Discovering all numbers within a specific range that conclude with particular digits

Question

Discovering all numbers within a specific range that conclude with particular digits

Seeking out numbers within a specific range that end with particular digits is my current quest.

Consider the following examples:

Identify all numbers up to 15 that end with 5. The expected result would be 2, as both 5 and 15 fulfill this criteria.
Determine all numbers up to 125 that conclude with 12. This should also yield a result of 2, since both 12 and 112 meet this requirement.

I initially utilized brute-force code:

def findnumber(n, m):
    count = 0
    for i in range(1, m+1):
        if i%10**len(str(n)) == n:
            count += 1
    return count

print(findnumber(5,15))

However, I am now looking to enhance the efficiency of this code.

The valid values are: 1 <= n, m <= 10**9

python

Answer 1

Answer №1

Modified based on the recommendation of Tomerikoo, utilizing the floordiv operator:

def find_number(n, m):
    base = 10 ** len(str(n))
    count = m // base
    if m % base >= n:
        count += 1
    return count

Answer 2

Modified based on the recommendation of Tomerikoo, utilizing the floordiv operator:

def find_number(n, m):
    base = 10 ** len(str(n))
    count = m // base
    if m % base >= n:
        count += 1
    return count

Answer 3

Answer №2

Presented below is a highly effective general algorithm: with a simple assumption that, for instance, n equals 7 and m equals 39, we only need to validate the numbers 7, 17, 27 as they are incremented by 10 which is calculated through offset = 10 ** (total digits of n).

The following code demonstrates this:

def calculate(n, m):
    count   = 0
    number  = n
    length  = len(str(n))
    offset  = 10 ** length
    
    while number <= m:
        count += 1
        number += offset

    return count


>>> calculate(7, 16)
1
>>> calculate(7, 19)
2
>>> calculate(7, 29)
3
>>> calculate(7, 129)
13
>>> calculate(7, 1129)
113

This method proves to be efficient in terms of minimizing computational operations.

Answer 4

Presented below is a highly effective general algorithm: with a simple assumption that, for instance, n equals 7 and m equals 39, we only need to validate the numbers 7, 17, 27 as they are incremented by 10 which is calculated through offset = 10 ** (total digits of n).

The following code demonstrates this:

def calculate(n, m):
    count   = 0
    number  = n
    length  = len(str(n))
    offset  = 10 ** length
    
    while number <= m:
        count += 1
        number += offset

    return count


>>> calculate(7, 16)
1
>>> calculate(7, 19)
2
>>> calculate(7, 29)
3
>>> calculate(7, 129)
13
>>> calculate(7, 1129)
113

This method proves to be efficient in terms of minimizing computational operations.

Answer 5

Answer №3

def find_number(n, m):
  counter = 0
  
  #Finding the first number that ends with n allows for faster computation
  found_the_first = False
  first_num = 1
  while(not found_the_first):
      if first_num % 10 == n:
          found_the_first = True
          break
      first_num += 1
  
  for j in range(first_num, m + 1, 10):
        counter += 1
  return counter

print(find_number(2, 20))

Answer 6

def find_number(n, m):
  counter = 0
  
  #Finding the first number that ends with n allows for faster computation
  found_the_first = False
  first_num = 1
  while(not found_the_first):
      if first_num % 10 == n:
          found_the_first = True
          break
      first_num += 1
  
  for j in range(first_num, m + 1, 10):
        counter += 1
  return counter

print(find_number(2, 20))

Answer 7

Answer №4

This method is much simpler and does not require any iteration.

In a sequence of 10 numbers, there will be one number that ends with the letter m, as shown here:

1 2 3 4 5 6 7 8 9 10 | 11 12 13 14 15

def find_specific_number(n, m):
    n_digits = 0
    n_copy = n
    while n_copy != 0:
        n_digits += 1
        n_copy //= 10

    digits = 10 ** n_digits
    count = m // digits
    if m % digits >= n:
        count += 1
    return count

print(find_specific_number(5, 15))
print(find_specific_number(35, 1400))

# Output:
# 2
# 14

Answer 8

This method is much simpler and does not require any iteration.

In a sequence of 10 numbers, there will be one number that ends with the letter m, as shown here:

1 2 3 4 5 6 7 8 9 10 | 11 12 13 14 15

def find_specific_number(n, m):
    n_digits = 0
    n_copy = n
    while n_copy != 0:
        n_digits += 1
        n_copy //= 10

    digits = 10 ** n_digits
    count = m // digits
    if m % digits >= n:
        count += 1
    return count

print(find_specific_number(5, 15))
print(find_specific_number(35, 1400))

# Output:
# 2
# 14

Discovering all numbers within a specific range that conclude with particular digits

Answer №1

Answer №2

Answer №3

Answer №4

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