You are given an integer array nums and a positive integer k. Return the number of subarrays where the maximum element of nums appears at least k times in that subarray. A subarray is a contiguous sequence of elements within an array.

countSubarrays.js

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var countSubarrays = function(nums, k) {
    // Initialize an array to store the indices of the maximum elements and a variable to store the maximum element
    let indices = [], max = 0;

    // Iterate over the nums array
    for (let i = 0; i < nums.length; i++) {
        // If the current number is greater than max, update max and reset indices
        if (nums[i] > max) {
            indices = [i];
            max = nums[i];
        } 
        // If the current number is equal to max, add its index to indices
        else if (nums[i] === max) {
            indices.push(i);
        }
    }

    // Initialize a variable to store the count of valid subarrays and a pointer to the left end of the current subarray
    let count = 0, leftPtr = 0;

    // Iterate over the indices array while there are at least k elements to the right of leftPtr
    while (leftPtr <= indices.length - k) {
        // Calculate the count of subarrays ending at the right end of the current subarray
        const currCount = indices[leftPtr] + 1;
        // Calculate the index of the right end of the current subarray
        const rightPtr = leftPtr + k - 1;
        // Calculate the index of the next maximum element or the end of nums if there is no next maximum element
        let nextPtr = rightPtr + 1 >= indices.length ? nums.length : indices[rightPtr + 1];
        // Calculate the count of subarrays starting at the next maximum element or the end of nums if there is no next maximum element
        let nextCount = nextPtr - indices[rightPtr] - 1;
        // Add the count of valid subarrays in the current subarray to count
        count += currCount + (currCount * nextCount);
        // Move the left end of the current subarray to the right
        leftPtr++;
    }

    // Return the count of valid subarrays
    return count;
};