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++

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// 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;
}

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Java

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// 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

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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#

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// 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

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Output:

4


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