Given an integer N, the task is to find the count of N-digit numbers with all distinct digits.
Input: N = 1
1, 2, 3, 4, 5, 6, 7, 8 and 9 are the 1-digit numbers
with all distinct digits.
Input: N = 3
Approach: If N > 10 i.e. there will be atleast one digit which will be repeating hence for such cases the answer will be 0 else for the values of N = 1, 2, 3, …, 9, a series will be formed as 9, 81, 648, 4536, 27216, 136080, 544320, … whose Nth term will be 9 * 9! / (10 – N)!.
Below is the implementation of the above approach:
Time Complexity: O(n)
- Numbers having Unique (or Distinct) digits
- Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
- Count numbers in given range such that sum of even digits is greater than sum of odd digits
- Count of distinct remainders when N is divided by all the numbers from the range [1, N]
- Count numbers with same first and last digits
- Count of numbers from range [L, R] that end with any of the given digits
- Count of numbers from range [L, R] whose sum of digits is Y
- Count different numbers possible using all the digits their frequency times
- Count of n digit numbers whose sum of digits equals to given sum
- Count numbers formed by given two digit with sum having given digits
- Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3
- Count different numbers that can be generated such that there digits sum is equal to 'n'
- Count Numbers with N digits which consists of odd number of 0's
- Count Numbers with N digits which consists of even number of 0’s
- Count of Numbers in Range where the number does not contain more than K non zero digits
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