Problem Description

In this problem, we're given an array of integers called nums, where every integer is either 0 or 1. Direct access to this array is not permitted; instead, we have access to an API ArrayReader that provides two functions. The query(a, b, c, d) function can be used to query the array with four different indices (with the condition 0 <= a < b < c < d < ArrayReader.length()) and returns the distribution of the four selected elements as follows:

• Returns 4 if all four elements are 0's or all are 1's.
• Returns 2 if three elements are the same (either three 0's and one 1, or three 1's and one 0).
• Returns 0 if there's an even split (two 0's and two 1's).

The length() function returns the size of the array.

Our goal is to find the index of any element with the most frequent value, which could be either 0 or 1. In case of a tie, where 0 and 1 are equally frequent, we return -1.

Intuition

To solve this problem, we start with the understanding that the query() function can give us an insight into the majority element by comparing the distribution of 0s and 1s among different subsets of four elements. Since we're allowed to call query() 2 * n times where n is the array length, we have enough queries at our disposal to deduce the majority element's index.

The solution includes querying the first four elements to establish the initial distribution (let's call this the baseline distribution). Afterwards, we loop through the rest of the elements, using query() with the first three indices fixed and the fourth one changing to include each new index from the rest of the array. If the distribution matches the baseline, the new element shares the majority value with the first three; if it doesn't match, it's different.

The first element not included in the initial query (index 4) is specially treated by checking its distribution with the next 3 consecutive elements. This helps us in confirming whether the initially queried elements were a majority or a split.

After looping through all elements, we compare the count of elements that matched the baseline distribution (a) with those that didn't (b). The index that occurs more is the majority. If both counts are equal, we have a tie, and we return -1.

The intuition is based on iteratively narrowing down the possible indices for the majority element by comparing the distribution of subsets of the array using the query() function.

Learn more about Math patterns.

Solution Approach

The implementation of the solution approach involves using a simple iteration to leverage the ArrayReader API effectively. Here's a breakdown of each part of the code and the algorithms, data structures, or patterns used:

1. Initialization: The n variable is initialized with the size of the array using reader.length(). Two counters, a and b, are initialized to keep track of the counts of indices that match or do not match the initial baseline distribution. Variable k is initialized to store an index where the value is different from the baseline if such an index is found.

2. Initial Query: We perform the first query(0, 1, 2, 3) to establish a baseline distribution stored in variable x. This tells us the distribution pattern of the first four elements.

3. Iterative Comparison: A for loop is used to iterate through the elements starting from index 4 to the last index n-1. In each iteration, we query a subset that includes the first three fixed indices (0, 1, 2) and the current index i.

   • If the return value of this query matches x, we increment counter a;
   • If the return value doesn't match x, we increment counter b and update k with the current index i.

4. Special Cases: After the loop, we need to verify the initial three elements (which are fixed in the previous query() calls) to ensure their distribution. This requires three additional queries, each omitting one of the first three indices alongside the fourth index 4. This helps in confirming whether the initially queried indices were a majority.

   • The results of these queries are compared against a different baseline distribution stored in y which is obtained by query(0, 1, 2, 4).
   • If the result of these comparisons is equal to y, increment a since the omitted element matches the initial query (suggesting it's a majority element); otherwise, increment b and update k respectively.

5. Determine Majority Index: Finally, we determine whether there is a majority by comparing the counts a and b.

   • If a equals b, we conclude there is a tie and return -1;
   • If a is greater, we return 3 which was part of the original subset and, therefore, in the majority.
   • If b is greater, the index k is returned, indicating that k is part of the frequent value.

Data structures used in this solution are primarily basic integer variables for counting. The pattern used is iterative comparisons with early stopping as soon as the majority element is confirmed. The algorithm doesn't require complex data structures as it makes effective use of the provided API to solve the problem.

Example Walkthrough

Let's assume we have a nums array of size 8, and we perform the following steps:

1. Initialization:

   • The array size n is 8 (determined using reader.length()).
   • We initialize two counters, a and b, to zero. Counter a will track indices matching the baseline distribution, while b tracks the others.
   • The k variable is initialized but not yet set.

2. Initial Query:

   • We run query(0, 1, 2, 3) and let's say it returns 4. This suggests all elements are the same (all 0s or all 1s), so our baseline distribution x is 4.

3. Iterative Comparison:

   • We iterate with a loop for indices 4 to 7 (the rest of the array).
   • We perform query(0, 1, 2, 4). Assume it returns 4, identical to our baseline, so we increment a.
   • Next, query(0, 1, 2, 5) might return 2, unlike our baseline. We increment b and set k to 5.
   • We continue with query(0, 1, 2, 6), returning 4, so we increment a.
   • Lastly, query(0, 1, 2, 7) returns 4, incrementing a again.

4. Special Cases:

   • We then create a separate baseline y with query(0, 1, 2, 4) (we have already done this and know it's 4).
   • We run query(1, 2, 3, 4) for special case handling, and if it returns 4, we increment a—let's assume it does;
   • Next, query(0, 2, 3, 4) and if it returns 4, we increment a—let's say it does;
   • Finally, query(0, 1, 3, 4) and if it returns 4, we increment a—let's assume it also does.

At the end of this process, let's assume a is 6, and b is 1. We now decide the majority element:

5. Determine Majority Index:
   • Since a is greater than b, we return index 3 because it belongs to the subset that matched our initial baseline the most.

This example confirmed that the majority values were present in the initial subset, and indexes that matched this distribution were in the majority. If b had been greater, we would return k, which in this walkthrough was 5. If a and b were equal, indicating an even split between 0s and 1s, we would return -1.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided guessMajority function primarily depends on the number of calls made to the reader.query method within a loop and outside of it.

• There is a fixed number of calls to reader.query made outside of the loop for the indices 0, 1, 2, 3, and four additional calls to compare with index 4.
• The loop iterates from index 4 to n - 1 (where n is the length of the array), making one call to reader.query per iteration.
• Hence, the number of query calls is 4 (initial) + 4 (comparisons with index 4) + (n - 4) for the loop.

Thus, the total number of query calls is 8 + (n - 4) = n + 4. Given that each query call is assumed to have a constant time complexity, the time complexity of the guessMajority function is O(n).

Space Complexity

The space complexity of the code is determined by the variables used in the guessMajority function.

• Constant extra space is used for the variables a, b, k, x, and y, independent of the input size n.
• No additional data structures that grow with the input size are used.

Therefore, the space complexity of the guessMajority function is O(1), which means it requires constant space.

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