Given two positive numbers **M** and **N**, the task is to find the count of all numbers having **M** as the last digit from the range **[1, N]**.

**Examples:**

Input:M = 5, N = 15Output:2Explanation:

Only 2 numbers(5 and 15) from the range [1, 15] ends with the digit ‘5’.Input:M = 1, N = 100Output:10

**Naive Approach: **The simplest approach is to iterate over the range **1** to **N** and check if the last digit is equal to **M** or not. If found to be true, then increment the count. Finally, print the count obtained. **Time complexity: **O(N) **Auxiliary Space:** O(1)**Efficient Approach: **To optimize the above approach, the idea is based on the fact that the count of numbers ending with every digit will be the same until the largest multiple of 10 which is less than **N** (say **x**). Therefore, its count will be (N / 10). Now, the task is reduced to compute the count of numbers ending with **M** which are between **x** and **N.**

Below are the steps:

- Initialize a variable to store the total count, say
**total_count**. - Add
**(N / 10)**to the total count. - Compute
**x**to store the largest multiple of 10 which is less than**N**using the formula:

x = (N / 10) * 10

- Now, calculate the count of numbers ending with
**M**lying in between**x**and**N**. - Add this count to the
**total_count**. Print the final value of**total_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` `// ending with M` `int` `getCount(` `int` `N, ` `int` `M)` `{` ` ` `// Stores the count of` ` ` `// numbers required` ` ` `int` `total_count = 0;` ` ` `// Calculate count upto` ` ` `// nearest power of 10` ` ` `total_count += (N / 10);` ` ` `// Computing the value of x` ` ` `int` `x = (N / 10) * 10;` ` ` `// Adding the count of numbers` ` ` `// ending at M from x to N` ` ` `if` `((N - x) >= M)` ` ` `{` ` ` `total_count = total_count + 1;` ` ` `}` ` ` `return` `total_count;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 100, M = 1;` ` ` `// Function Call` ` ` `cout << getCount(N, M);` ` ` `return` `0;` `}` |

## Java

`// Java program to implement` `// the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to count the numbers` `// ending with M` `static` `int` `getCount(` `int` `N, ` `int` `M)` `{` ` ` `// Stores the count of` ` ` `// numbers required` ` ` `int` `total_count = ` `0` `;` ` ` `// Calculate count upto` ` ` `// nearest power of 10` ` ` `total_count += (N / ` `10` `);` ` ` `// Computing the value of x` ` ` `int` `x = (N / ` `10` `) * ` `10` `;` ` ` `// Adding the count of numbers` ` ` `// ending at M from x to N` ` ` `if` `((N - x) >= M)` ` ` `{` ` ` `total_count = total_count + ` `1` `;` ` ` `}` ` ` `return` `total_count;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `100` `, M = ` `1` `;` ` ` `// Function call` ` ` `System.out.print(getCount(N, M));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python3 Program to implement` `# the above approach` `# Function to count the numbers` `# ending with M` `def` `getCount(N, M):` ` ` `# Stores the count of` ` ` `# numbers required` ` ` `total_count ` `=` `0` ` ` `# Calculate count upto` ` ` `# nearest power of 10` ` ` `total_count ` `+` `=` `N ` `/` `/` `10` ` ` `# Computing the value of x` ` ` `x ` `=` `(N ` `/` `/` `10` `) ` `*` `10` ` ` `# Adding the count of numbers` ` ` `# ending at M from x to N` ` ` `if` `((N ` `-` `x) >` `=` `M):` ` ` `total_count ` `=` `total_count ` `+` `1` ` ` `return` `total_count` `# Driver Code` `N ` `=` `100` `M ` `=` `1` `# Function call` `print` `(getCount(N, M))` `# This code is contributed by Shivam Singh` |

## C#

`// C# program to implement` `// the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to count the` ` ` `// numbers ending with M` ` ` `static` `int` `getCount(` `int` `N, ` `int` `M)` ` ` `{` ` ` `// Stores the count of` ` ` `// numbers required` ` ` `int` `total_count = 0;` ` ` `// Calculate count upto` ` ` `// nearest power of 10` ` ` `total_count += (N / 10);` ` ` `// Computing the value of x` ` ` `int` `x = (N / 10) * 10;` ` ` `// Adding the count of numbers` ` ` `// ending at M from x to N` ` ` `if` `((N - x) >= M)` ` ` `{` ` ` `total_count = total_count + 1;` ` ` `}` ` ` `return` `total_count;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `int` `N = 100, M = 1;` ` ` `// Function call` ` ` `Console.Write(getCount(N, M));` ` ` `}` `}` `// This code is contributed by shikhasingrajput` |

## Javascript

`<script>` `// Javascript Program to implement` `// the above approach` `// Function to count the numbers` `// ending with M` `function` `getCount(N, M){` ` ` `// Stores the count of` ` ` `// numbers required` ` ` `let total_count = 0` ` ` `// Calculate count upto` ` ` `// nearest power of 10` ` ` `total_count += Math.floor(N / 10)` ` ` `// Computing the value of x` ` ` `let x = Math.floor(N / 10) * 10` ` ` `// Adding the count of numbers` ` ` `// ending at M from x to N` ` ` `if` `((N - x) >= M){` ` ` `total_count = total_count + 1` ` ` `}` ` ` `return` `total_count` `}` `// Driver Code` `let N = 100` `let M = 1` `// Function call` `document.write(getCount(N, M))` `// This code is contributed by Saurabh Jaiswal` `</script>` |

**Output:**

10

**Time complexity: **O(1) **Auxiliary Space:** O(1)

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