Count the Digit Python Codewars

Take an integer n (n >= 0) and a digit d (0 <= d <= 9) as an integer.

Square all numbers k (0 <= k <= n) between 0 and n.

Count the numbers of digits d used in the writing of all the k**2.

Call nb_dig (or nbDig or ...) the function taking n and d as parameters and returning this count.

Examples:

n = 10, d = 1 
the k*k are 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
We are using the digit 1 in: 1, 16, 81, 100. The total count is then 4.

nb_dig(25, 1) returns 11 since
the k*k that contain the digit 1 are:
1, 16, 81, 100, 121, 144, 169, 196, 361, 441.
So there are 11 digits 1 for the squares of numbers between 0 and 25.

Note that 121 has twice the digit 1.

def nb_dig(n, d):
    count = 0
    for i in range(n+1):
        sq = i*i
        for i in str(sq):
            if i == str(d):
                count += 1
    return count

print(nb_dig(5750,0))

def nb_dig(n, d):
    return sum(str(i*i).count(str(d)) for i in range(n+1))

Training on Count the Digit | Codewars

Related Posts