Given a number n and a digit d, count all occurrences of d in range from 0 to n.
Input : n = 25 d = 2 Output : 9 The occurrences are 2, 12, 20, 21 22 (Two occurrences), 23, 24, 25 Input : n = 25 d = 3 Output :3 The occurrences are 3, 13, 23 Input : n = 32 d = 3 Output : 6 The occurrences are 3, 13, 23, 30, 31, 32
The first occurrence of d cannot be before number d. So we start to iterate from d and do following check again and again. We jump the number by 10 most of the time (in step 2) except for the cases mentioned in steps 2.a and 2.b.
Step 1: Check whether the last digit of the number is equal to the d, if it is then increment the count.
a) If the number is completely divisible by 10 then increment both the count and number (For example if we reach the number 30 which is completely divisible by 10 and d=3, then we have to count all numbers from 31-39 that’s why we increment count by 1 and number by 1)
b) else if the first digit of the number is one less than d then it means that we have come to the row where we have to increment the number by 10 and subtract the d from it. For example if we reach 23 for d=3 then we increment the number by 23+10-3 = 30)
c) else increment the number by 10 only. (For example: d=3, itr=3, then increment by 10 i.e.13, 23)
Step 3:- Return the count.
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