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Count the numbers with N digits and whose suffix is divisible by K

  • Difficulty Level : Expert
  • Last Updated : 25 Nov, 2021

Given two positive integers N and K, the task is to count the number of positive integers D such that D has N digits and any of the suffixes of its decimal representation is divisible by K. 

Examples:  

Input: N = 1, K = 2 
Output:
Explanation: 
There are 4 such integers in which any of the suffix is divisible by K: 
{2, 4, 6, 8}

Input: N = 2, K = 2 
Output: 45 
Explanation: 
There are 45, Two digit integers in which any of the suffix is divisible by 2: 
Some of such integers is given below – 
{10, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23…} 
Notice that, 21 and 23 numbers are not divisible by 2. But their suffix 2 is divisible by 2. 



Naive Approach: Iterate over all the integers of the N digit and for each integer check that any suffix of the number is divisible by K, If yes, then increment the count of such numbers by 1.
Approach: The idea is to use the concept of Dynamic Programming by increasing the suffix length and placing the digits from 0 to 9 recursively. Below is the illustration of the steps:  

  • Function Definition: This problem can be solved recursively in which, at each step, we can choose the digits for the suffix for the N digit number. So, the Function Definition of the recursive solution will be:
// Recursive Function to count of values
// whose suffixes of length pos 
// have remainder rem with K
recur(pos, rem)
  • Base Case: The base case for this problem is when, for any index, the remainder of the suffix with K becomes 0, then all the other digits can be placed with all the possible integers from 0 to 9.
f(pos, 0) = 9 * (10^(n-i-1))
  • Recursive Case: At each step of recursion we increase the suffix length by one by placing all integers from 0-9, change the remainder with K accordingly and move to the next step.
for num in range [0-9]:
     f(pos, rem) += f(pos+1, (rem*10 + num)%k)

Below is the implementation of the above approach: 

C++




// C++ implementation to Count the
// numbers with N digits and whose
// suffix is divisible by K
 
#include <bits/stdc++.h>
 
using namespace std;
 
int mod = 1000000007;
int dp[1005][105][2];
int powers[1005];
int powersModk[1005];
 
// Suffix of length pos with
// remainder rem and Z representing
// whether the suffix has a
// non zero digit until now
int calculate(int pos, int rem,
           int z, int k, int n)
{
    // Base case
    if (rem == 0 && z) {
         
        // If count of digits
        // is less than n
        if (pos != n)
             
            // Placing all possible
            // digits in remaining
            // positions
            return (powers[n - pos -
                        1] * 9) % mod;
        else
            return 1;
    }
     
    // If remainder non zero
    // and suffix has n digits
    if (pos == n)
        return 0;
     
    // If the subproblem
    // is already solved
    if (dp[pos][rem][z] != -1)
        return dp[pos][rem][z];
 
    int count = 0;
 
    // Placing all digits at MSB
    // of suffix and increasing
    // it's length by 1
    for (int i = 0; i < 10; i++) {
        if (i == 0)
            count = (count + (calculate(
             pos + 1, (rem + (i *
                 powersModk[pos]) % k) %
                    k, z, k, n))) % mod;
 
        // Non zero digit is placed
        else
            count = (count + (calculate(
                pos + 1, (rem + (i *
                powersModk[pos]) % k) %
                k, 1, k, n))) % mod;
    }
 
    // Store and return the
    // solution to this subproblem
    return dp[pos][rem][z] = count;
}
 
// Function to Count the numbers
// with N digits and whose suffix
// is divisible by K
int countNumbers(int n, int k)
{
 
    // Since we need powers of 10
    // for counting, it's better to
    // pre store them along with their
    // modulo with 1e9 + 7 for counting
    int st = 1;
    for (int i = 0; i <= n; i++) {
        powers[i] = st;
        st *= 10;
        st %= mod;
    }
 
