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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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