Given a integer represented as a string, we need to get the sum of all possible substrings of this string.
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 + sumofdigit + sumofdigit … + 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 = 1 = 1 sumofdigit = 2 + 12 = 14 sumofdigit = 3 + 23 + 123 = 149 sumofdigit = 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 = 4 + 34 + 234 + 1234 = 4 + 30 + 4 + 230 + 4 + 1230 + 4 = 4*4 + 10*(3 + 23 +123) = 4*4 + 10*(sumofdigit) 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.
Time Complexity : O(n) where n is length of input string.
Auxiliary Space : O(n)
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O(1) space approach
Approach is same as described above. What we have observed that at current index we are dependent on current sum + previous index sum so instead of storing in a dp array we can store it in two variables current and prev.
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