Вывести второе минимальное число в списке

print(__import__("heapq").nsmallest(2, map(int, input().split()))[-1])

a = [1, 2, 3, 4, 5, 6]
k = min(a)
for i in range(len(a)):
    if a[i] == k:
        a.pop(i)
        break
print(min(a))

a = [1, 1, 1, 2, 3, 4, 5, 6]

In [1]: from random import randrange
 
In [2]: a = [randrange(-1_000, 1_000) for _ in range(10)]
 
In [3]: from heapq import nsmallest
 
In [4]: %timeit nsmallest(2, a)[-1]
4.69 µs ± 79.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
 
In [5]: %timeit sorted(a)[1]
618 ns ± 3.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
 
In [6]: print(*a)
-765 -671 -730 -578 649 -620 -631 512 -764 839

In [7]: a = [randrange(-1_000, 1_000) for _ in range(100_000)]
 
In [8]: %timeit nsmallest(2, a)[-1]
7.3 ms ± 171 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
 
In [9]: %timeit sorted(a)[1]
28.1 ms ± 176 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

a = [1, 1, 1, 2, 3, 4, 5, 6]

a = sorted([3, 4, 1, 1, 1, 2, 5, 6])
for i in a[1:]:
    if i != a[0]:
        print(i)
        break

In [1]: from random import randrange
 
In [2]: a = [randrange(-1_000, 1_000) for _ in range(10)]
 
In [3]: from heapq import nsmallest
 
In [4]: %timeit nsmallest(2, a)[-1]
3.17 µs ± 61.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
 
In [5]: %timeit sorted(a)[1]
347 ns ± 4.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
 
In [6]: print(*a)
523 -196 -456 989 -534 -713 506 -48 -580 853
 
In [7]: a = [randrange(-1_000, 1_000) for _ in range(100_000)]
 
In [8]: %timeit nsmallest(2, a)[-1]
4.41 ms ± 171 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
 
In [9]: %timeit sorted(a)[1]
26.9 ms ± 451 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)