Given a string consisting of integers 0 to 9. The task is to count the number of substrings which when convert into integer are divisible by 6. Substring does not contain leading zeroes.
Input : s = "606". Output : 5 Substrings "6", "0", "6", "60", "606" are divisible by 6. Input : s = "4806". Output : 5 "0", "6", "48", "480", "4806" are substring which are divisible by 6.
Method 2:(Dynamic Programming) As discussed in Check if a large number is divisible by 6 or not. A number is divisible by 6 if last digit is divisible by 2 and sum of digits is divisible by 3.
The idea is to use Dynamic Programming, which enables us to compute answer quickly and efficiently by tracking previously computed answers and using these stored answer instead of recomputing values.
Let f(i, m) be the number of strings starting at index i and sum of their digits modulo 3 (so far) is m and number it represents is even. So, our answer would be
Let x be the ith digit in the string. From f(i, m) we need to find all the even substrings that start in i + 1.
Also, we will get an extra substring if (x + m) itself is divisible by 3 and x is even. So, we get recurrence relation
// We initially pass m (sum modulo 3 so far) as 0 f(i, m) = ((x + m)%3 == 0 and x%2 == 0) + f(i + 1, (m + x)%3) // Recursive
By memorizing the states, we get O(n) solution.
Below is implementation of this approach:
Time Complexity: O(n).
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