loop finding "bouncy" numbers

Question

Consider the following definitions of positive numbers:

A number is nondecreasing if its digits never get smaller as you go from left to right. For example, 12345 and 3388 are nondecreasing.

A number is nonincreasing if its digits never larger as you go from left to right. For example, 987542 and 881 are nonincreasing.

A number is bouncy if it is neither nondecreasing nor nonincreasing. For example, 12134 and 98462 are bouncy. Write a Python function bouncy that consumes a positive natural number (called n) and produces the percentage of numbers between 1 and n, inclusive, which are bouncy. The result should be produced as a natural number between 0 and 100, inclusive. Use round to convert the floating point percentage to an integer.

def bouncy(input):
list1 = [0 for i in range(input)]
list1[0] = 0
for x in range(1, input-1):
    if x < 100:
        list1[x] = list1[x - 1]
    else:
        n=x
        a = [0 for i in range(x)]
        i = 0
        while n > 0:
            a[i]=n % 10
            n/= 10
            i+=1
        flag = 1
        for k in range(1, len(a) - 2):
            if not ((a[k - 1] < a[k] < a[k + 1]) or (a[k - 1] > a[k] > a[k + 1])):
                flag = 0
                break
        if flag == 0:
            list1[x]==list[x-1]+ 1
return list1[input-1]

when i ran my code, it displays builtins.IndexError: list assignment index out of range.

Anyone got an idea?

Answer 1

You don't have to do any of that. Just turn the number into a string. If it's sorted it's nondecreasing, if it's reverse sorted it's nonincreasing, otherwise it's bouncy.

def bouncy(n):
    return round(sum(list(i) not in (sorted(i), sorted(i, reverse=True)) for i in map(str, range(1, n+1)))/n*100)

This map()s each number in the range to a string, then checks whether a list() of that string is not found in a sorted() version of that string (either increasing or decreasing). Then it adds together how many numbers match that, divides by n, multiplies by 100, round()s that, and returns it.