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

Last Updated : 16 Apr, 2024

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

Examples:

`Input: n = 5Output: Sum of digits in numbers from 1 to 5 = 15Input: n = 12Output: Sum of digits in numbers from 1 to 12 = 51Input: n = 328Output: 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.
Implementation:

C++ ```// 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; } ``` C ```// A Simple C program to compute sum of digits in numbers from 1 to n #include<stdio.h> //Function declaration int sumOfDigits(int x); int sumOfDigitsFrom1ToN(int n); // 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; printf("Sum of digits in numbers from 1 to %d is : %d",n,sumOfDigitsFrom1ToN(n)); return 0; } ``` Java ```// A Simple JAVA program to compute sum of // digits in numbers from 1 to n import java.io.*; class GFG { // Returns sum of all digits in numbers // from 1 to n static 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 static int sumOfDigits(int x) { int sum = 0; while (x != 0) { sum += x % 10; x = x / 10; } return sum; } // Driver Program public static void main(String args[]) { int n = 328; System.out.println("Sum of digits in numbers" +" from 1 to " + n + " is " + sumOfDigitsFrom1ToN(n)); } } /*This code is contributed by Nikita Tiwari.*/ ``` Python3 ```# A Simple Python program to compute sum # of digits in numbers from 1 to n # Returns sum of all digits in numbers # from 1 to n def sumOfDigitsFrom1ToN(n) : result = 0 # initialize result # One by one compute sum of digits # in every number from 1 to n for x in range(1, n+1) : result = result + sumOfDigits(x) return result # A utility function to compute sum of # digits in a given number x def sumOfDigits(x) : sum = 0 while (x != 0) : sum = sum + x % 10 x = x // 10 return sum # Driver Program n = 328 print("Sum of digits in numbers from 1 to", n, "is", sumOfDigitsFrom1ToN(n)) # This code is contributed by Nikita Tiwari. ``` C# ```// A Simple C# program to compute sum of // digits in numbers from 1 to n using System; public class GFG { // Returns sum of all digits in numbers // from 1 to n static int sumOfDigitsFrom1ToN(int n) { // initialize result int result = 0; // 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 static int sumOfDigits(int x) { int sum = 0; while (x != 0) { sum += x % 10; x = x / 10; } return sum; } // Driver Program public static void Main() { int n = 328; Console.WriteLine("Sum of digits" + " in numbers from 1 to " + n + " is " + sumOfDigitsFrom1ToN(n)); } } // This code is contributed by shiv_bhakt. ``` Javascript ```<script> // A Simple Javascript program to compute sum of // digits in numbers from 1 to n // Returns sum of all digits in numbers // from 1 to n function sumOfDigitsFrom1ToN(n) { // initialize result let result = 0; // One by one compute sum of digits // in every number from 1 to n for (let x = 1; x <= n; x++) result += sumOfDigits(x); return result; } // A utility function to compute sum // of digits in a given number x function sumOfDigits(x) { let sum = 0; while (x != 0) { sum += x % 10; x = parseInt(x / 10, 10); } return sum; } let n = 328; document.write("Sum of digits" + " in numbers from 1 to " + n + " is " + sumOfDigitsFrom1ToN(n)); // This code is contributed by divyeshrabadiya07. </script> ``` PHP ```<?php // A Simple php program to compute sum //of digits in numbers from 1 to n // Returns sum of all digits in // numbers from 1 to n function sumOfDigitsFrom1ToN(\$n) { \$result = 0; // initialize result // One by one compute sum of digits // in every number from 1 to n for (\$x = 1; \$x <= \$n; \$x++) \$result += sumOfDigits(\$x); return \$result; } // A utility function to compute sum // of digits in a given number x function sumOfDigits(\$x) { \$sum = 0; while (\$x != 0) { \$sum += \$x %10; \$x = \$x /10; } return \$sum; } // Driver Program \$n = 328; echo "Sum of digits in numbers from" . " 1 to " . \$n . " is " . sumOfDigitsFrom1ToN(\$n); // This code is contributed by ajit ?> ```

Output
```Sum of digits in numbers from 1 to 328 is 3241
```

Time complexity: O(N*len(N)), where len(X) gives the no. of digits in X.
Auxiliary Space: O(1)

Efficient Solution:

Above is a naive solution. We can do it more efficiently by finding a pattern.
Let us take a few examples.

