`heapq.heapreplace()`¶

heapq.heapreplace(heap, item) → value. Pop and return the current smallest value, and add the new item.¶

This is more efficient than heappop() followed by heappush(), and can be more appropriate when using a fixed-size heap. Note that the value returned may be larger than item! That constrains reasonable uses of this routine unless written as part of a conditional replacement: