Sum of Digits till a number which is given as input - algorithm

Question

If a number is given as an input find sum of all the digits of number till that number
For example 11 is input then answer is 1+2....+9+(1+0)+(1+1)
The Brute-force method would be to calculate sum of digits of all the numbers that are less than a number.I have implemented that method iam wondering if there is any other way to do it without actually calculating sum of digits of every number

Answer 1

You can do that faster (in O(log n) operations). Let S(n) be the sum of the digits of all numbers 0 <= k < n. Then
S(10*n) = 10*S(n) + 45*n
because among the numbers less than 10*n, each k < n appears as the initial part of a number 10 times, with last digits 0, 1, ..., 9. So that contributes 45 for the sum of the last digits, and 10 times the sum of the digits of k.
Reversing that, we find
S(n) = 10*S(n/10) + 45*(n/10) + (n%10)*DS(n/10) + (n%10)*((n%10)-1)/2
where DS(k) is the plain digit sum of k. The first two terms come from the above, the remaining two come from the sum of the digits of n - n%10, ..., n - n%10 + (n%10 + 1).
Start is S(n) = 0 for n <= 1.
To include the upper bound, call it as S(n+1).

Answer 2

Let us take few examples.
sum(9) = 1 + 2 + 3 + 4 ........... + 9
= 9*10/2
= 45
sum(99) = 45 + (10 + 45) + (20 + 45) + ..... (90 + 45)
= 45*10 + (10 + 20 + 30 ... 90)
= 45*10 + 10(1 + 2 + ... 9)
= 45*10 + 45*10
= sum(9)*10 + 45*10
sum(999) = sum(99)*10 + 45*100
In general, we can compute sum(10d – 1) using below formula
sum(10d - 1) = sum(10d-1 - 1) * 10 + 45*(10d-1)
In below implementation, the above formula is implemented using dynamic programming as there are overlapping subproblems.
The above formula is one core step of the idea. Below is complete algorithm
Algorithm: sum(n)
1) Find number of digits minus one in n. Let this value be 'd'.
For 328, d is 2.
2) Compute some of digits in numbers from 1 to 10d - 1. Let this
sum be w. For 328, we compute sum of digits from 1 to 99 using
above formula.
3) Find Most significant digit (msd) in n. For 328, msd is 3.
4) Overall sum is sum of following terms
a) Sum of digits in 1 to "msd * 10d - 1". For 328, sum of
digits in numbers from 1 to 299.
For 328, we compute 3*sum(99) + (1 + 2)*100. Note that sum of
sum(299) is sum(99) + sum of digits from 100 to 199 + sum of digits
from 200 to 299.
Sum of 100 to 199 is sum(99) + 1*100 and sum of 299 is sum(99) + 2*100.
In general, this sum can be computed as w*msd + (msd*(msd-1)/2)*10d
b) Sum of digits in msd * 10d to n. For 328, sum of digits in
300 to 328.
For 328, this sum is computed as 3*29 + recursive call "sum(28)"
In general, this sum can be computed as msd * (n % (msd*10d) + 1)
+ sum(n % (10d))
Below is C++ implementation of above aglorithm.
// C++ program to compute sum of digits in numbers from 1 to n
#include<bits/stdc++.h>
using namespace std;
// Function to computer sum of digits in numbers from 1 to n
// Comments use example of 328 to explain the code
int sumOfDigitsFrom1ToN(int n)
{
// base case: if n<10 return sum of
// first n natural numbers
if (n<10)
return n*(n+1)/2;
// d = number of digits minus one in n. For 328, d is 2
int d = log10(n);
// computing sum of digits from 1 to 10^d-1,
// d=1 a[0]=0;
// d=2 a[1]=sum of digit from 1 to 9 = 45
// d=3 a[2]=sum of digit from 1 to 99 = a[1]*10 + 45*10^1 = 900
// d=4 a[3]=sum of digit from 1 to 999 = a[2]*10 + 45*10^2 = 13500
int *a = new int[d+1];
a[0] = 0, a[1] = 45;
for (int i=2; i<=d; i++)
a[i] = a[i-1]*10 + 45*ceil(pow(10,i-1));
// computing 10^d
int p = ceil(pow(10, d));
// Most significant digit (msd) of n,
// For 328, msd is 3 which can be obtained using 328/100
int msd = n/p;
// EXPLANATION FOR FIRST and SECOND TERMS IN BELOW LINE OF CODE
// First two terms compute sum of digits from 1 to 299
// (sum of digits in range 1-99 stored in a[d]) +
// (sum of digits in range 100-199, can be calculated as 1*100 + a[d]
// (sum of digits in range 200-299, can be calculated as 2*100 + a[d]
// The above sum can be written as 3*a[d] + (1+2)*100
// EXPLANATION FOR THIRD AND FOURTH TERMS IN BELOW LINE OF CODE
// The last two terms compute sum of digits in number from 300 to 328
// The third term adds 3*29 to sum as digit 3 occurs in all numbers
// from 300 to 328
// The fourth term recursively calls for 28
return msd*a[d] + (msd*(msd-1)/2)*p +
msd*(1+n%p) + sumOfDigitsFrom1ToN(n%p);
}
// Driver Program
int main()
{
int n = 328;
cout << "Sum of digits in numbers from 1 to " << n << " is "
<< sumOfDigitsFrom1ToN(n);
return 0;
}
Output
Sum of digits in numbers from 1 to 328 is 3241