 # What is the logic behind this problem "Number of ways" ?

#1

How do I solve this problem ? Number of ways . I read the editorial here but didnt quite understand why cnt[i+2] is being added when ss==s . The editorialist’s solution can be found here.

#2

I think it will become clear with an example.

Let’s assume the given array is `a = [0 0 0 0 0]`.

“Let’s create an array `cnt[]`, where `cnt*` equals 1, if the sum of elements from i-th to n-th equals `S/3` and 0 — otherwise.” -where `S` is the sum of all the elements of the array `a`. Constructing the `cnts` array, we get
`a = [0 0 0 0 0]`
`cnts = [1 1 1 1 1]`

"Now, to calculate the answer we have to find the sum `cnt[j] + cnt[j+1] + ... + cnt[n]` faster then O(n). There are a lot of required ways to do this, but the easiest one is to create a new additional array `sums[]` where in j-th element will be `cnt[j] + cnt[j+1] + ... + cnt[n]`"
`a = [0 0 0 0 0]`
`cnts = [1 1 1 1 1]`
`sums = [5 4 3 2 1]`

“Thus, we receive very simple solution: for each prefix of initial array 1…i with the sum that equals we need to add to the answer sums[i+2].”

We are required to break the given array into 3 pieces. If we are at the i-th element and the sum of elements from 0 to i is equal to `S/3` then the segment from 0 to i is a valid first piece of the array. And for all the position j where `cnts[j]` is 1, the segment from j to the end is a valid third piece of the array. Since we require a second piece of the array, there must be a segment of non-zero length between these 2 segments, which is why the “i+2” comes in. It ensures that the second segment is of length ≥ 1. Hence if you choose the given prefix 0 to i with sum `S/3` as the first piece, then the number of ways to choose the third piece is equal to the number of positions ≥ i+2 where the `cnts[]` value is 1. This is nothing but `sums[i+2]`.

Getting back to the example, only 3 prefixes within acceptable range have sum `S/3` marked with `|` and their +2 positions marked with `^`.
`a = [0 0 0 0 0]`
`cnts = [1 1 1 1 1]`
`sums = [5 4 3 2 1]`
`. |---^`
`. |---^`
`. |---^`

The answer is hence 3+2+1=6, and the actual 6 ways are
`[0 | 0 | 0 0 0]`
`[0 | 0 0 | 0 0]`
`[0 | 0 0 0 | 0]`
`[0 0 | 0 | 0 0]`
`[0 0 | 0 0 | 0]`
`[0 0 0 | 0 | 0]`

#3

Thanks a lot mate.

#4

No problem 