Count of numbers in range which are divisible by M and have digit D at odd places

Given two numbers M, D and an array arr[] which represents the range [L, R], the task is to count the numbers in the range [L, R] which is divisible by M and the digit D occurs at every odd positions.

Note: The numbers might be large. So, the range array {L, R} is given in the form of a string.

Examples:

Input: arr[] = {20, 32}, M = 2, D = 2
Output: 4
Explanation:
There are 4 numbers in the range [20, 32] which are divisible by 2 and 2 at the odd (first) position. They are:
20, 24, 26, 28.

Input: arr[] = {40, 60}, M = 2, D = 5
Output: 5
Explanation:
There are 5 numbers in the range [40, 60] which are divisible by 2 and 5 at the odd (first) position. They are:
50, 52, 54, 56, 58.



Naive Approach: The naive approach for this problem is to take every number between L to R and check whether the digit D occurs at odd positions or not and whether the number is divisible by M or not. The time complexity of this solution will be O(N*K) where N is the count of numbers in the range [L, R] and K is the maximum number of digits in any number in the range [L, R].

Efficient Approach: The idea is to use the concept of digit-dp and find the number of combinations of the numbers that solves the problem. In order to find the count of numbers satisfying the condition, we find the count of numbers satisfying the condition up to the highest number R in the range. This includes all the 1 digit, 2 digit, 3 digit numbers. After finding this count, we simply subtract the count of numbers satisfying up to L – 1 from the above value to get the final answer.

  1. On assuming the numbers as a sequence of digits, we try to form an N digit number which follows the conditions at every iteration where N is the number of digits in the given upper range R. Let the count be C.
  2. This count C includes the 1-digit numbers, 2-digit numbers, 3-digit numbers … up to N-digit numbers where N is the number of digits in R.
  3. Similarly, we find the count of numbers satisfying the condition up to L – 1. Let this count be D.
  4. The required answer needed is the difference between both the counts C – D.
  5. The count of numbers satisfying the given condition is found out by the concept of digit dp.
  6. A three-dimensional table is formed and filled using this concept.

Below is the implementation of the above approach:

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// C++ program to find the count of numbers
// in the range [L, R] which are divisible
// by M and have digit D at the odd places
  
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
  
// Variables to store M, N, D
ll m, n, d;
  
// Vector to store the digit number
// in the form of digits
vector<int> v;
ll const k = 1e9 + 7;
  
// Dp table to compute the answer
ll dp[2001][2001][2];
  
// Function to add the individual
// digits into the vector
void init(string s)
{
    memset(dp, -1, sizeof(dp));
    v.clear();
  
    // Iterating through the number
    // and adding the digits into
    // the vector
    for (int i = 0; i < s.size(); i++) {
        v.push_back(s[i] - '0');
    }
    n = s.size();
}
  
// Function to subtract 1 from a number
// represented in a form of a string
string number_minus_one(string a)
{
    string s = a.substr(1);
    string s1 = "";
  
    // Iterating through the number
    for (int i = 0; i < s.size() - 1; i++)
        s1 += '0';
  
    // If the first digit is 1, then make it 0
    // and add 9 at the end of the string.
    if (a[0] == 1 and s == s1) {
        ll l = s.size();
        a = "";
        a += '0';
        for (int i = 0; i < l; i++)
            a += '9';
    }
    else {
        for (int i = a.size() - 1; i >= 0; i--) {
  
            // If we need to subtract 1 from 0,
            // then make it 9 and subtract 1
            // from the previous digits
            if (a[i] == '0')
                a[i] = '9';
  
            // Else, simply subtract 1
            else {
                a[i] = (((a[i] - '0') - 1) + '0');
                break;
            }
        }
    }
    return a;
}
  
// Function to find the count of numbers
// in the range [L, R] which are divisible
// by M and have digit D at the odd places
ll fun(ll pos, ll sum, ll f)
{
  
    // Base case
    if (pos == n) {
        if (sum == 0) {
  
            // If we have built N-digit number
            // and the number is divisible
            // by m then we have got
            // one possible answer.
            return 1;
        }
        return 0;
    }
  
    // If the answer has already been computed,
    // then return the answer
    if (dp[pos][sum][f] != -1)
        return dp[pos][sum][f];
    ll lmt = 9;
  
    // The possible digits which we can
    // place at the pos position.
    if (!f)
        lmt = v[pos];
  
    ll ans = 0;
  
    // Iterating through all the digits
    for (ll i = 0; i <= lmt; i++) {
        if (i == d and pos % 2 == 1)
            ans += 0;
        else if (i != d and pos % 2 == 0)
            ans += 0;
        else {
            ll new_f = f;
  
            // If we have placed all the digits
            // up to pos-1 equal to their
            // limit and currently we are placing
            // a digit which is smaller than this
            // position's limit then we can place
            // 0 to 9 at all the next positions
            // to make the number smaller than R
            if (f == 0 and i < lmt)
                new_f = 1;
  
            // Calculating the number upto pos mod m.
            ll new_sum = sum;
  
            // Combinations of numbers as there are
            // 10 digits in the range [0, 9]
            new_sum *= 10;
            new_sum += i;
            new_sum %= m;
  
            // Recursively call the function
            // for the next position
            ans += fun(pos + 1, new_sum, new_f);
            ans %= k;
        }
    }
  
    // Returning the final answer
    return dp[pos][sum][f] = ans;
}
  
// Function to call the function
// for every query
void operations(string L, string R)
{
    init(R);
    ll ans = fun(0, 0, 0);
    L = number_minus_one(L);
    init(L);
    ans -= fun(0, 0, 0);
    if (ans < 0)
        ans += k;
    cout << ans << "\n";
}
  
// Driver code
int main()
{
    m = 2, d = 2;
    ll Q = 1;
    string arr[][2] = { { "20", "32" } };
  
    for (ll i = 0; i < Q; i++) {
        operations(arr[i][0], arr[i][1]);
    }
  
    return 0;
}
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Output:
4

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