Count numbers (smaller than or equal to N) with given digit sum

Given a number N and a sum S, find the count of numbers upto N that have digit sum equal to S.

Examples:
Input : N = 100, S = 4
Output : 5
Upto 100 only 5 numbers(4, 13, 22, 31, 40)
can produce 4 as their sum of digits.

Input : N = 1000, S = 1
Output : 4
Upto 1000 only 4 numbers(1, 10, 100 and 1000) 
can produce 1 as their sum of digits.

We can do a digit DP having state (current index, whether the currently constructed number of i digits is equal or less than the number formed by first i digits of N, the sum of digits of currently constructed number).

Let dp[i][tight][sum_so_far] denotes the count of numbers whose first i digits have been considered and tight denotes whether the currently constructed number is equal or less than number formed by first i digits of N. If tight is true, then it means that currently constructed number is equal to the number constructed by first i digits of N. If it is false then it means that currently constructed number is less than the number constructed by first i digits of N. sum_so_far denotes the sum of digits of currently constructed number.

Base Case:
If i = number of digits in N, sum_so_far = Sum, then answer = 1 else answer is 0.

Transitions:
For filling (i+1)th digit we can consider following –
If tight is true, then it means that our constructed number is still equal to number formed by first i digits of N. We can try our current possible digit value from 0 to (i+1)th digit of N. If we try the digit more than (i+1)th digit, then the constructed number will become greater than number formed by first i digits of N, which will violate the property that our constructed number should be <= N.
If tight is false, then it means that number constructed from first i – 1 digits has become less than number constructed from the first i – 1 digit of N, So it means that our number can never exceed N, so we can chose the any digit from 0 to 9.
nsum_so_far can be obtained by adding sum_so_far and current digit(currdigit).

Finally we will return answer which is count of numbers upto N that have digit sum equal to S.

C++

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#include <bits/stdc++.h>
using namespace std;
  
// N can be max 10^18 and hence digitsum will be 162 maximum.
long long dp[18][2][162];
  
long long solve(int i, bool tight, int sum_so_far,
                int Sum, string number, int len)
{
    if (i == len) {
  
        // If sum_so_far equals to given sum then 
        // return 1 else 0
        if (sum_so_far == Sum)
            return 1;
        else
            return 0;
    }
  
    long long& ans = dp[i][tight][sum_so_far];
    if (ans != -1) {
        return ans;
    }
  
    ans = 0;
    bool ntight;
    int nsum_so_far;
    for (char currdigit = '0'; currdigit <= '9'; currdigit++) {
  
        // Our constructed number should not become
        // greater than N.
        if (!tight && currdigit > number[i]) {
            break;
        }
  
        // If tight is true then it will also be true for (i+1) digit.
        ntight = tight || currdigit < number[i];
        nsum_so_far = sum_so_far + (currdigit - '0');
        ans += solve(i + 1, ntight, nsum_so_far, Sum, number, len);
    }
  
    return ans;
}
  
// Driver code
int main()
{
    long long count = 0;
    long long sum = 4;
    string number = "100";
    memset(dp, -1, sizeof dp);
    cout << solve(0, 0, 0, sum, number, number.size());
    return 0;
}

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Java

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// Java program to count 2s from 0 to n 
class GFG 
{
  
// N can be max 10^18 and hence 
// digitsum will be 162 maximum. 
static long dp[][][] = new long[18][2][162]; 
  
static long solve(int i, boolean tight, int sum_so_far, 
                int Sum, String number, int len) 
    if (i == len) 
    
  
        // If sum_so_far equals to given sum then 
        // return 1 else 0 
        if (sum_so_far == Sum) 
            return 1
        else
            return 0
    
  
    long ans = dp[i][1][sum_so_far]; 
    if (ans != -1
    
        return ans; 
    
  
    ans = 0
    boolean ntight; 
    int nsum_so_far; 
    for (char currdigit = '0'; currdigit <= '9'; currdigit++) 
    
