Skip to content
Related Articles

Related Articles

Improve Article

Count of integers from the range [0, N] whose digit sum is a multiple of K

  • Last Updated : 15 Jun, 2021
Geek Week

Given two integers N and K, the task is to calculate the number of integers in the range [0, N] whose digit sum is a multiple of K. The answer could be large, so print the answer modulo 109 +7.

Examples:  

Input: N = 10, K = 5 
Output:
0 and 5 are the only possible integers.

Input: N = 30, K = 4 
Output: 7  

Naive Approach: For small value of N, loop through the range [0, N] and check if the sum of the digits of the numbers are multiples of K or not.
 



Efficient Approach: The idea is to use digit dp to solve this problem. Subproblems iterating through all index values from the left or most significant digit(MSD) in the given integer will be solved and for each index, store the number of ways such that (sum of digits upto current index) mod K to be zero. The dp states will be: 

dp[idx][sum][tight] 
idx = position, it tells about the index value from left in the given integer 
sum = sum of digits mod k, This parameter will store the (sum of digits mod k) in the generated integer from most significant digit(MSD) to p 
tight = flag if the current value is crossing the range (1, n) or not 
For unrestricted range tight = 0 
For restricted range tight = 1 

Let’s say we are at the MSD having index idx. So initially the sum will be 0. At every position, set a limit that is always in the range [0, 9]
Therefore, fill the digit at index by the digits in its range from 0 to limit and fetch the answer from the next state having index = idx + 1 and new_tight for next state is calculated separately. The dp state definition will be: 

dp[idx][sum][tight] += dp[idx + 1][(sum + d) % k][new_tight] 
for d in [0, limit]  

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 100005
#define MOD 1000000007
 
// To store the states of the dp
int dp[MAX][101][2];
 
// Function to return the count of numbers
// from the range [0, n] whose digit sum
// is a multiple of k using bottom-up dp
int countNum(int idx, int sum, int tight,
             vector<int> num, int len, int k)
{
    if (len == idx) {
        if (sum == 0)
            return 1;
        else
            return 0;
    }
    if (dp[idx][sum][tight] != -1)
        return dp[idx][sum][tight];
    int res = 0, limit;
 
    // The digit in this index can
    // only be from [0, num[idx]]
    if (tight == 0) {
        limit = num[idx];
    }
 
    // The digit in this index can
    // be anything from [0, 9]
    else {
        limit = 9;
    }
    for (int i = 0; i <= limit; i++) {
 
        // new_tight is the flag value
        // for the next position
        int new_tight = tight;
        if (tight == 0 && i < limit)
            new_tight = 1;
        res += countNum(idx + 1,
                        (sum + i) % k, new_tight,
                        num, len, k);
        res %= MOD;
    }
 
    // res can't be negative
    if (res < 0)
        res += MOD;
    return dp[idx][sum][tight] = res;
}
 
// Function to process the string to
// a vector of digits from MSD to LSD
vector<int> process(string s)
{
    vector<int> num;
    for (int i = 0; i < s.length(); i++) {
        num.push_back(s[i] - '0');
    }
    return num;
}
 
// Driver code
int main()
{
 
    // For large input number n
    string n = "98765432109876543210";
 
    // Total number of digits in n
    int len = n.length();
 
    int k = 58;
 
    // Clean dp table
    memset(dp, -1, sizeof(dp));
 
    // Process the string to a vector
    // of digits from MSD to LSD
    vector<int> num = process(n);
 
    cout << countNum(0, 0, 0, num, len, k);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
static final int MAX = 100005;
static final int MOD = 1000000007;
 
// To store the states of the dp
static int [][][]dp = new int[MAX][101][2];
 
// Function to return the count of numbers
// from the range [0, n] whose digit sum
// is a multiple of k using bottom-up dp
static int countNum(int idx, int sum, int tight,
            Vector<Integer> num, int len, int k)
{
    if (len == idx)
    {
        if (sum == 0)
            return 1;
        else
            return 0;
    }
    if (dp[idx][sum][tight] != -1)
        return dp[idx][sum][tight];
    int res = 0, limit;
 