    // Since at each recursive step
    // we increase the suffix length by 1
    // by placing digits at its leftmost
    // position, we need powers of 10
    // modded with k, in order to fpos
    // the new remainder efficiently
    st = 1;
    for (int i = 0; i <= n; i++) {
        powersModk[i] = st;
        st *= 10;
        st %= mod;
    }
 
    // Initialising dp table values -1
    // represents subproblem hasn't
    // been solved yet
    memset(dp, -1, sizeof(dp));
 
    return calculate(0, 0, 0, k, n);
}
 
// Driver Code
int main()
{
    int N = 2;
    int K = 2;
 
    cout << countNumbers(N, K);
 
    return 0;
}

Java




// Java implementation to Count the
// numbers with N digits and whose
// suffix is divisible by K
import java.util.*;
import java.util.Arrays;
 
class GFG
{
     
    static int mod = 1000000007;
    static int dp[][][] = new int[1005][105][2];
    static int powers[] = new int[1005];
    static int powersModk[] = new int[1005];
     
    // Suffix of length pos with
    // remainder rem and Z representing
    // whether the suffix has a
    // non zero digit until now
    static int calculate(int pos, int rem,
            int z, int k, int n)
    {
        // Base case
        if (rem == 0 && z!=0) {
             
            // If count of digits
            // is less than n
            if (pos != n)
                 
                // Placing all possible
                // digits in remaining
                // positions
                return (powers[n - pos -
                            1] * 9) % mod;
            else
                return 1;
        }
         
        // If remainder non zero
        // and suffix has n digits
        if (pos == n)
            return 0;
         
        // If the subproblem
        // is already solved
        if (dp[pos][rem][z] != -1)
            return dp[pos][rem][z];
     
        int count = 0;
     
        // Placing all digits at MSB
        // of suffix and increasing
        // it's length by 1
        for (int i = 0; i < 10; i++) {
            if (i == 0)
                count = (count + (calculate(
                pos + 1, (rem + (i *
                    powersModk[pos]) % k) %
                        k, z, k, n))) % mod;
     
            // Non zero digit is placed
            else
                count = (count + (calculate(
                    pos + 1, (rem + (i *
                    powersModk[pos]) % k) %
                    k, 1, k, n))) % mod;
        }
     
        // Store and return the
        // solution to this subproblem
        return dp[pos][rem][z] = count;
    }
     
    // Function to Count the numbers
    // with N digits and whose suffix
    // is divisible by K
    static int countNumbers(int n, int k)
    {
     
        // Since we need powers of 10
        // for counting, it's better to
        // pre store them along with their
        // modulo with 1e9 + 7 for counting
        int st = 1;
        int i;
        for (i = 0; i <= n; i++) {
            powers[i] = st;
            st *= 10;
            st %= mod;
        }
     
        // Since at each recursive step
        // we increase the suffix length by 1
        // by placing digits at its leftmost
        // position, we need powers of 10
        // modded with k, in order to fpos
        // the new remainder efficiently
        st = 1;
        for (i = 0; i <= n; i++) {
            powersModk[i] = st;
            st *= 10;
            st %= mod;
        }
     
        // Initialising dp table values -1
        // represents subproblem hasn't
        // been solved yet
        for (int[][] row: dp)
        {
            for (int[] innerRow: row)
            {
                Arrays.fill(innerRow, -1);
            }
        };
     
        return calculate(0, 0, 0, k, n);
    }
 
    // Driver Code
    public static void main(String []args)
    {
        int N = 2;
        int K = 2;
     
        System.out.print(countNumbers(N, K));
    }
}
 
// This code is contributed by chitranayal

Python3




# Python3 implementation to Count the
# numbers with N digits and whose
# suffix is divisible by K
 
mod = 1000000007
dp = [[[-1 for i in range(2)] for i in range(105)] for i in range(1005)]
powers = [0]*1005
powersModk = [0]*1005
 
