Given a positive integer **N** and an integer **D** representing a digit, the task is to count the numbers in the range**[1, N]** such that at least one digit in octal representation of the number is **d**.

**Examples:**

Input:N = 20, D = 7Output:2Explanation:

The numbers in the range [1, 20] having digit 7 in their octal representation are 7 and 15.

Therefore, the required output is 2.

Input:N = 40, D = 5Output:6Explanation:

The numbers in the range [1, 40] having digit 5 in their octal representation are 5, 13, 21, 29, 37, and 40.

Therefore, the required output is 6

**Approach:** Follow the steps below to solve the problem:

- Iterate over the range
**[1, N]**. For every**i**number check if at least one digit in octal representation of the number is^{th}**d**or not. If found to be true, then increment the count. - Finally, print the count obtained.

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to count the numbers in given ` `// range whose octal representation` `// contains atleast digit, d` `void` `countNumbers(` `int` `n, ` `int` `d)` `{` ` ` `// Store count of numbers up to n ` ` ` `// whose octal representation` ` ` `// contains digit d` ` ` `int` `total = 0;` ` ` `// Iterate over the range[1, n]` ` ` `for` `(` `int` `i = 1; i <= n; i++) {` ` ` `int` `x = i;` ` ` `// Calculate digit of i in ` ` ` `// octal representation` ` ` `while` `(x > 0) {` ` ` `// Check if octal representation` ` ` `// of x contains digit d` ` ` `if` `(x % 8 == d) {` ` ` `// Update total` ` ` `total++;` ` ` `break` `;` ` ` `}` ` ` `// Update x` ` ` `x = x / 8;` ` ` `}` ` ` `}` ` ` `// Print the answer` ` ` `cout << total;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given N and D` ` ` `int` `n = 20, d = 7;` ` ` `// Counts and prints numbers` `// up to N having D as a digit` `// in its octal representation` ` ` `countNumbers(n, d);` ` ` `return` `0;` `}` |

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## Java

`// Java program to implement ` `// the above approach ` `import` `java.io.*;` `import` `java.util.*;` ` ` `class` `GFG{` ` ` `// Function to count the numbers in given ` `// range whose octal representation` `// contains atleast digit, d` `static` `void` `countNumbers(` `int` `n, ` `int` `d)` `{` ` ` ` ` `// Store count of numbers up to n ` ` ` `// whose octal representation` ` ` `// contains digit d` ` ` `int` `total = ` `0` `;` ` ` ` ` `// Iterate over the range[1, n]` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i++)` ` ` `{` ` ` `int` `x = i;` ` ` ` ` `// Calculate digit of i in ` ` ` `// octal representation` ` ` `while` `(x > ` `0` `) ` ` ` `{` ` ` ` ` `// Check if octal representation` ` ` `// of x contains digit d` ` ` `if` `(x % ` `8` `== d) ` ` ` `{` ` ` ` ` `// Update total` ` ` `total++;` ` ` `break` `;` ` ` `}` ` ` ` ` `// Update x` ` ` `x = x / ` `8` `;` ` ` `}` ` ` `}` ` ` ` ` `// Print the answer` ` ` `System.out.println(total);` `}` ` ` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Given N and D` ` ` `int` `n = ` `20` `, d = ` `7` `;` ` ` ` ` `// Counts and prints numbers` ` ` `// up to N having D as a digit` ` ` `// in its octal representation` ` ` `countNumbers(n, d);` `}` `}` `// This code is contributed by sanjoy_62` |

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## Python3

`# Python3 program to implement` `# the above approach` `# Function to count the numbers in given` `# range whose octal representation` `# contains atleast digit, d` `def` `countNumbers(n, d):` ` ` ` ` `# Store count of numbers up to n` ` ` `# whose octal representation` ` ` `# contains digit d` ` ` `total ` `=` `0` ` ` `# Iterate over the range[1, n]` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `):` ` ` `x ` `=` `i` ` ` `# Calculate digit of i in` ` ` `# octal representation` ` ` `while` `(x > ` `0` `):` ` ` `# Check if octal representation` ` ` `# of x contains digit d` ` ` `if` `(x ` `%` `8` `=` `=` `d):` ` ` `# Update total` ` ` `total ` `+` `=` `1` ` ` `break` ` ` `# Update x` ` ` `x ` `=` `x ` `/` `/` `8` ` ` `# Prthe answer` ` ` `print` `(total)` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Given N and D` ` ` `n , d ` `=` `20` `, ` `7` ` ` `# Counts and prints numbers` ` ` `# up to N having D as a digit` ` ` `# in its octal representation` ` ` `countNumbers(n, d)` `# This code is contributed by mohit kumr 29` |

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## C#

`// C# program to implement` `// the above approach` `using` `System;` ` ` `class` `GFG{` ` ` `// Function to count the numbers in given ` `// range whose octal representation` `// contains atleast digit, d` `static` `void` `countNumbers(` `int` `n, ` `int` `d)` `{` ` ` ` ` `// Store count of numbers up to n ` ` ` `// whose octal representation` ` ` `// contains digit d` ` ` `int` `total = 0;` ` ` ` ` `// Iterate over the range[1, n]` ` ` `for` `(` `int` `i = 1; i <= n; i++)` ` ` `{` ` ` `int` `x = i;` ` ` ` ` `// Calculate digit of i in ` ` ` `// octal representation` ` ` `while` `(x > 0) ` ` ` `{` ` ` ` ` `// Check if octal representation` ` ` `// of x contains digit d` ` ` `if` `(x % 8 == d) ` ` ` `{` ` ` ` ` `// Update total` ` ` `total++;` ` ` `break` `;` ` ` `}` ` ` ` ` `// Update x` ` ` `x = x / 8;` ` ` `}` ` ` `}` ` ` ` ` `// Print the answer` ` ` `Console.WriteLine(total);` `} ` ` ` `// Driver Code` `static` `void` `Main()` `{` ` ` ` ` `// Given N and D` ` ` `int` `n = 20, d = 7;` ` ` ` ` `// Counts and prints numbers` ` ` `// up to N having D as a digit` ` ` `// in its octal representation` ` ` `countNumbers(n, d); ` `}` `}` `// This code is contributed by susmitakundugoaldanga` |

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**Output:**

2

**Time Complexity: **O(N * log_{8}N)**Auxiliary Space:** O(1)

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