`sum(9) = 1 + 2 + 3 + 4 ........... + 9       = 9*10/2        = 45sum(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 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)

`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 the implementation of the above algorithm.

C++ ```// 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; } ``` Java ```// JAVA program to compute sum of digits // in numbers from 1 to n import java.io.*; import java.math.*; class GFG{ // Function to computer sum of digits in // numbers from 1 to n. Comments use // example of 328 to explain the code static 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 = (int)(Math.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 * (int)(Math.ceil(Math.pow(10, i-1))); // computing 10^d int p = (int)(Math.ceil(Math.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 public static void main(String args[]) { int n = 328; System.out.println("Sum of digits in numbers " + "from 1 to " +n + " is " + sumOfDigitsFrom1ToN(n)); } } /*This code is contributed by Nikita Tiwari.*/ ``` Python3 ```# PYTHON 3 program to compute sum of digits # in numbers from 1 to n import math # Function to computer sum of digits in # numbers from 1 to n. Comments use example # of 328 to explain the code def sumOfDigitsFrom1ToN( 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 d = (int)(math.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""" a = [0] * (d + 1) a[0] = 0 a[1] = 45 for i in range(2, d+1) : a[i] = a[i-1] * 10 + 45 * (int)(math.ceil(math.pow(10,i-1))) # computing 10^d p = (int)(math.ceil(math.pow(10, d))) # Most significant digit (msd) of n, # For 328, msd is 3 which can be obtained # using 328/100 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 (int)(msd * a[d] + (msd*(msd-1) // 2) * p + msd * (1 + n % p) + sumOfDigitsFrom1ToN(n % p)) # Driver Program n = 328 print("Sum of digits in numbers from 1 to", n ,"is",sumOfDigitsFrom1ToN(n)) # This code is contributed by Nikita Tiwari. ``` C# ```// C# program to compute sum of digits // in numbers from 1 to n using System; public class GFG { // Function to computer sum of digits in // numbers from 1 to n. Comments use // example of 328 to explain the code static 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 = (int)(Math.Log(n) / Math.Log(10)); // 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 * (int)(Math.Ceiling(Math.Pow(10, i-1))); // computing 10^d int p = (int)(Math.Ceiling(Math.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 public static void Main() { int n = 328; Console.WriteLine("Sum of digits in numbers " + "from 1 to " +n + " is " + sumOfDigitsFrom1ToN(n)); } } // This code is contributed by shiv_bhakt. ``` Javascript ```<script> // Javascript program to compute sum of digits // in numbers from 1 to n // Function to computer sum of digits in // numbers from 1 to n. Comments use // example of 328 to explain the code function sumOfDigitsFrom1ToN(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 let d = parseInt(Math.log(n) / Math.log(10), 10); // 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 let a = new Array(d+1); a[0] = 0; a[1] = 45; for (let i = 2; i <= d; i++) a[i] = a[i-1] * 10 + 45 * parseInt(Math.ceil(Math.pow(10, i-1)), 10); // computing 10^d let p = parseInt(Math.ceil(Math.pow(10, d)), 10); // Most significant digit (msd) of n, // For 328, msd is 3 which can be obtained // using 328/100 let msd = parseInt(n / p, 10); // 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)); } let n = 328; document.write("Sum of digits in numbers from 1 to " + n + " is " + sumOfDigitsFrom1ToN(n)); </script> ``` PHP ```<?php // PHP program to compute sum of digits // in numbers from 1 to n // Function to computer sum of digits in // numbers from 1 to n. Comments use // example of 328 to explain the code function sumOfDigitsFrom1ToN(\$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 \$d = (int)(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 \$a[\$d + 1] = array(); \$a[0] = 0; \$a[1] = 45; for (\$i = 2; \$i <= \$d; \$i++) \$a[\$i] = \$a[\$i - 1] * 10 + 45 * (int)(ceil(pow(10, \$i - 1))); // computing 10^d \$p = (int)(ceil(pow(10, \$d))); // Most significant digit (msd) of n, // For 328, msd is 3 which can be obtained // using 328/100 \$msd = (int)(\$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 * (int)(\$msd - 1) / 2) * \$p + \$msd * (1 + \$n % \$p) + sumOfDigitsFrom1ToN(\$n % \$p)); } // Driver Code \$n = 328; echo ("Sum of digits in numbers " ), "from 1 to " , \$n , " is ", sumOfDigitsFrom1ToN(\$n); // This code is contributed by Sachin ?> ```

