# 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

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; ` `} ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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. ` `?> ` |

*chevron_right*

*filter_none*

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Find the highest occurring digit in prime numbers in a range
- Count of Numbers in Range where first digit is equal to last digit of the number
- Number of occurrences of 2 as a digit in numbers from 0 to n
- Find the remainder when First digit of a number is divided by its Last digit
- Count the number of occurrences of a particular digit in a number
- Count numbers with unit digit k in given range
- Biggest integer which has maximum digit sum in range from 1 to n
- Count of Numbers in a Range where digit d occurs exactly K times
- Integers from the range that are composed of a single distinct digit
- Count occurrences of a prime number in the prime factorization of every element from the given range
- Find the last digit of given series
- Find the Number which contain the digit d
- Find Last Digit of a^b for Large Numbers
- Find nth number that contains the digit k or divisible by k.
- Program to find last digit of n'th Fibonnaci Number