Count Subarrays Where Max Element Appears at Least K Times

First compute the global maximum M. Sweep right across the array, incrementing a counter when nums[right] equals M. While the counter is at least k, move left forward, decrementing when nums[left] is M. After this shrinking, left is the number of starting positions that produce a subarray ending at right with at least k occurrences of M, so add left to the answer. The invariant is that starts before left are valid and starts at or after left are not yet valid. Use a 64-bit answer because subarray counts can be quadratic. Check k = 1, fewer than k maximums, and repeated maxima. The two pointers each move at most n times, yielding O(n) time and O(1) extra space.