# Suffix of length pos with
# remainder rem and Z representing
# whether the suffix has a
# non zero digit until now
def calculate(pos, rem, z, k, n):
    # Base case
    if (rem == 0 and z):
 
        # If count of digits
        # is less than n
        if (pos != n):
 
            # Placing all possible
            # digits in remaining
            # positions
            return (powers[n - pos -1] * 9) % mod
        else:
            return 1
 
    # If remainder non zero
    # and suffix has n digits
    if (pos == n):
        return 0
 
    # If the subproblem
    # is already solved
    if (dp[pos][rem][z] != -1):
        return dp[pos][rem][z]
 
    count = 0
 
    # Placing all digits at MSB
    # of suffix and increasing
    # it's length by 1
    for i in range(10):
        if (i == 0):
            count = (count + (calculate(
            pos + 1, (rem + (i *
                powersModk[pos]) % k) %
                    k, z, k, n))) % mod
 
        # Non zero digit is placed
        else:
            count = (count + (calculate(
                pos + 1, (rem + (i *
                powersModk[pos]) % k) %
                k, 1, k, n))) % mod
 
    # Store and return the
    # solution to this subproblem
    dp[pos][rem][z] = count
    return count
 
# Function to Count the numbers
# with N digits and whose suffix
# is divisible by K
def countNumbers(n, k):
 
    # Since we need powers of 10
    # for counting, it's better to
    # pre store them along with their
    # modulo with 1e9 + 7 for counting
    st = 1
    for i in range(n + 1):
        powers[i] = st
        st *= 10
        st %= mod
 
    # Since at each recursive step
    # we increase the suffix length by 1
    # by placing digits at its leftmost
    # position, we need powers of 10
    # modded with k, in order to fpos
    # the new remainder efficiently
    st = 1
    for i in range(n + 1):
        powersModk[i] = st
        st *= 10
        st %= mod
 
    # Initialising dp table values -1
    # represents subproblem hasn't
    # been solved yet
    # memset(dp, -1, sizeof(dp))
 
    return calculate(0, 0, 0, k, n)
 
# Driver Code
if __name__ == '__main__':
    N = 2
    K = 2
 
    print(countNumbers(N, K))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation to Count the
// numbers with N digits and whose
// suffix is divisible by K
using System;
 
class GFG
{
      
    static int mod = 1000000007;
    static int [,,]dp = new int[1005, 105, 2];
    static int []powers = new int[1005];
    static int []powersModk = new int[1005];
      
    // Suffix of length pos with
    // remainder rem and Z representing
    // whether the suffix has a
    // non zero digit until now
    static int calculate(int pos, int rem,
            int z, int k, int n)
    {
        // Base case
        if (rem == 0 && z != 0) {
              
            // If count of digits
            // is less than n
            if (pos != n)
                  
                // Placing all possible
                // digits in remaining
                // positions
                return (powers[n - pos -
                            1] * 9) % mod;
            else
                return 1;
        }
          
        // If remainder non zero
        // and suffix has n digits
        if (pos == n)
            return 0;
          
        // If the subproblem
        // is already solved
        if (dp[pos, rem, z] != -1)
            return dp[pos,rem,z];
      
        int count = 0;
      
        // Placing all digits at MSB
        // of suffix and increasing
        // it's length by 1
        for (int i = 0; i < 10; i++) {
            if (i == 0)
                count = (count + (calculate(
                pos + 1, (rem + (i *
                    powersModk[pos]) % k) %
                        k, z, k, n))) % mod;
      
            // Non zero digit is placed
            else
                count = (count + (calculate(
                    pos + 1, (rem + (i *
                    powersModk[pos]) % k) %
                    k, 1, k, n))) % mod;
        }
      
        // Store and return the
        // solution to this subproblem
        return dp[pos,rem,z] = count;
    }
      
    // Function to Count the numbers
    // with N digits and whose suffix
    // is divisible by K
    static int countNumbers(int n, int k)
    {
      
