Count Subarrays With Fixed Bounds

Scan once while retaining lastBad, the latest index with a value outside [minK, maxK], plus lastMin and lastMax, the latest positions of the required bounds. At index i, every start after lastBad is range-valid, and to include both bounds it must be at or before min(lastMin, lastMax). Therefore the number of valid subarrays ending at i is max(0, min(lastMin,lastMax) - lastBad). Add that value at every step. Initialize all three indices to -1. This direct index invariant handles minK == maxK naturally and avoids maintaining a costly multiset. Test values outside the range, missing one bound, and repeated bounds. Time is O(n), space is O(1), and the answer should use 64-bit storage.