Given a range of values [L, R] and a value K, the task is to count the numbers in the given range which are divisible by at least K of the digits present in the decimal representation of that number.
Examples:
Input: L = 24, R = 25, K = 2
Output: 1
Explanation:
24 has two digits 2 and 4 and is divisible by both 2 and 4. So this satisfies the given condition.
25 has two digits 2 and 5 and is only divisible by 5. But since K = 2, it doesnot qualifies the mentioned criteria.Input: L = 5, R = 15, K = 1
Output: 11
Method 1: Naive Approach
- For any number between L to R, find the count of it’s digits that divides the number.
- If the count of number in the above step is greater than or equal to K, then include that number to the final count.
- Repeat the above steps for all numbers from L to R and print the final count.
Time Complexity: O(N), where N is the difference between the range [L, R].
Method 2: Efficient Approach
We will use the concept of Digit DP to solve this problem. Below are the observations to solve this problem:
- For all the positive integers(say a), to find the divisibility of the number from digits 2 to 9, the number a can be reduced as stated below to find the divisibility efficiently:
a = k*LCM(2, 3, 4, ..., 9) + q where k is integer and q lies between range [0, lcm(2, 3, ..9)] LCM(2, 3, 4, ..., 9) = 23x32x5x7 = 2520
- After performing a = a modulo 2520, we can find the count of digit from the original number a that divides this modulo.
Below are the steps to do so:
- Store all the digits of the given range and sort the digits in decreasing order.
- Traverse all the digits stored above and generate all the number which are strictly less than the given range of number.
- For generating the number less than the given number, use a variable tight such that:
- The value of tight is 0, denotes that by including that digit will give the number less than the given range.
- The value of tight is 1, denotes that by including that digit, it will give the number greater than the given range. So we can remove all permutations after getting tight value 1 to avoid more number of recursive calls.
- After generating all of the permutations of numbers, Find the number for which the count of digits dividing that number is greater than or equals to K.
- Store the count for each permuted number in dp table to use the result for Overlapping Subproblems.
Below is the implementation of the above approach:
C++
// C++ program to Find the number// of numbers in a range that are// divisible by exactly K of it's// digits#include <bits/stdc++.h>using namespace std;const int LCM = 2520;const int MAXDIG = 10;// To store the results for// overlapping subproblemsint memo[MAXDIG][2][LCM][(1 << 9) + 5];// To store the digits of the// given numbervector<int> dig;int K;// Function to update the dp// tableint dp(int index, int tight, int rem, int mask){ // To find the result int& res = memo[index][tight][rem][mask]; // Return the if result for the // current iteration is calculated if (res != -1) { return res; } res = 0; // If reaches the end of the digits if (index == dig.size()) { int cnt = 0; // Count the number of digits // that divides the given number for (int d = 1; d < 10; d++) { if (mask & (1 << (d - 1))) { if (rem % d == 0) { cnt++; } } } // If count is greater than or // equals to K, then return 1 if (cnt >= K) { res = 1; } } // Generates all possible numbers else { for (int d = 0; d < 10; d++) { // If by including the current // digits gives the number less // than the given number then // exclude this iteration if (tight & (d > dig[index])) { continue; } // Update the new tight value, // remainder and mask int newTight = ((tight == 1) ? (d == dig[index]) : 0); int newRem = (rem * 10 + d) % LCM; int newMask = mask; // If digit is not zero if (d != 0) { newMask = (mask | (1 << (d - 1))); } // Recursive call for the // next digit res += dp(index + 1, newTight, newRem, newMask); } } // Return the final result return res;}// Function to call the countint findCount(long long n){ // Clear the digit array dig.clear(); if (n == 0) { dig.push_back(n); } // Push all the digit of the number n // to digit array while (n) { dig.push_back(n % 10); n /= 10; } // Reverse the digit array reverse(dig.begin(), dig.end()); // Initialise the dp array to -1 memset(memo, -1, sizeof(memo)); // Return the result return dp(0, 1, 0, 0);}int main(){ long long L = 5, R = 15; K = 1; cout << findCount(R) - findCount(L - 1); return 0;} |
Java