  
        // Our constructed number should not become 
        // greater than N. 
        if (!tight && currdigit > number.charAt(i))
        
            break
        
  
        // If tight is true then it will 
        // also be true for (i+1) digit. 
        ntight = tight || currdigit < number.charAt(i); 
        nsum_so_far = sum_so_far + (currdigit - '0'); 
        ans += solve(i + 1, ntight, nsum_so_far, 
                        Sum, number, len); 
    
    return ans; 
  
// Driver code 
public static void main(String[] args) 
{
    long count = 0
    int sum = 4
    String number = "100"
    for(int i = 0; i < 18; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            for(int k = 0; k < 162; k++)
            dp[i][j][k] = -1;
        }
    }
    System.out.println( solve(0, false, 0, sum, 
                        number, number.length())); 
    }
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 implementation of the given approach. 
  
def solve(i, tight, sum_so_far, Sum, number, length): 
  
    if i == length: 
  
        # If sum_so_far equals to given 
        # sum then return 1 else 0 
        if sum_so_far == Sum
            return 1
        else:
            return 0
      
    ans = dp[i][tight][sum_so_far] 
    if ans != -1
        return ans 
      
    ans = 0
    for currdigit in range(0, 10): 
  
        currdigitstr = str(currdigit)
          
        # Our constructed number should 
        # not become greater than N. 
        if not tight and currdigitstr > number[i]: 
            break
  
        # If tight is true then it will also be true for (i+1) digit. 
        ntight = tight or currdigitstr < number[i] 
        nsum_so_far = sum_so_far + currdigit 
        ans += solve(i + 1, ntight, nsum_so_far, Sum, number, length) 
      
    return ans 
  
# Driver code 
if __name__ == "__main__":
  
    count, Sum = 0, 4
    number = "100"
    dp = [[[-1 for i in range(162)] for j in range(2)] for k in range(18)]
    print(solve(0, 0, 0, Sum, number, len(number)))
  
# This code is contributed by Rituraj Jain

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C#

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// C# program to count 2s from 0 to n 
using System;
  
class GFG 
  
    // N can be max 10^18 and hence 
    // digitsum will be 162 maximum. 
    static long [ , , ]dp = new long[18,2,162]; 
  
    static long solve(int i, bool tight, int sum_so_far, 
                    int Sum, String number, int len) 
    
        if (i == len) 
        
  
            // If sum_so_far equals to given sum then 
            // return 1 else 0 
            if (sum_so_far == Sum) 
                return 1; 
            else
                return 0; 
        
  
        long ans = dp[i,1,sum_so_far]; 
        if (ans != -1) 
        
            return ans; 
        
  
        ans = 0; 
        bool ntight; 
        int nsum_so_far; 
        for (char currdigit = '0'; currdigit <= '9'; currdigit++) 
        
  
            // Our constructed number should not become 
            // greater than N. 
            if (!tight && currdigit > number[i]) 
            
                break
            
  
            // If tight is true then it will 
            // also be true for (i+1) digit. 
            ntight = tight || currdigit < number[i]; 
            nsum_so_far = sum_so_far + (currdigit - '0'); 
            ans += solve(i + 1, ntight, nsum_so_far, 
                            Sum, number, len); 
        
        return ans; 
    
  
    // Driver code 
    public static void Main(String[] args) 
    
        int sum = 4; 
        String number = "100"
        for(int i = 0; i < 18; i++) 
        
            for(int j = 0; j < 2; j++) 
            
                for(int k = 0; k < 162; k++) 
                dp[i,j,k] = -1; 
            
        
        Console.WriteLine( solve(0, false, 0, sum, 
                            number, number.Length)); 
        
  
// This code has been contributed by Rajput-Ji

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Output:

5

Time Complexity: 2(tight) * Sum * log 10(N) * 10(transitions) = 20*log 10(N)*Sum.



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I am a second year BTech (CSE) engineering student at GB Pant Goverment Engineering College Delhi(IPU)

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