Sum of all substrings of a string representing a number

Given a integer represented as a string, we need to get the sum of all possible substrings of this string.

Examples:

Input  : num = “1234”
Output : 1670
Sum = 1 + 2 + 3 + 4 + 12 + 23 +
       34 + 123 + 234 + 1234 
    = 1670

Input  : num = “421”
Output : 491
Sum = 4 + 2 + 1 + 42 + 21 + 421
    = 491

We can solve this problem using dynamic programming. We can write summation of all substrings on basis of digit at which they are ending in that case,
Sum of all substrings = sumofdigit[0] + sumofdigit[1] + sumofdigit[2] … + sumofdigit[n-1] where n is length of string.

Where sumofdigit[i] stores sum of all substring ending at ith index digit, in above example,

Example : num = "1234"
sumofdigit[0] = 1 = 1
sumofdigit[1] = 2 + 12  = 14
sumofdigit[2] = 3 + 23  + 123 = 149
sumofdigit[3] = 4 + 34  + 234 + 1234  = 1506
Result = 1670

Now we can get the relation between sumofdigit values and can solve the question iteratively. Each sumofdigit can be represented in terms of previous value as shown below,

For above example,
sumofdigit[3] = 4 + 34 + 234 + 1234
		   = 4 + 30 + 4 + 230 + 4 + 1230 + 4
		   = 4*4 + 10*(3 + 23 +123)
		   = 4*4 + 10*(sumofdigit[2])
In general, 
sumofdigit[i]  =  (i+1)*num[i] + 10*sumofdigit[i-1]

Using above relation we can solve the problem in linear time. In below code a complete array is taken to store sumofdigit, as each sumofdigit value requires just previous value, we can solve this problem without allocating complete array also.

//  C++ program to print sum of all substring of
// a number represented as a string
#include <bits/stdc++.h>
using namespace std;

// Utility method to covert character digit to
// integer digit
int toDigit(char ch)
{
    return (ch - '0');
}

// Returns sum of all substring of num
int sumOfSubstrings(string num)
{
    int n = num.length();

    //  allocate memory equal to length of string
    int sumofdigit[n];

    //  initialize first value with first digit
    sumofdigit[0] = toDigit(num[0]);
    int res = sumofdigit[0];

    //  loop over all digits of string
    for (int i=1; i<n; i++)
    {
        int numi = toDigit(num[i]);

        // update each sumofdigit from previous value
        sumofdigit[i] = (i+1) * numi +
                        10 * sumofdigit[i-1];

        // add current value to the result
        res += sumofdigit[i];
    }

    return res;
}

//  Driver code to test above methods
int main()
{
    string num = "1234";
    cout << sumOfSubstrings(num) << endl;
    return 0;
}

Output:

1670

Time Complexity : O(n) where n is length of input string.
Auxiliary Space : O(n)

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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