# Compute sum of digits in all numbers from 1 to n

Given a number x, find sum of digits in all numbers from 1 to n.
Examples:

```Input: n = 5
Output: Sum of digits in numbers from 1 to 5 = 15

Input: n = 12
Output: Sum of digits in numbers from 1 to 12 = 51

Input: n = 328
Output: Sum of digits in numbers from 1 to 328 = 3241
```

Naive Solution:
A naive solution is to go through every number x from 1 to n, and compute sum in x by traversing all digits of x. Below is C++ implementation of this idea.

`// A Simple C++ program to compute sum of digits in numbers from 1 to n`
`#include<iostream>`
`using` `namespace` `std;`
`int` `sumOfDigits(``int` `);`
`// Returns sum of all digits in numbers from 1 to n`
`int` `sumOfDigitsFrom1ToN(``int` `n)`
`{`
`    ``int` `result = 0; ``// initialize result`
`    ``// One by one compute sum of digits in every number from`
`    ``// 1 to n`
`    ``for` `(``int` `x=1; x<=n; x++)`
`        ``result += sumOfDigits(x);`
`    ``return` `result;`
`}`
`// A utility function to compute sum of digits in a`
`// given number x`
`int` `sumOfDigits(``int` `x)`
`{`
`    ``int` `sum = 0;`
`    ``while` `(x != 0)`
`    ``{`
`        ``sum += x %10;`
`        ``x   = x /10;`
`    ``}`
`    ``return` `sum;`
`}`
`// 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`

Efficient Solution:
Above is a naive solution. We can do it more efficiently by finding a pattern.

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`

The efficient algorithm has one more advantage that we need to compute the array ‘a[]’ only once even when we are given multiple inputs.

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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