    // The digit in this index can
    // only be from [0, num[idx]]
    if (tight == 0)
    {
        limit = num.get(idx);
    }
 
    // The digit in this index can
    // be anything from [0, 9]
    else
    {
        limit = 9;
    }
    for (int i = 0; i <= limit; i++)
    {
 
        // new_tight is the flag value
        // for the next position
        int new_tight = tight;
        if (tight == 0 && i < limit)
            new_tight = 1;
        res += countNum(idx + 1,
                        (sum + i) % k, new_tight,
                        num, len, k);
        res %= MOD;
    }
 
    // res can't be negative
    if (res < 0)
        res += MOD;
    return dp[idx][sum][tight] = res;
}
 
// Function to process the String to
// a vector of digits from MSD to LSD
static Vector<Integer> process(String s)
{
    Vector<Integer> num = new Vector<Integer>();
    for (int i = 0; i < s.length(); i++)
    {
        num.add(s.charAt(i) - '0');
    }
    return num;
}
 
// Driver code
public static void main(String[] args)
{
 
    // For large input number n
    String n = "98765432109876543210";
 
    // Total number of digits in n
    int len = n.length();
 
    int k = 58;
 
    // Clean dp table
    for(int i = 0; i < MAX; i++)
    {
        for(int j = 0; j < 101; j++)
        {
            for(int l = 0; l < 2; l++)
            dp[i][j][l] = -1;
        }
    }
 
    // Process the String to a vector
    // of digits from MSD to LSD
    Vector<Integer> num = process(n);
 
    System.out.print(countNum(0, 0, 0, num, len, k));
 
}
}
 
// This code is contributed by 29AjayKumar

Python 3




# Python 3 implementation of the approach
MAX = 10005
MOD = 1000000007
 
# Function to return the count of numbers
# from the range [0, n] whose digit sum
# is a multiple of k using bottom-up dp
def countNum(idx, sum, tight, num, len1, k):
    if (len1 == idx):
        if (sum == 0):
            return 1
        else:
            return 0
    if (dp[idx][sum][tight] != -1):
        return dp[idx][sum][tight]
    res = 0
 
    # The digit in this index can
    # only be from [0, num[idx]]
    if (tight == 0):
        limit = num[idx]
 
    # The digit in this index can
    # be anything from [0, 9]
    else:
        limit = 9
    for i in range(limit + 1):
         
        # new_tight is the flag value
        # for the next position
        new_tight = tight
        if (tight == 0 and i < limit):
            new_tight = 1
        res += countNum(idx + 1,(sum + i) % k,
                      new_tight, num, len1, k)
        res %= MOD
 
    # res can't be negative
    if (res < 0):
        res += MOD
    dp[idx][sum][tight] = res
    return dp[idx][sum][tight]
 
# Function to process the string to
# a vector of digits from MSD to LSD
def process(s):
    num = []
    for i in range(len(s)):
        num.append(ord(s[i]) - ord('0'))
    return num
 
# Driver code
if __name__ == '__main__':
     
    # For large input number n
    n = "98765432109876543210"
 
    # Total number of digits in n
    len1 = len(n)
 
    k = 58
     
    # To store the states of the dp
    dp = [[[-1 for i in range(2)]
               for j in range(101)]
               for k in range(MAX)]
 
    # Process the string to a vector
    # of digits from MSD to LSD
    num = process(n)
 
    print(countNum(0, 0, 0, num, len1, k))
 
# This code is contributed by Surendra_Gangwar

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
static readonly int MAX = 10005;
static readonly int MOD = 1000000007;
 
// To store the states of the dp
static int [,,]dp = new int[MAX, 101, 2];
 
