Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M

Given a range [L, R] and two positive integers N and M. The task is to count the numbers in the range containing only non-zero digits whose sum of digits is equal to N and the number is divisible by M.

Examples:

Input: L = 1, R = 100, N = 8, M = 2
Output: 4
Only 8, 26, 44 and 62 are valid numbers

Input: L = 1, R = 200, N = 4, M = 11
Output: 2
Only 22 and 121 are valid numbers

Prerequisites : Digit DP

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Firstly, if we are able to count the required numbers up to R i.e. in the range [0, R], we can easily reach our answer in the range [L, R] by solving for from zero to R and then subtracting the answer we get after solving for from zero to L – 1. Now, we need to define the DP states.
DP States:

Since we can consider our number as a sequence of digits, one state is the position at which we are currently in. This position can have values from 0 to 18 if we are dealing with the numbers up to 10¹⁸. In each recursive call, we try to build the sequence from left to right by placing a digit from 0 to 9.
Second state is the sum of the digits we have placed so far.
Third state is the remainder which defines the modulus of the number we have made so far modulo M.
Another state is the boolean variable tight which tells the number we are trying to build has already become smaller than R so that in the upcoming recursive calls we can place any digit from 0 to 9. If the number has not become smaller, the maximum limit of digit we can place is digit at the current position in R.

For the number to have only non-zero digits, we maintain a variable nonz whose value if 1 tells the first digit in the number we have placed is a non-zero digit and thus, now we can’t place any zero digit in upcoming calls. Otherwise, we can place a zero digit as a leading zero so as to make number of digits in current number smaller than number of digits in upper limit.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
const int M = 20; 
  
// states - position, sum, rem, tight 
// sum can have values upto 162, if we 
// are dealing with numbers upto 10^18 
// when all 18 digits are 9, then sum 
// is 18 * 9 = 162 
int dp[M][165][M][2]; 
  
// n is the sum of digits and number should 
// be divisible by m 
int n, m; 
  
// Function to return the count of 
// required numbers from 0 to num 
int count(int pos, int sum, int rem, int tight, 
          int nonz, vector<int> num) 
{ 
    // Last position 
    if (pos == num.size()) { 
        if (rem == 0 && sum == n) 
            return 1; 
        return 0; 
    } 
  
    // If this result is already computed 
    // simply return it 
    if (dp[pos][sum][rem][tight] != -1) 
        return dp[pos][sum][rem][tight]; 
  
    int ans = 0; 
  
    // Maximum limit upto which we can place 
    // digit. If tight is 1, means number has 
    // already become smaller so we can place 
    // any digit, otherwise num[pos] 
    int limit = (tight ? 9 : num[pos]); 
  
    for (int d = 0; d <= limit; d++) { 
  
        // If the current digit is zero 
        // and nonz is 1, we can't place it 
        if (d == 0 && nonz) 
            continue; 
        int currSum = sum + d; 
        int currRem = (rem * 10 + d) % m; 
        int currF = tight || (d < num[pos]); 
        ans += count(pos + 1, currSum, currRem, 
                     currF, nonz || d, num); 
    } 
    return dp[pos][sum][rem][tight] = ans; 
} 
  
// Function to convert x into its digit vector 
// and uses count() function to return the 
// required count 
int solve(int x) 
{ 
    vector<int> num; 
    while (x) { 
        num.push_back(x % 10); 
        x /= 10; 
    } 
    reverse(num.begin(), num.end()); 
  
    // Initialize dp 
    memset(dp, -1, sizeof(dp)); 
    return count(0, 0, 0, 0, 0, num); 
} 
  
// Driver code 
int main() 
{ 
    int L = 1, R = 100; 
    n = 8, m = 2; 
    cout << solve(R) - solve(L); 
    return 0; 
} 

Java

// Java implementation of the approach  
import java.util.*; 
  
class GFG 
{ 
      
static int M = 20;  
  
// states - position, sum, rem, tight  
// sum can have values upto 162, if we  
// are dealing with numbers upto 10^18  
// when all 18 digits are 9, then sum  
// is 18 * 9 = 162  
static int dp[][][][] = new int [M][165][M][2];  
  
// n is the sum of digits and number should  
// be divisible by m  
static int n, m;  
  
// Function to return the count of  
// required numbers from 0 to num  
static int count(int pos, int sum, int rem, int tight,  
        int nonz, Vector<Integer> num)  
{  
    // Last position  
    if (pos == num.size()) 
    {  
        if (rem == 0 && sum == n)  
            return 1;  
        return 0;  
    }  
  
    // If this result is already computed  
    // simply return it  
    if (dp[pos][sum][rem][tight] != -1)  
        return dp[pos][sum][rem][tight];  
  
    int ans = 0;  
  
    // Maximum limit upto which we can place  
    // digit. If tight is 1, means number has  
    // already become smaller so we can place  
    // any digit, otherwise num[pos]  
    int limit = (tight != 0 ? 9 : num.get(pos));  
  
    for (int d = 0; d <= limit; d++) 
    {  
  
        // If the current digit is zero  
        // and nonz is 1, we can't place it  
        if (d == 0 && nonz != 0)  
            continue;  
        int currSum = sum + d;  
        int currRem = (rem * 10 + d) % m;  
        int currF = (tight != 0 || (d < num.get(pos))) ? 1 : 0;  
        ans += count(pos + 1, currSum, currRem,  
                    currF, (nonz != 0 || d != 0) ? 1 : 0, num);  
    }  
    return dp[pos][sum][rem][tight] = ans;  
}  
  
