m = len(x)
n = len(x)//2
left = 2*n + 1
right = 2*n + 2
if left < m and x[left] <= x[n]:
smallest = left
else:
smallest = n
if right < m and x[right] <= x[smallest]:
smallest = right
if not largest == n:
x[n], x[smallest] = x[smallest], x[n]
heapify(x)
by the way the last call is just a recursive call of the function

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