Given a number n, count number of occurrences of digits 0, 2 and 4 including n.

Example1:

n = 10
output: 4

n = 22
output: 11

My Code:

n = 22

def count_digit(n):
    count = 0
    for i in range(n 1):
        if '2' in str(i):
            count  = 1
        if '0' in str(i):
            count  = 1
        if '4' in str(i):
            count  = 1
    return count

count_digit(n)

Code Output: 10

Desired Output: 11

Constraints: 1 <= N <= 10^5

Note: The solution should not cause outOfMemoryException or Time Limit Exceeded for large numbers.

CodePudding user response：

There are numbers in which the desired number is repeated, such as 20 or 22, so instead of adding 1 you must add 2

>>> 
>>> string = ','.join(map(str,range(23)))
>>> 
>>> string
'0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22'
>>> 
>>> string.count('0')   string.count('2')   string.count('4')
11
>>> 



n = 22

def count_digit(n):
    count = 0
    for i in map(str,range(n 1)):

        count =i.count('0')
        count =i.count('2')
        count =i.count('3')
    return count
print(count_digit(n))

CodePudding user response：

You can increment your count like this:

def count_digit(n):
    count = 0
    for i in range(n   1):
        if '2' in str(i):
            count  = str(i).count('2')
        if '0' in str(i):
            count  = str(i).count('0')
        if '4' in str(i):
            count  = str(i).count('4')
    return count

In that way, edge cases like 22, 44, and so on are covered!