Number of integers in a range [L, R] which are divisible by exactly K of it’s digits

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:

  1. Store all the digits of the given range and sort the digits in decreasing order.
  2. Traverse all the digits stored above and generate all the number which are strictly less than the given range of number.
  3. 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.
  4. 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.
  5. 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:

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// 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 subproblems
int memo[MAXDIG][2][LCM][(1 << 9) + 5];
  
// To store the digits of the
// given number
vector<int> dig;
int K;
  
// Function to update the dp
// table
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))) {
                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 count
int 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());
  
    // Intialise 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;
}

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Output:

11



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