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 <= 9digits[i].length == 1digits[i]is a digit from'1'to'9'.- All the values in
digitsare unique. digitsis 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
};
Input
digits = ["1","3","5","7"], n = 100
Output
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
Input
digits = ["1","3","5","7"], n = 100
Output
20