Given three positive integers L, R and K.The task is to find the count of all the numbers from the range [L, R] that contains at least one digit which divides the number K.
Examples:
Input: L = 5, R = 11, K = 10
Output: 3
5, 10 and 11 are only such numbers.Input: L = 32, R = 38, K = 13
Output: 0
Approach: Initialise count = 0 and for every element in the range [L, R], check if it contains at least one digit that divides K. If yes then increment the count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if num// contains at least one digit// that divides kbool digitDividesK(int num, int k){ while (num) { // Get the last digit int d = num % 10; // If the digit is non-zero // and it divides k if (d != 0 and k % d == 0) return true; // Remove the last digit num = num / 10; } // There is no digit in num // that divides k return false;}// Function to return the required// count of elements from the given// range which contain at least one// digit that divides kint findCount(int l, int r, int k){ // To store the result int count = 0; // For every number from the range for (int i = l; i <= r; i++) { // If any digit of the current // number divides k if (digitDividesK(i, k)) count++; } return count;}// Driver codeint main(){ int l = 20, r = 35; int k = 45; cout << findCount(l, r, k); return 0;} |
Java
// Java implementation of the approach class GFG{ // Function that returns true if num // contains at least one digit // that divides k static boolean digitDividesK(int num, int k) { while (num != 0) { // Get the last digit int d = num % 10; // If the digit is non-zero // and it divides k if (d != 0 && k % d == 0) return true; // Remove the last digit num = num / 10; } // There is no digit in num // that divides k return false; } // Function to return the required // count of elements from the given // range which contain at least one // digit that divides k static int findCount(int l, int r, int k) { // To store the result int count = 0; // For every number from the range for (int i = l; i <= r; i++) { // If any digit of the current // number divides k if (digitDividesK(i, k)) count++; } return count; } // Driver code public static void main(String []args) { int l = 20, r = 35; int k = 45; System.out.println(findCount(l, r, k)); } }// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach# Function that returns true if num# contains at least one digit# that divides kdef digitDividesK(num, k): while (num): # Get the last digit d = num % 10 # If the digit is non-zero # and it divides k if (d != 0 and k % d == 0): return True # Remove the last digit num = num // 10 # There is no digit in num # that divides k return False# Function to return the required# count of elements from the given# range which contain at least one# digit that divides kdef findCount(l, r, k): # To store the result count = 0 # For every number from the range for i in range(l, r + 1): # If any digit of the current # number divides k if (digitDividesK(i, k)): count += 1 return count# Driver codel = 20r = 35k = 45print(findCount(l, r, k))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System;class GFG{ // Function that returns true if num // contains at least one digit // that divides k static bool digitDividesK(int num, int k) { while (num != 0) { // Get the last digit int d = num % 10; // If the digit is non-zero // and it divides k if (d != 0 && k % d == 0) return true; // Remove the last digit num = num / 10; } // There is no digit in num // that divides k return false; } // Function to return the required // count of elements from the given // range which contain at least one // digit that divides k static int findCount(int l, int r, int k) { // To store the result int count = 0; // For every number from the range for (int i = l; i <= r; i++) { // If any digit of the current // number divides k if (digitDividesK(i, k)) count++; } return count; } // Driver code public static void Main() { int l = 20, r = 35; int k = 45; Console.WriteLine(findCount(l, r, k)); } }// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the approach// Function that returns true if num// contains at least one digit// that divides kfunction digitDividesK(num, k){ while (num) { // Get the last digit let d = num % 10; // If the digit is non-zero // and it divides k if (d != 0 && k % d == 0) return true; // Remove the last digit num = parseInt(num / 10); } // There is no digit in num // that divides k return false;}// Function to return the required// count of elements from the given// range which contain at least one// digit that divides kfunction findCount(l, r, k){ // To store the result let count = 0; // For every number from the range for(let i = l; i <= r; i++) { // If any digit of the current // number divides k if (digitDividesK(i, k)) count++; } return count;}// Driver codelet l = 20, r = 35;let k = 45;document.write(findCount(l, r, k));// This code is contributed by souravmahato348</script> |
Output:
10
Time Complexity: O((r-l)*(log10(num)))
Auxiliary Space: O(1)
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