    
    return -1;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We make O(n) queries and each query takes constant time

✓ Linear Growth

Space Complexity

O(n)

We need to store inferred values for reconstruction

⚡ Linearithmic Space

Constraints

5 ≤ nums.length ≤ 10⁵
nums[i] is either 0 or 1
You can call query() at most 2 × nums.length times
0 ≤ a < b < c < d < ArrayReader.length() for all query calls

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force Query Strategy — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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