Algorithm


Problem Name: 902. Numbers At Most N Given Digit Set

Given an array of digits which is sorted in non-decreasing order. You can write numbers using each digits[i] as many times as we want. For example, if digits = ['1','3','5'], we may write numbers such as '13', '551', and '1351315'.

Return the number of positive integers that can be generated that are less than or equal to a given integer n.

 

Example 1:

Input: digits = ["1","3","5","7"], n = 100
Output: 20
Explanation: 
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.

Example 2:

Input: digits = ["1","4","9"], n = 1000000000
Output: 29523
Explanation: 
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits array.

Example 3:

Input: digits = ["7"], n = 8
Output: 1

 

Constraints:

  • 1 <= digits.length <= 9
  • digits[i].length == 1
  • digits[i] is a digit from '1' to '9'.
  • All the values in digits are unique.
  • digits is sorted in non-decreasing order.
  • 1 <= n <= 109
 

Code Examples

#1 Code Example with Javascript Programming

Code - Javascript Programming


const atMostNGivenDigitSet = function(digits, n) {
  const str = '' + n, { pow } = Math
  const len = str.length, dsize = digits.length
  let res = 0
  
  for(let i = 1; i  <  len; i++) {
    res += pow(dsize, i)
  }
  
  for(let i = 0; i  <  len; i++) {
    let sameNum = false
    for(const d of digits) {
      if(+d < +str[i]) {
        res += pow(dsize, len - i - 1)
      } else if(+d === +str[i]) sameNum = true
    }
    if(sameNum === false> return res
  }
  
  return res + 1
};
Copy The Code & Try With Live Editor

Input

x
+
cmd
digits = ["1","3","5","7"], n = 100

Output

x
+
cmd
20

#2 Code Example with Python Programming

Code - Python Programming


class Solution:
    def atMostNGivenDigitSet(self, D, N):
        def less(c):
            return len([char for char in D if char < c])
        d, cnt, l = len(D), 0, len(str(N))
        # For numbers which have less digits than N, simply len(D) ** digits_length different numbers can be created
        for i in range(1, l):
            cnt += d ** i
        """
        We should also consider edge cases where previous digits match with related digits in N. In this case, we can make a number with             previous digits + (digits less than N[i]) + D ** remaining length
        """
        for i, c in enumerate(str(N)):
            cnt += less(c) * (d ** (l - i - 1))
            if c not in D: break
            if i == l - 1: cnt += 1
        return cnt
Copy The Code & Try With Live Editor

Input

x
+
cmd
digits = ["1","3","5","7"], n = 100

Output

x
+
cmd
20
Advertisements

Demonstration


Previous
#901 Leetcode Online Stock Span Solution in C, C++, Java, JavaScript, Python, C# Leetcode
Next
#903 Leetcode Valid Permutations for DI Sequence Solution in C, C++, Java, JavaScript, Python, C# Leetcode