Output
```Sum of digits in numbers from 1 to 328 is 3241
```

Time complexity: O((len(N))2)
Auxiliary Space: O(len(N))

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++ ```// 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 = max(((int)(log10(n))), 1); 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); } // This code is contributed by ajaykr00kj // This code is modified by Susobhan Akhuli ``` Java ```// JAVA program to compute sum of digits // in numbers from 1 to n import java.io.*; import java.math.*; class GFG { // Function to computer sum of digits in // numbers from 1 to n static int sumOfDigitsFrom1ToN(int n) { int d = Math.max(((int)(Math.log10(n))), 1); 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 * (int)(Math.ceil( Math.pow(10, i - 1))); return sumOfDigitsFrom1ToNUtil(n, a); } static int sumOfDigitsFrom1ToNUtil(int n, int a[]) { if (n < 10) return (n * (n + 1) / 2); int d = (int)(Math.log10(n)); int p = (int)(Math.ceil(Math.pow(10, d))); int msd = n / p; return (msd * a[d] + (msd * (msd - 1) / 2) * p + msd * (1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a)); } // Driver Program public static void main(String args[]) { int n = 328; System.out.println("Sum of digits in numbers " + "from 1 to " + n + " is " + sumOfDigitsFrom1ToN(n)); } } /*This code is contributed by Narendra Jha.*/ // This code is modified by Susobhan Akhuli ``` Python3 ```import math # Function to compute sum of digits in # numbers from 1 to n def sumOfDigitsFrom1ToN(n): d = max(int(math.log(n, 10)), 1) a = [0]*(d + 1) a[0] = 0 a[1] = 45 for i in range(2, d + 1): a[i] = a[i - 1] * 10 + 45 * \ int(math.ceil(pow(10, i - 1))) return sumOfDigitsFrom1ToNUtil(n, a) def sumOfDigitsFrom1ToNUtil(n, a): if (n < 10): return (n * (n + 1)) // 2 d = int(math.log(n,10)) p = int(math.ceil(pow(10, d))) msd = n // p return (msd * a[d] + (msd * (msd - 1) // 2) * p + msd * (1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a)) # Driver code n = 328 print("Sum of digits in numbers from 1 to",n,"is",sumOfDigitsFrom1ToN(n)) ``` C# ```// C# program to compute sum of digits // in numbers from 1 to n using System; class GFG { // Function to computer sum of digits in // numbers from 1 to n static int sumOfDigitsFrom1ToN(int n) { int d = Math.Max(((int)(Math.Log10(n))), 1); 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 * (int)(Math.Ceiling( Math.Pow(10, i - 1))); return sumOfDigitsFrom1ToNUtil(n, a); } static int sumOfDigitsFrom1ToNUtil(int n, int[] a) { if (n < 10) return (n * (n + 1) / 2); int d = (int)(Math.Log10(n)); int p = (int)(Math.Ceiling(Math.Pow(10, d))); int msd = n / p; return (msd * a[d] + (msd * (msd - 1) / 2) * p + msd * (1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a)); } // Driver code public static void Main(String[] args) { int n = 328; Console.WriteLine("Sum of digits in numbers " + "from 1 to " + n + " is " + sumOfDigitsFrom1ToN(n)); } } // This code contributed by Rajput-Ji // This code is modified by Susobhan Akhuli ``` Javascript ```<script> // JavaScript program to compute sum of digits // in numbers from 1 to n // Function to computer sum of digits in // numbers from 1 to n function sumOfDigitsFrom1ToNUtil( n,a) { if (n < 10) return (n * (n + 1) / 2); var d = (parseInt)(Math.log10(n)); var p = (Math.ceil(Math.pow(10, d))); var msd =(parseInt) (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 function sumOfDigitsFrom1ToN( n) { var d = Math.max(((parseInt)(Math.log10(n))), 1); var a=new Array(d + 1).fill(0); a[0] = 0; a[1] = 45; for(var i = 2; i <= d; i++) a[i] = a[i - 1] * 10 + 45 * (parseInt)(Math.ceil(Math.pow(10, i - 1))); return sumOfDigitsFrom1ToNUtil(n, a); } var n = 328; document.write( "Sum of digits in numbers from 1 to " + n + " is "+ sumOfDigitsFrom1ToN(n)) // This code is modified by Susobhan Akhuli </script> ```

Output
```Sum of digits in numbers from 1 to 328 is 3241
```

Time complexity: O(len(N))
Auxiliary Space: O(len(N))