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

Namachila November 19, 2025

0 1 3 minutes read

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

Time Complexity: O(r)

Auxiliary Space: O(1), since no extra space has been taken.

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