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

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:

4

Short Python Implementation :

Python3

# User Input
l, r, n, m = 1, 100, 8, 2

# Initialize Result
output = []

# Traverse through all numbers
for x in range(l, r+1):

    # Check for all conditions in every number
    if sum([int(k) for k in str(x)]) == n and x % m == 0 and '0' not in str(x): # Check conditions
        output.append(x)

print(len(output))

# This code is contributed by mailprakashindia