Compute sum of digits in all numbers from 1 to n

brahmajit-mohapatra-f8fe5582 · 9 June 2021 08:16

Hello Everybody,

Given a number n, find the 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 the sum in x by traversing all digits of x. Below is the implementation of this idea.

// A Simple C++ program to compute sum of digits in numbers from 1 to n

#include<bits/stdc++.h>

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 = 910/2 = 45 sum(99) = 45 + (10 + 45) + (20 + 45) + … (90 + 45) = 4510 + (10 + 20 + 30 … 90) = 4510 + 10(1 + 2 + … 9) = 4510 + 45*10 = sum(9)10 + 4510 sum(999) = sum(99)10 + 45100

In general, we can compute sum(10d – 1) using the below formula

sum(10d - 1) = sum(10d-1 - 1) * 10 + 45*(10d-1)

In the 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 the complete algorithm

Algorithm: sum(n)

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 3sum(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) + 1100 and sum of 299 is sum(99) + 2100. In general, this sum can be computed as wmsd + (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 329 + recursive call “sum(28)” In general, this sum can be computed as msd * (n % (msd*10d) + 1) + sum(n % (10d))

Below is the implementation of the above algorithm.

// 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.

Improvement:
The above implementation takes O(d2) time as each recursive call calculates dp[] array once again. The first call takes O(d), the second call takes O(d-1), the third call O(d-2), and so on. We don’t need to recalculate dp[] array in each recursive call. Below is the modified implementation which works in O(d) time. Where d is a number of digits in the input number.

// C++ program to compute sum of digits

// in numbers from 1 to n

#include<bits/stdc++.h>

using namespace std;

int sumOfDigitsFrom1ToNUtil( int n, int a[])

{

if (n < 10)

return (n * (n + 1) / 2);

int d = ( int )( log10 (n));

int p = ( int )( ceil ( pow (10, d)));

int msd = n / p;

return (msd * a[d] + (msd * (msd - 1) / 2) * p +

msd * (1 + n % p) +

sumOfDigitsFrom1ToNUtil(n % p, a));

}

// Function to computer sum of digits in

// numbers from 1 to n

int sumOfDigitsFrom1ToN( int n)

{

int d = ( int )( log10 (n));

int a[d + 1];

a[0] = 0; a[1] = 45;

for ( int i = 2; i <= d; i++)

a[i] = a[i - 1] * 10 + 45 *

( int )( ceil ( pow (10, i - 1)));

return sumOfDigitsFrom1ToNUtil(n, a);

}

// Driver code

int main()

{

int n = 328;

cout << "Sum of digits in numbers from 1 to "

<< n << " is " << sumOfDigitsFrom1ToN(n);

}

Output:

Sum of digits in numbers from 1 to 328 is 3241