We are given two integers n and d, we need to count all n digit numbers that do not have digit d.
Input : n = 2, d = 7 Output : 72 All two digit numbers that don't have 7 as digit are 10, 11, 12, 13, 14, 15, 16, 18, ..... Input : n = 3, d = 9 Output : 648
A simple solution is to traverse through all d digit numbers. For every number, check if it has x as digit or not.
Efficient approach In this method, we check first if excluding digit d is zero or non-zero. If it is zero then, we have 9 numbers (1 to 9) as first number otherwise we have 8 numbers(excluding x digit and 0). Then for all other digits, we have 9 choices i.e (0 to 9 excluding d digit). We simple call digitNumber function with n-1 digits as first number we already find it can be 8 or 9 and simple multiply it. For other numbers we check if digits are odd or even if it is odd we call digitNumber function with (n-1)/2 digits and multiply it by 9 oterwise we call digitNumber function with n/2 digits and store them in result and take result square.
Number from 1 to 7 excluding digit 9. digits multiple number 1 8 8 2 8*9 72 3 8*9*9 648 4 8*9*9*9 5832 5 8*9*9*9*9 52488 6 8*9*9*9*9*9 472392 7 8*9*9*9*9*9*9 4251528
As we can see, in each step we are half the number of digits. Suppose we have 7 digits in this we call function from main with 6(7-1) digits. when we half the digits we left with 3(6/2) digits. Because of this we have to multiply result due to 3 digits with itself to get result for 6 digits. Similarly for 3 digits we have odd digits, we have odd digits. We find result with 1 ((3-1)/2) digits and find result square and multiply it with 9, because we find result for d-1 digits.
Time Complexity : O(log n).
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