// Function to return the count of numbers
// from the range [0, n] whose digit sum
// is a multiple of k using bottom-up dp
static int countNum(int idx, int sum, int tight,
              List<int> num, int len, int k)
{
    if (len == idx)
    {
        if (sum == 0)
            return 1;
        else
            return 0;
    }
     
    if (dp[idx, sum, tight] != -1)
        return dp[idx, sum, tight];
    int res = 0, limit;
 
    // The digit in this index can
    // only be from [0, num[idx]]
    if (tight == 0)
    {
        limit = num[idx];
    }
 
    // The digit in this index can
    // be anything from [0, 9]
    else
    {
        limit = 9;
    }
    for (int i = 0; i <= limit; i++)
    {
 
        // new_tight is the flag value
        // for the next position
        int new_tight = tight;
        if (tight == 0 && i < limit)
            new_tight = 1;
        res += countNum(idx + 1,
                       (sum + i) % k, new_tight,
                        num, len, k);
        res %= MOD;
    }
 
    // res can't be negative
    if (res < 0)
        res += MOD;
    return dp[idx, sum, tight] = res;
}
 
// Function to process the String to
// a vector of digits from MSD to LSD
static List<int> process(String s)
{
    List<int> num = new List<int>();
    for (int i = 0; i < s.Length; i++)
    {
        num.Add(s[i] - '0');
    }
    return num;
}
 
// Driver code
public static void Main(String[] args)
{
 
    // For large input number n
    String n = "98765432109876543210";
 
    // Total number of digits in n
    int len = n.Length;
 
    int k = 58;
 
    // Clean dp table
    for(int i = 0; i < MAX; i++)
    {
        for(int j = 0; j < 101; j++)
        {
            for(int l = 0; l < 2; l++)
            dp[i, j, l] = -1;
        }
    }
 
    // Process the String to a vector
    // of digits from MSD to LSD
    List<int> num = process(n);
 
    Console.Write(countNum(0, 0, 0, num, len, k));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the approach
 
var MAX = 100005;
var MOD = 1000000007;
 
// To store the states of the dp
var dp = Array.from(Array(MAX), () => Array(101));
for(var i =0; i<MAX; i++)
        for(var j =0; j<101; j++)
            dp[i][j] = new Array(2).fill(-1);
 
// Function to return the count of numbers
// from the range [0, n] whose digit sum
// is a multiple of k using bottom-up dp
function countNum(idx, sum, tight, num, len, k)
{
    if (len == idx) {
        if (sum == 0)
            return 1;
        else
            return 0;
    }
    if (dp[idx][sum][tight] != -1)
        return dp[idx][sum][tight];
    var res = 0, limit;
 
    // The digit in this index can
    // only be from [0, num[idx]]
    if (tight == 0) {
        limit = num[idx];
    }
 
    // The digit in this index can
    // be anything from [0, 9]
    else {
        limit = 9;
    }
    for (var i = 0; i <= limit; i++) {
 
        // new_tight is the flag value
        // for the next position
        var new_tight = tight;
        if (tight == 0 && i < limit)
            new_tight = 1;
        res += countNum(idx + 1,
                        (sum + i) % k, new_tight,
                        num, len, k);
        res %= MOD;
    }
 
    // res can't be negative
    if (res < 0)
        res += MOD;
    return dp[idx][sum][tight] = res;
}
 
// Function to process the string to
// a vector of digits from MSD to LSD
function process(s)
{
    var num = [];
    for (var i = 0; i < s.length; i++) {
        num.push(s[i].charCodeAt(0) - '0'.charCodeAt(0));
    }
    return num;
}
 
// Driver code
// For large input number n
var n = "98765432109876543210";
// Total number of digits in n
var len = n.length;
var k = 58;
// Process the string to a vector
// of digits from MSD to LSD
var num = process(n);
document.write( countNum(0, 0, 0, num, len, k));
 
</script>
Output: 
635270835

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :