# Count of integers in given range having their last K digits are equal

Given a range from **L** to **R** and an integer **K**, the task is to count the number of integers in the given range such that their last **K** digits are equal.

**Example: **

Input:L = 49, R = 101, K=2Output:6Explanation:There are 6 possible integers t.e., 55, 66, 77, 88, 99 and 100 such that their last K(i.e., 2) digits are equal.

Input:L = 10, R = 20, K=2Output:1

**Efficient Approach: **It can be observed that the count of integers **i** in the range **1** to **X** having the last **K** digits equal to an integer** z** (i.e., i % 10^{K} = z) are **(X – z)/10 ^{K} + 1**. Using this observation the above problem can be solved using the below steps:

- Suppose
**intCount(X, K)**represents the count of integers from**1**to**X**having the last**K**digits as equal. - To calculate
**intCount(X, K)**, iterate over all possibilities of**z**having**K**digits (i.e.,*{00…0, 11…1, 22…2, 33…3, 44…4, …., 99…9 }*) in the formula (X – z)/10^{K}+1 and maintain their sum which is the required value. - Therefore, the count of integers in range
**L**to**R**can be obtained as**intCount(R, K) – intCount(L-1, K)**.

Below is the implementation of the above approach:

## C++

`// C++ Program of the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the count of` `// integers from 1 to X having the` `// last K digits as equal` `int` `intCount(` `int` `X, ` `int` `K)` `{` ` ` `// Stores the total count of integers` ` ` `int` `ans = 0;` ` ` `// Loop to iterate over all` ` ` `// possible values of z` ` ` `for` `(` `int` `z = 0; z < ` `pow` `(10, K);` ` ` `z += (` `pow` `(10, K) - 1) / 9) {` ` ` `// Terminate the loop when z > X` ` ` `if` `(z > X)` ` ` `break` `;` ` ` `// Add count of integers with` ` ` `// last K digits equal to z` ` ` `ans += ((X - z) / ` `pow` `(10, K) + 1);` ` ` `}` ` ` `// Return count` ` ` `return` `ans;` `}` `// Function to return the count of` `// integers from L to R having the` `// last K digits as equal` `int` `intCountInRange(` `int` `L, ` `int` `R, ` `int` `K)` `{` ` ` `return` `(intCount(R, K)` ` ` `- intCount(L - 1, K));` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `L = 49;` ` ` `int` `R = 101;` ` ` `int` `K = 2;` ` ` `// Function Call` ` ` `cout << intCountInRange(L, R, K);` ` ` `return` `0;` `}` |

## Java

`// Java Program of the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to return the count of` `// integers from 1 to X having the` `// last K digits as equal` `static` `int` `intCount(` `int` `X, ` `int` `K)` `{` ` ` `// Stores the total count of integers` ` ` `int` `ans = ` `0` `;` ` ` `// Loop to iterate over all` ` ` `// possible values of z` ` ` `for` `(` `int` `z = ` `0` `; z < Math.pow(` `10` `, K);` ` ` `z += (Math.pow(` `10` `, K) - ` `1` `) / ` `9` `) {` ` ` `// Terminate the loop when z > X` ` ` `if` `(z > X)` ` ` `break` `;` ` ` `// Add count of integers with` ` ` `// last K digits equal to z` ` ` `ans += ((X - z) / Math.pow(` `10` `, K) + ` `1` `);` ` ` `}` ` ` `// Return count` ` ` `return` `ans;` `}` `// Function to return the count of` `// integers from L to R having the` `// last K digits as equal` `static` `int` `intCountInRange(` `int` `L, ` `int` `R, ` `int` `K)` `{` ` ` `return` `(intCount(R, K)` ` ` `- intCount(L - ` `1` `, K));` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `L = ` `49` `;` ` ` `int` `R = ` `101` `;` ` ` `int` `K = ` `2` `;` ` ` `// Function Call` ` ` `System.out.print(intCountInRange(L, R, K));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python3 program for the above approach` `# Function to return the count of` `# integers from 1 to X having the` `# last K digits as equal` `def` `intCount(X, K):` ` ` ` ` `# Stores the total count of integers` ` ` `ans ` `=` `0` ` ` `# Loop to iterate over all` ` ` `# possible values of z` ` ` `for` `z ` `in` `range` `(` `0` `, ` `int` `(` `pow` `(` `10` `, K)),` ` ` `int` `((` `pow` `(` `10` `, K) ` `-` `1` `) ` `/` `9` `)):` ` ` `# Terminate the loop when z > X` ` ` `if` `(z > X):` ` ` `break` ` ` `# Add count of integers with` ` ` `# last K digits equal to z` ` ` `ans ` `+` `=` `int` `((X ` `-` `z) ` `/` `int` `(` `pow` `(` `10` `, K)) ` `+` `1` `)` ` ` ` ` `# Return count` ` ` `return` `ans` `# Function to return the count of` `# integers from L to R having the` `# last K digits as equal` `def` `intCountInRange(L, R, K):` ` ` ` ` `return` `(intCount(R, K) ` `-` `intCount(L ` `-` `1` `, K))` `# Driver Code` `L ` `=` `49` `R ` `=` `101` `K ` `=` `2` `# Function Call` `print` `(intCountInRange(L, R, K))` `# This code is contributed by sanjoy_62` |

