Second Largest O(n)

second_largest.py

def second_largest(n):
    first = n[0]
    second = n[1]
    if second > first:
        first, second = second, first
    for i in range(2, len(n)):
        if n[i] > first:
            first, second = n[i], first
            if second > first:
                first, second = second, first
        elif (n[i] > second) and (n[i] < first):
            second = n[i]
    return second
    
    
x = [1, 2, 3, 4, 10, 9, 8, 7, 6, 5]
print(second_largest(x))