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 k bool 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 k 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 int 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 k def 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 k def 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 code l = 20 r = 35 k = 45 print (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 k function 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 k function 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 code let 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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