// Function to convert x into its digit vector  
// and uses count() function to return the  
// required count  
static int solve(int x)  
{  
    Vector<Integer> num = new Vector<Integer>();  
    while (x != 0)  
    {  
        num.add(x % 10);  
        x /= 10;  
    }  
    Collections.reverse(num);  
  
    // Initialize dp  
    for(int i = 0; i < M; i++) 
        for(int j = 0; j < 165; j++) 
            for(int k = 0; k < M; k++) 
                for(int l = 0; l < 2; l++) 
                    dp[i][j][k][l]=-1; 
      
    return count(0, 0, 0, 0, 0, num);  
}  
  
// Driver code  
public static void main(String args[])  
{  
    int L = 1, R = 100;  
    n = 8; m = 2;  
    System.out.print( solve(R) - solve(L));  
}  
} 
  
// This code is contributed by Arnab Kundu 

Python3

# Python3 implementation of the approach  
  
# Function to return the count of  
# required numbers from 0 to num  
def count(pos, Sum, rem, tight, nonz, num):  
  
    # Last position  
    if pos == len(num):  
        if rem == 0 and Sum == n:  
            return 1
        return 0
  
    # If this result is already computed  
    # simply return it 
    if dp[pos][Sum][rem][tight] != -1:  
        return dp[pos][Sum][rem][tight]  
      
    ans = 0
  
    # Maximum limit upto which we can place  
    # digit. If tight is 1, means number has  
    # already become smaller so we can place  
    # any digit, otherwise num[pos] 
    if tight: 
        limit = 9
    else:  
        limit = num[pos] 
      
    for d in range(0, limit + 1):  
  
        # If the current digit is zero  
        # and nonz is 1, we can't place it  
        if d == 0 and nonz: 
            continue
              
        currSum = Sum + d  
        currRem = (rem * 10 + d) % m  
        currF = int(tight or (d < num[pos]))  
        ans += count(pos + 1, currSum, currRem,  
                     currF, nonz or d, num)  
      
    dp[pos][Sum][rem][tight] = ans 
    return ans 
  
# Function to convert x into its digit  
# vector and uses count() function to  
# return the required count  
def solve(x):  
  
    num = [] 
    global dp 
      
    while x > 0:  
        num.append(x % 10)  
        x //= 10
      
    num.reverse()  
  
    # Initialize dp  
    dp = [[[[-1, -1] for i in range(M)]  
                     for j in range(165)]  
                     for k in range(M)] 
    return count(0, 0, 0, 0, 0, num)  
  
# Driver code  
if __name__ == "__main__": 
  
    L, R = 1, 100
      
    # n is the sum of digits and number  
    # should be divisible by m 
    n, m, M = 8, 2, 20
      
    # States - position, sum, rem, tight  
    # sum can have values upto 162, if we  
    # are dealing with numbers upto 10^18  
    # when all 18 digits are 9, then sum  
    # is 18 * 9 = 162  
    dp = [] 
  
    print(solve(R) - solve(L))  
      
# This code is contributed by Rituraj Jain 

C#

// C# implementation of the approach  
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
      
static int M = 20;  
  
// states - position, sum, rem, tight  
// sum can have values upto 162, if we  
// are dealing with numbers upto 10^18  
// when all 18 digits are 9, then sum  
// is 18 * 9 = 162  
static int [,,,]dp = new int [M, 165, M, 2];  
  
// n is the sum of digits and number should  
// be divisible by m  
static int n, m;  
  
// Function to return the count of  
// required numbers from 0 to num  
static int count(int pos, int sum, int rem, int tight,  
                            int nonz, List<int> num)  
{  
    // Last position  
    if (pos == num.Count) 
    {  
        if (rem == 0 && sum == n)  
            return 1;  
        return 0;  
    }  
  
    // If this result is already computed  
    // simply return it  
    if (dp[pos,sum,rem,tight] != -1)  
        return dp[pos,sum,rem,tight];  
  
    int ans = 0;  
  
    // Maximum limit upto which we can place  
    // digit. If tight is 1, means number has  
    // already become smaller so we can place  
    // any digit, otherwise num[pos]  
    int limit = (tight != 0 ? 9 : num[pos]);  
  
    for (int d = 0; d <= limit; d++) 
    {  
  
        // If the current digit is zero  
        // and nonz is 1, we can't place it  
        if (d == 0 && nonz != 0)  
            continue;  
        int currSum = sum + d;  
        int currRem = (rem * 10 + d) % m;  
        int currF = (tight != 0 || (d < num[pos])) ? 1 : 0;  
        ans += count(pos + 1, currSum, currRem,  
                    currF, (nonz != 0 || d != 0) ? 1 : 0, num);  
    }  
    return dp[pos, sum, rem, tight] = ans;  
}  
  
// Function to convert x into its digit vector  
// and uses count() function to return the  
// required count  
static int solve(int x)  
{  
    List<int> num = new List<int>();  
    while (x != 0)  
    {  
        num.Add(x % 10);  
        x /= 10;  
    }  
    num.Reverse();  
  
    // Initialize dp  
    for(int i = 0; i < M; i++) 
        for(int j = 0; j < 165; j++) 
            for(int k = 0; k < M; k++) 
                for(int l = 0; l < 2; l++) 
                    dp[i, j, k, l] = -1; 
      
    return count(0, 0, 0, 0, 0, num);  
}  
  
// Driver code  
public static void Main(String []args)  
{  
    int L = 1, R = 100;  
    n = 8; m = 2;  
    Console.Write( solve(R) - solve(L));  
}  
} 
  
// This code has been contributed by 29AjayKumar 

Output:

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C++

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