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 numbersint 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 codeint main() { int l = 11, r = 33; cout << countNums(l, r) ;} // This code is contributed by AnkitRai01 |
Java
// Java implementation of the approachclass 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 numbersdef 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 codeif __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:
8
Time Complexity: O(r)
Auxiliary Space: O(1), since no extra space has been taken.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
