Count Subarrays Where Max Element Appears at Least K Times

class Solution {
public:
    long long computeSubarrays(vector<int>& elements, int threshold) {
        int highestValue = *max_element(elements.begin(), elements.end());
        vector<int> maxIndices;
        long long result = 0;

        for (int i = 0; i < elements.size(); i++) {
            if (elements[i] == highestValue) {
                maxIndices.push_back(i);
            }

            int count = maxIndices.size();
            if (count >= threshold) {
                result += maxIndices[count - threshold] + 1;
            }
        }

        return result;
    }
};

The provided code consists of a method computeSubarrays that computes the number of subarrays in which the maximum element appears at least k times. The method is written in C++ and belongs to the Solution class.

• Start by extracting the highest value in the array using the max_element function. This identifies the target value for which the appearances in the subarrays are scrutinized.

• Initialize a vector<int> named maxIndices to store indices where the highest value occurs in the elements vector.

• Iterate over the vector elements. Each time you encounter the highestValue, append the current index i to maxIndices.

• Maintain a running count of occurrences of the maximum element by keeping track of the size of maxIndices.

• If the count of these occurrences meets or exceeds the threshold k, add to the result the number of possible starting positions of subarrays that include the k-th last appearance of highestValue as the maximum. This is calculated as maxIndices[count - threshold] + 1.

• Finally, return the total count stored in result. This value represents the number of subarrays where the maximum element appears at least k times.

By following this approach, the function efficiently determines and counts suitable subarrays meeting the given condition using a single pass through the input array and leveraging the properties of vector indexing.

class Solution {

    public long findSubarrayCountsWithMaxElements(int[] array, int threshold) {

        int highestValue = Integer.MIN_VALUE;
        for (int num : array) {
            if (num > highestValue) {
                highestValue = num;
            }
        }

        ArrayList<Integer> maxElementPositions = new ArrayList<>();
        long result = 0;

        for (int pos = 0; pos < array.length; pos++) {
            if (array[pos] == highestValue) {
                maxElementPositions.add(pos);
            }

            int frequency = maxElementPositions.size();
            if (frequency >= threshold) {
                result += maxElementPositions.get(frequency - threshold) + 1;
            }
        }
        return result;
    }
}

This Java solution solves the problem of counting subarrays where the maximum element appears at least 'K' times. It accomplishes this by first identifying the highest value in the array. The method then keeps track of all positions where this maximum element occurs.

Here are the key steps of the implementation process:

1. Initialize highestValue to the smallest possible integer to ensure any element in the array can update it.
2. Traverse through the array to identify the maximum value.
3. Use a dynamic array, maxElementPositions, to store the indices of the array where the maximum value appears.
4. Iterate through the array again,
   • For each position, if the element equals the maximum value, add the position to maxElementPositions.
   • Check if the count of maximum values collected so far meets or exceeds the threshold 'K'.
   • If it does, calculate the contribution of the current position to the total count of valid subarrays.
5. The results from steps that meet the required condition are tallied to return the total count.

This approach ensures each subarray where the highest element occurs at least 'K' times is considered, leveraging efficient array traversal and dynamic list operations for indexing. The final result represents the total count of such subarrays.

class Solution:
    def countSubarrays(self, numbers: List[int], threshold: int) -> int:
        highest_value = max(numbers)
        max_index_list = []
        result_count = 0

        for index, value in enumerate(numbers):

            if value == highest_value:
                max_index_list.append(index)

            frequency = len(max_index_list)
            if frequency >= threshold:
                result_count += max_index_list[-threshold] + 1

        return result_count

To tackle the given problem of counting subarrays in which the maximum element appears at least k times, you need to understand the approach used in the Python solution provided:

• First, identify the highest value in the array using the max() function. This is crucial because subarrays are defined by the frequency of this maximum value.

• Store the indices of all occurrences of this maximum element in a list called max_index_list. This helps in determining possible subarray starting points.

• To count the subarrays meeting the condition, iterate through the numbers array:

  • For each element that equals the highest value, add its index to max_index_list.
  • If the count of indices in max_index_list is at least k, increment the result_count by the position that allows the creation of a new valid subarray. This position is calculated as max_index_list[-k] + 1, representing the index of the starting element of the new subarray which satisfies the condition.

• Finally, return the result_count which now holds the number of valid subarrays.

This solution leverages indexing and list operations effectively to count subarrays without needing to individually check each potential subarray, thus optimizing performance.