        // Since we need powers of 10
        // for counting, it's better to
        // pre store them along with their
        // modulo with 1e9 + 7 for counting
        int st = 1;
        int i;
        for (i = 0; i <= n; i++) {
            powers[i] = st;
            st *= 10;
            st %= mod;
        }
      
        // Since at each recursive step
        // we increase the suffix length by 1
        // by placing digits at its leftmost
        // position, we need powers of 10
        // modded with k, in order to fpos
        // the new remainder efficiently
        st = 1;
        for (i = 0; i <= n; i++) {
            powersModk[i] = st;
            st *= 10;
            st %= mod;
        }
      
        // Initialising dp table values -1
        // represents subproblem hasn't
        // been solved yet
        for(i = 0; i < 1005; i++){
            for(int j = 0; j < 105; j++){
                for(int l = 0; l < 2; l++)
                    dp[i, j, l] = -1;
            }
        }
   
        return calculate(0, 0, 0, k, n);
    }
  
    // Driver Code
    public static void Main(String []args)
    {
        int N = 2;
        int K = 2;
      
        Console.Write(countNumbers(N, K));
    }
}
  
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript implementation to Count the
// numbers with N digits and whose
// suffix is divisible by K
 
var mod = 1000000007;
var dp = Array.from(Array(1005), ()=>Array(105));
 
for(var i = 0; i< 1005; i++)
{
    for(var j =0; j<105; j++)
    {
        dp[i][j] = Array(2).fill(-1);
    }
}
var powers = Array(1005);
var powersModk= Array(1005);
 
// Suffix of length pos with
// remainder rem and Z representing
// whether the suffix has a
// non zero digit until now
function calculate(pos, rem, z, k, n)
{
    // Base case
    if (rem == 0 && z) {
         
        // If count of digits
        // is less than n
        if (pos != n)
             
            // Placing all possible
            // digits in remaining
            // positions
            return (powers[n - pos -
                        1] * 9) % mod;
        else
            return 1;
    }
     
    // If remainder non zero
    // and suffix has n digits
    if (pos == n)
        return 0;
     
    // If the subproblem
    // is already solved
    if (dp[pos][rem][z] != -1)
        return dp[pos][rem][z];
 
    var count = 0;
 
    // Placing all digits at MSB
    // of suffix and increasing
    // it's length by 1
    for (var i = 0; i < 10; i++) {
        if (i == 0)
            count = (count + (calculate(
             pos + 1, (rem + (i *
                 powersModk[pos]) % k) %
                    k, z, k, n))) % mod;
 
        // Non zero digit is placed
        else
            count = (count + (calculate(
                pos + 1, (rem + (i *
                powersModk[pos]) % k) %
                k, 1, k, n))) % mod;
    }
 
    // Store and return the
    // solution to this subproblem
    return dp[pos][rem][z] = count;
}
 
// Function to Count the numbers
// with N digits and whose suffix
// is divisible by K
function countNumbers(n, k)
{
 
    // Since we need powers of 10
    // for counting, it's better to
    // pre store them along with their
    // modulo with 1e9 + 7 for counting
    var st = 1;
    for (var i = 0; i <= n; i++) {
        powers[i] = st;
        st *= 10;
        st %= mod;
    }
 
    // Since at each recursive step
    // we increase the suffix length by 1
    // by placing digits at its leftmost
    // position, we need powers of 10
    // modded with k, in order to fpos
    // the new remainder efficiently
    st = 1;
    for (var i = 0; i <= n; i++) {
        powersModk[i] = st;
        st *= 10;
        st %= mod;
    }
 
 
    return calculate(0, 0, 0, k, n);
}
 
// Driver Code
var N = 2;
var K = 2;
document.write( countNumbers(N, K));
 
 
</script>
Output: 
45

 

Time Complexity: O(N * K)

Auxiliary Space: O(1005 * 105 * 2)
 




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