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

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 <bits/stdc++.h> ` `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; ` `} ` |

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## 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. ` |

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## 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 ` |

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## 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. ` |

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## PHP

`<?php ` `// PHP program to count appearances of ` `// a digit 'd' in range from [0..n] ` ` ` `function` `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` `++; ` ` ` ` ` `// When the first digit ` ` ` `// is equal to d then ` ` ` `if` `(` `$itr` `!= 0 && ` `floor` `(` `$itr` `/ 10) == ` `$d` `) ` ` ` `{ ` ` ` ` ` `// increment result as ` ` ` `// well as number ` ` ` `$result` `++; ` ` ` `$itr` `++; ` ` ` `} ` ` ` ` ` `// In case of reverse of ` ` ` `// number such as 12 and 21 ` ` ` `else` `if` `(` `floor` `(` `$itr` `/ 10) == ` `$d` `- 1) ` ` ` `$itr` `= ` `$itr` `+ (10 - ` `$d` `); ` ` ` `else` ` ` `$itr` `= ` `$itr` `+ 10; ` ` ` `} ` ` ` `return` `$result` `; ` `} ` ` ` ` ` `// Driver code ` ` ` `$n` `= 11; ` ` ` `$d` `= 1; ` ` ` `echo` `getOccurence(` `$n` `, ` `$d` `); ` ` ` `// This code is contributed by nitin mittal. ` `?> ` |

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Output:

Output:

4

This article is contributed by **Rakesh Kumar**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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