Find the occurrences of digit d in the range [0..n]

Given a number n and a digit d, count all occurrences of d in range from 0 to n.

Examples:

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.
Step 2:
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.

C++

 // C++ program to count appearances of // a digit 'd' in range from [0..n] #include using namespace std;    int getOccurence(int n, int d) {     int result = 0; // Initialize result        // Count appearances in numbers starting     // from d.     int itr = d;     while (itr <= n)     {         // When the last digit is equal to d         if (itr%10 == d)             result++;            // When the first digit is equal to d then         if (itr != 0 && itr/10 == d)         {             // increment result as well as number             result++;             itr++;         }            // In case of reverse of number such as 12 and 21         else if (itr/10 == d-1)             itr = itr + (10 - d);         else             itr = itr+10;     }     return result; }    // Driver code int main(void) {     int n = 11, d = 1;     cout << getOccurence(n, d);     return 0; }

Java

 // java program to count appearances of // a digit 'd' in range from [0..n] import java.*;    public class GFG {            static int getOccurence(int n, int d)     {                    // Initialize result         int result = 0;                // Count appearances in numbers         // starting from d.         int itr = d;                    while (itr <= n)         {                            // When the last digit is             // equal to d             if (itr % 10 == d)                 result++;                    // When the first digit is             // equal to d then             if (itr != 0 && itr/10 == d)             {                                    // increment result as                 // well as number                 result++;                 itr++;             }                    // In case of reverse of number              // such as 12 and 21             else if (itr/10 == d-1)                 itr = itr + (10 - d);             else                 itr = itr + 10;         }                    return result;     }               // Driver code     public static void main (String[] args)     {         int n = 11, d = 1;                System.out.println(getOccurence(n, d) );     } }    // This code is contributed by Sam007.

Python3

 # Python3 program to count appearances  # of a digit 'd' in range from [0..n]  import math; def getOccurence(n, d):             # Initialize result      result = 0;         # Count appearances in numbers      # starting from d.      itr = d;      while(itr <= n):                     # When the last digit is equal to d          if (itr % 10 == d):              result += 1;             # When the first digit is equal to d then          if (itr != 0 and math.floor(itr / 10) == d):                            # increment result as well as number              result += 1;              itr += 1;             # In case of reverse of number          # such as 12 and 21          elif (math.floor(itr / 10) == d - 1):              itr = itr + (10 - d);          else:             itr = itr + 10;         return result;     # Driver code  n = 11;  d = 1;  print(getOccurence(n, d));     # This code is contributed by mits

C#

 // C# program to count appearances of // a digit 'd' in range from [0..n] using System;            public class GFG {            static int getOccurence(int n, int d)     {                    // Initialize result         int result = 0;                // Count appearances in numbers         // starting from d.         int itr = d;         while (itr <= n)         {                            // When the last digit is             // equal to d             if (itr % 10 == d)                 result++;                    // When the first digit is             // equal to d then             if (itr != 0 && itr/10 == d)             {                                    // increment result as                  // well as number                 result++;                 itr++;             }                    // In case of reverse of number             // such as 12 and 21             else if (itr/10 == d-1)                 itr = itr + (10 - d);             else                 itr = itr + 10;         }                    return result;     }            // Driver code     public static void Main()      {         int n = 11, d = 1;                    Console.Write(getOccurence(n, d));     } }    // This code is contributed by Sam007.

PHP



Output:

4

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