## C#

`// C# Program of the above approach` `using` `System;` `class` `GFG{` `// Function to return the count of` `// integers from 1 to X having the` `// last K digits as equal` `static` `int` `intCount(` `int` `X, ` `int` `K)` `{` ` ` `// Stores the total count of integers` ` ` `int` `ans = 0;` ` ` `// Loop to iterate over all` ` ` `// possible values of z` ` ` `for` `(` `int` `z = 0; z < Math.Pow(10, K);` ` ` `z += ((` `int` `)Math.Pow(10, K) - 1) / 9) {` ` ` `// Terminate the loop when z > X` ` ` `if` `(z > X)` ` ` `break` `;` ` ` `// Add count of integers with` ` ` `// last K digits equal to z` ` ` `ans += ((X - z) / (` `int` `)Math.Pow(10, K) + 1);` ` ` `}` ` ` `// Return count` ` ` `return` `ans;` `}` `// Function to return the count of` `// integers from L to R having the` `// last K digits as equal` `static` `int` `intCountInRange(` `int` `L, ` `int` `R, ` `int` `K)` `{` ` ` `return` `(intCount(R, K)` ` ` `- intCount(L - 1, K));` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `L = 49;` ` ` `int` `R = 101;` ` ` `int` `K = 2;` ` ` `// Function Call` ` ` `Console.Write(intCountInRange(L, R, K));` `}` `}` `// This code is contributed by shivanisinghss2110` |

## Javascript

`<script>` `// JavaScript program for the above approach` `// Function to return the count of` `// integers from 1 to X having the` `// last K digits as equal` `function` `intCount(X, K)` `{` ` ` ` ` `// Stores the total count of integers` ` ` `let ans = 0;` ` ` `// Loop to iterate over all` ` ` `// possible values of z` ` ` `for` `(let z = 0; z < Math.pow(10, K);` ` ` `z += Math.floor((Math.pow(10, K) - 1) / 9))` ` ` `{` ` ` ` ` `// Terminate the loop when z > X` ` ` `if` `(z > X)` ` ` `break` `;` ` ` `// Add count of integers with` ` ` `// last K digits equal to z` ` ` `ans += Math.floor(((X - z) /` ` ` `Math.pow(10, K) + 1));` ` ` `}` ` ` `// Return count` ` ` `return` `ans;` `}` `// Function to return the count of` `// integers from L to R having the` `// last K digits as equal` `function` `intCountInRange(L, R, K)` `{` ` ` `return` `(intCount(R, K) -` ` ` `intCount(L - 1, K));` `}` `// Driver Code` `let L = 49;` `let R = 101;` `let K = 2;` `// Function Call` `document.write(intCountInRange(L, R, K));` `// This code is contributed by Potta Lokesh` `</script>` |

**Output**

6

**Time Complexity:*** O(log K)***Space Complexity: ***O(1)*