# Count of numbers from range [L, R] that end with any of the given digits

Given a range **[L, R]**, the task is to find the count of numbers from the range whose last digit is either **2**, **3** or **9**.**Examples:**

Input:L = 1, R = 3Output:2

2 and 3 are the only valid numbers.Input:L = 11, R = 33Output:8

**Approach:** Initialize a counter count = 0 and run a loop for every element from the range, if the current element ends with any of the given digits then increment the count.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <iostream>` `using` `namespace` `std;` `// Function to return the count` `// of the required numbers` `int` `countNums(` `int` `l, ` `int` `r)` `{` ` ` `int` `cnt = 0;` ` ` `for` `(` `int` `i = l; i <= r; i++)` ` ` `{` ` ` `// Last digit of the current number` ` ` `int` `lastDigit = (i % 10);` ` ` `// If the last digit is equal to` ` ` `// any of the given digits` ` ` `if` `((lastDigit % 10) == 2 || (lastDigit % 10) == 3` ` ` `|| (lastDigit % 10) == 9)` ` ` `{` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` `return` `cnt;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `l = 11, r = 33;` ` ` `cout << countNums(l, r) ;` `}` ` ` `// This code is contributed by AnkitRai01` |

## Java

`// Java implementation of the approach` `class` `GFG {` ` ` `// Function to return the count` ` ` `// of the required numbers` ` ` `static` `int` `countNums(` `int` `l, ` `int` `r)` ` ` `{` ` ` `int` `cnt = ` `0` `;` ` ` `for` `(` `int` `i = l; i <= r; i++) {` ` ` `// Last digit of the current number` ` ` `int` `lastDigit = (i % ` `10` `);` ` ` `// If the last digit is equal to` ` ` `// any of the given digits` ` ` `if` `((lastDigit % ` `10` `) == ` `2` `|| (lastDigit % ` `10` `) == ` `3` ` ` `|| (lastDigit % ` `10` `) == ` `9` `) {` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` `return` `cnt;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `l = ` `11` `, r = ` `33` `;` ` ` `System.out.print(countNums(l, r));` ` ` `}` `}` |

## Python3

`# Python3 implementation of the approach` `# Function to return the count` `# of the required numbers` `def` `countNums(l, r) :` ` ` `cnt ` `=` `0` `;` ` ` `for` `i ` `in` `range` `(l, r ` `+` `1` `) :` ` ` `# Last digit of the current number` ` ` `lastDigit ` `=` `(i ` `%` `10` `);` ` ` `# If the last digit is equal to` ` ` `# any of the given digits` ` ` `if` `((lastDigit ` `%` `10` `) ` `=` `=` `2` `or` `(lastDigit ` `%` `10` `) ` `=` `=` `3` ` ` `or` `(lastDigit ` `%` `10` `) ` `=` `=` `9` `) :` ` ` ` ` `cnt ` `+` `=` `1` `;` ` ` `return` `cnt;` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `l ` `=` `11` `; r ` `=` `33` `;` ` ` ` ` `print` `(countNums(l, r)) ;` `# This code is contributed by AnkitRai01` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to return the count` ` ` `// of the required numbers` ` ` `static` `int` `countNums(` `int` `l, ` `int` `r)` ` ` `{` ` ` `int` `cnt = 0;` ` ` `for` `(` `int` `i = l; i <= r; i++)` ` ` `{` ` ` `// Last digit of the current number` ` ` `int` `lastDigit = (i % 10);` ` ` `// If the last digit is equal to` ` ` `// any of the given digits` ` ` `if` `((lastDigit % 10) == 2 ||` ` ` `(lastDigit % 10) == 3 ||` ` ` `(lastDigit % 10) == 9)` ` ` `{` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` `return` `cnt;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `l = 11, r = 33;` ` ` `Console.Write(countNums(l, r));` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Javascript

`<script>` ` ` `// JavaScript implementation of the approach` ` ` ` ` `// Function to return the count` ` ` `// of the required numbers` ` ` `function` `countNums(l, r)` ` ` `{` ` ` `let cnt = 0;` ` ` ` ` `for` `(let i = l; i <= r; i++) {` ` ` ` ` `// Last digit of the current number` ` ` `let lastDigit = (i % 10);` ` ` ` ` `// If the last digit is equal to` ` ` `// any of the given digits` ` ` `if` `((lastDigit % 10) == 2 || (lastDigit % 10) == 3` ` ` `|| (lastDigit % 10) == 9) {` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` ` ` `return` `cnt;` ` ` `}` ` ` ` ` `let l = 11, r = 33;` ` ` `document.write(countNums(l, r));` ` ` `</script>` |

**Output:**

8

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