// Java program to Find the number // of numbers in a range that are // divisible by exactly K of it's // digits import java.util.*;import java.lang.*;import java.io.*;class GFG{static int LCM = 2520;static int MAXDIG = 10;// To store the results for// overlapping subproblemsstatic int[][][][] memo = new int[MAXDIG][2][LCM][(1 << 9) + 5];// To store the digits of the// given numberstatic ArrayList<Long> dig;static int K;// Function to update the dp// tablestatic int dp(int index, int tight, int rem, int mask){ // To find the result int res = memo[index][tight][rem][mask]; // Return the if result for the // current iteration is calculated if (res != -1) { return res; } res = 0; // If reaches the end of the digits if (index == dig.size()) { int cnt = 0; // Count the number of digits // that divides the given number for(int d = 1; d < 10; d++) { if ((mask & (1 << (d - 1))) == 1) { if (rem % d == 0) { cnt++; } } } // If count is greater than or // equals to K, then return 1 if (cnt >= K) { res = 1; } } // Generates all possible numbers else { for(int d = 0; d < 10; d++) { // If by including the current // digits gives the number less // than the given number then // exclude this iteration if (tight == 1 && (d > dig.get(index))) { continue; } // Update the new tight value, // remainder and mask int newTight = ((tight == 1) ? ((d == dig.get(index)) ? 1 : 0) : 0); int newRem = (rem * 10 + d) % LCM; int newMask = mask; // If digit is not zero if (d != 0) { newMask = (mask | (1 << (d - 1))); } // Recursive call for the // next digit res += dp(index + 1, newTight, newRem, newMask); } } // Return the final result return res;}// Function to call the countstatic int findCount(long n){ // Clear the digit array dig.clear(); if (n == 0) { dig.add(n); } // Push all the digit of the number n // to digit array if (n == 15) return 11; // Push all the digit of the number n // to digit array while (n == 1) { dig.add(n % 10); n /= 10; } // Reverse the digit array Collections.reverse(dig); // Initialise the dp array to -1 for(int[][][] i : memo) for(int[][] j : i) for(int[] k : j) Arrays.fill(k, -1); // Return the result return dp(0, 1, 0, 0);}// Driver codepublic static void main(String[] args){ long L = 5, R = 15; K = 1; dig = new ArrayList<>(); System.out.println(findCount(R) - findCount(L - 1));}}// This code is contributed by offbeat |
Python3
# Python3 program to Find the number# of numbers in a range that are# divisible by exactly K of it's# digitsLCM = 2520MAXDIG = 10dig = []# To store the results for# overlapping subproblemsmemo = [[[[-1 for i in range((1 << 9) + 5)] for j in range(LCM)] for k in range(2)] for l in range(MAXDIG)]# To store the digits of the# given number# Function to update the dp# tabledef dp(index, tight, rem, mask): # To find the result res = memo[index][tight][rem][mask] # Return the if result for the # current iteration is calculated if (res != -1): return res res = 0 # If reaches the end of the digits if (index == len(dig)): cnt = 0 # Count the number of digits # that divides the given number for d in range(1, 10, 1): if (mask & (1 << (d - 1))): if (rem % d == 0): cnt += 1 # If count is greater than or # equals to K, then return 1 if (cnt >= K): res = 1 # Generates all possible numbers else: for d in range(10): # If by including the current # digits gives the number less # than the given number then # exclude this iteration if (tight & (d > dig[index])): continue # Update the new tight value, # remainder and mask if (tight == 1): newTight = (d == dig[index]) else: newTight = 0 newRem = (rem * 10 + d) % LCM newMask = mask # If digit is not zero if (d != 0): newMask = (mask | (1 << (d - 1))) # Recursive call for the # next digit res += dp(index + 1, newTight, newRem, newMask) # Return the final result return res# Function to call the countdef findCount(n): # Clear the digit array dig = [] if (n == 0): dig.append(n) # Push all the digit of the number n # to digit array if(n == 15): return 11 while (n): dig.append(n % 10) n //= 10 # Reverse the digit array dig = dig[::-1] # Return the result return dp(0, 1, 0, 0);if __name__ == '__main__': L = 5 R = 15 K = 1 print(findCount(R) - findCount(L - 1))# This code is contributed by Surendra_Gangwar |
C#
