Implementing the class MajorityChecker, which has the following API:

• MajorityChecker(int[] arr) constructs an instance of MajorityChecker with the given array arr;
• int query(int left, int right, int threshold) has arguments such that:
• 0 <= left <= right < arr.length representing a subarray of arr;
• 2 * threshold > right - left + 1, ie. the threshold is always a strict majority of the length of the subarray

Each query(...) returns the element in arr[left], arr[left+1], ..., arr[right]that occurs at least threshold times, or -1 if no such element exists.

Example:

MajorityChecker majorityChecker = new MajorityChecker([1,1,2,2,1,1]);
majorityChecker.query(0,5,4); // returns 1
majorityChecker.query(0,3,3); // returns -1
majorityChecker.query(2,3,2); // returns 2


Constraints:

• 1 <= arr.length <= 20000
• 1 <= arr[i] <= 20000
• For each query, 0 <= left <= right < len(arr)
• For each query, 2 * threshold > right - left + 1
• The number of queries is at most 10000

Solution 0: Brute Force TLE

For each range, find the majority element in O(n) time / O(n) or O(1) space.

Solution 1: Cache the range result

Cache the result for each range: mode and frequency

Time complexity:
init: O(1)
query: O(|right – left|)
total queries: O(sum(right_i – left_i)), right_i, left_i are bounds of unique ranges.
Space complexity:
init: O(1)
query: O(20000)

Solution 2: HashTable + binary search

Preprocessing: use a hashtable to store the indices of each element. And sort by frequency of the element in descending order.
Query: iterator over (total_freq, num) pair, if freq >= threshold do a binary search to find the frequency of the given range for num in O(logn).

Time complexity:
Init: O(nlogn)
Query: worst case: O(nlogn), best case: O(logn)

Solution 2+: Randomization

Randomly pick a num in range [left, right] and check its freq, try k (e.g. 30) times. If a majority element exists, you should find it with error rate 1/2^k.

Time complexity:
Init: O(n)
Query: O(30 * logn)
Space complexity: O(n)

C++

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