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# Count numbers with unit digit k in given range in C++

We are given an interval [first,last]. The goal is to find the count of numbers that have a unit digit k and lie between range [first,last].

We will do this by traversing from i=first to i=last. For each number i compare its unit digit with the k, if they are the same increment the count.

Let’s understand with examples.

**Input** − first=8 last=40 , k=8

**Output** − Count of numbers with unit digit k − 4

**Explanation** −

Numbers between 8 and 40 with unit digit = 8 8,18, 28, 38

**Input** − first=100 last=200 , k=9

**Output** − Count of numbers with unit digit k − 10

**Explanation** −

Numbers between 100 and 200 with unit digit = 9 109, 119, 129, 139, 149, 159, 169, 179, 189, 199. Total:10

## Approach used in the below program is as follows

We take two integers first and last to define range [first, last].

Function getCount(int fst, int lst, int k) takes range variables and k and returns the count of numbers between fst and lst and have unit digit as k.

Take the initial count as 0.

Using for loop start from i=fst to i=lst, for each i calculate unit digit as ldigit=i%10.

If ldigit==k, increment count.

Return count as result.

## Example

#include <bits/stdc++.h> using namespace std; int getCount(int fst,int lst,int k){ int count=0; for(int i=fst;i<=lst;i++){ int ldigit=i%10; //to get last digit if(ldigit==k) //if both are equal increment count { ++count; } } return count; } int main(){ int first = 5, last = 30; int K=5; cout<<"Numbers with unit digit K in range:"<<getCount(first, last, K); return 0; }

## Output

If we run the above code it will generate the following output −

Numbers with unit digit K in range:3

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