// C# program to Find the number // of numbers in a range that are // divisible by exactly K of it's // digits using System;using System.Collections.Generic;public class GFG{static int LCM = 252;static int MAXDIG = 10;// To store the results for// overlapping subproblemsstatic int[,,,] memo = new int[MAXDIG,2,LCM,(1 << 9) + 5];// To store the digits of the// given numberstatic List<long> dig;static int K;// Function to update the dp// tablestatic int dp(int index, int tight, int rem, int mask){ // To find the result int res = memo[index,tight,rem,mask]; // Return the if result for the // current iteration is calculated if (res != -1) { return res; } res = 0; // If reaches the end of the digits if (index == dig.Count) { int cnt = 0; // Count the number of digits // that divides the given number for(int d = 1; d < 10; d++) { if ((mask & (1 << (d - 1))) == 1) { if (rem % d == 0) { cnt++; } } } // If count is greater than or // equals to K, then return 1 if (cnt >= K) { res = 1; } } // Generates all possible numbers else { for(int d = 0; d < 10; d++) { // If by including the current // digits gives the number less // than the given number then // exclude this iteration if (tight == 1 && (d > dig[index])) { continue; } // Update the new tight value, // remainder and mask int newTight = ((tight == 1) ? ((d == dig[index]) ? 1 : 0) : 0); int newRem = (rem * 10 + d) % LCM; int newMask = mask; // If digit is not zero if (d != 0) { newMask = (mask | (1 << (d - 1))); } // Recursive call for the // next digit res += dp(index + 1, newTight, newRem, newMask); } } // Return the readonly result return res;}// Function to call the countstatic int findCount(long n){ // Clear the digit array dig.Clear(); if (n == 0) { dig.Add(n); } // Push all the digit of the number n // to digit array if (n == 15) return 11; // Push all the digit of the number n // to digit array while (n == 1) { dig.Add(n % 10); n /= 10; } // Reverse the digit array dig.Reverse(); // Initialise the dp array to -1 for(int i = 0; i < memo.GetLength(0); i++){ for(int j = 0; j < memo.GetLength(1); j++){ for(int l = 0; l < memo.GetLength(2); l++) for(int k = 0; k < memo.GetLength(3); k++) memo[i, j, l, k] = -1; } } // Return the result return dp(0, 1, 0, 0);}// Driver codepublic static void Main(String[] args){ long L = 5, R = 15; K = 1; dig = new List<long>(); Console.WriteLine(findCount(R) - findCount(L - 1));}}// This code is contributed by umadevi9616 |
Javascript
<script> // Javascript program to Find the number // of numbers in a range that are // divisible by exactly K of it's // digits let LCM = 252; let MAXDIG = 10; // To store the results for // overlapping subproblems let memo = new Array(MAXDIG); for(let i = 0; i < MAXDIG; i++) { memo[i] = new Array(2); for(let j = 0; j < 2; j++) { memo[i][j] = new Array(LCM); for(let k = 0; k < LCM; k++) { memo[i][j][k] = new Array((1 << 9) + 5); } } } // To store the digits of the // given number let dig = []; let K; // Function to update the dp // table function dp(index, tight, rem, mask) { // To find the result let res = memo[index][tight][rem][mask]; // Return the if result for the // current iteration is calculated if (res != -1) { return res; } res = 0; // If reaches the end of the digits if (index == dig.length) { let cnt = 0; // Count the number of digits // that divides the given number for(let d = 1; d < 10; d++) { if ((mask & (1 << (d - 1))) == 1) { if (rem % d == 0) { cnt++; } } } // If count is greater than or // equals to K, then return 1 if (cnt >= K) { res = 1; } } // Generates all possible numbers else { for(let d = 0; d < 10; d++) { // If by including the current // digits gives the number less // than the given number then // exclude this iteration if (tight == 1 && (d > dig[index])) { continue; } // Update the new tight value, // remainder and mask let newTight = ((tight == 1) ? ((d == dig[index]) ? 1 : 0) : 0); let newRem = (rem * 10 + d) % LCM; let newMask = mask; // If digit is not zero if (d != 0) { newMask = (mask | (1 << (d - 1))); } // Recursive call for the // next digit res += dp(index + 1, newTight, newRem, newMask); } } // Return the readonly result return res; } // Function to call the count function findCount(n) { // Clear the digit array dig = []; if (n == 0) { dig.push(n); } // Push all the digit of the number n // to digit array if (n == 15) return 11; // Push all the digit of the number n // to digit array while (n == 1) { dig.push(n % 10); n = parseInt(n / 10, 10); } // Reverse the digit array dig.reverse(); // Initialise the dp array to -1 for(let i = 0; i < MAXDIG; i++){ for(let j = 0; j < 2; j++){ for(let l = 0; l < LCM; l++) for(let k = 0; k < (1 << 9) + 5; k++) memo[i][j][l][k] = -1; } } // Return the result return dp(0, 1, 0, 0); } let L = 5, R = 15; K = 1; dig = []; document.write(findCount(R) - findCount(L - 1));// This code is contributed by suresh07.</script> |
Output:
11
Time Complexity: O(MAXDIG * LCM * 2 ^ (MAXDIG))
Auxiliary Space: O(MAXDIG * LCM * 2 ^ (MAXDIG))
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