## PROBLEM LINK:

**Author:** Misha Chorniy

**Tester:** Lewin Gan

**Editorialist:** Adarsh Kumar

## DIFFICULTY:

Easy-Medium

## PREREQUISITES:

Sieve of Eratosthenes

## PROBLEM:

You are given an array A with size N. For each K between 1 and N (inclusive), find out if it is possible to partition (split) the array A into K contiguous subarrays such that the sum of elements within each of these subarrays is the same.

## EXPLANATION:

Let S[i] = \sum_{j=1}^{i} A[j]. Let’s make some observation here now:

- All the values of $S[i]$ will be distinct because all $A[j] > 0$.
- $S[N]$ must be divisible by $K$ if we want to partition the array into $K$ contiguous part.
- Sum of each part will be equal to $\frac{S[N]}{K} = X$ (say).
- For every multiple of $X$ i.e. for every $X*i$ ∀ $1 \le i \le K$ there must exist some $j$ such that $S[j]=X.i$.

Iterate over all the divisors of S[N] that is \le N. For every such divisor K we will check if all of the multiples of \frac{S[N]}{K} are present in the prefix sum array or not. Let’s make a cnt[] array which stores how many multiples of \frac{S[N]}{K} are present in the prefix sum array in cnt[K]. Any K will be good iff cnt[K]==K.

From the above observations, S[j]=\frac{S[N]}{K}.i which means \frac{S[j]}{S[N]}=\frac{i}{K}.

We are going to iterate over S[j], compute the irreducible fraction \frac{S[j]}{S[N]} =\frac{P}{Q}. Now we need to increase cnt[Q.i] by 1 for all those Q.i that are divisors of S[N] and \le N, because for every Q.i you get a distinct P.i because of the property that all prefix sums are distinct.

Doing the later step for every j can be costly. Hence, we will use sieve for this purpose. When iterating over j, just increment cnt[Q] by 1 if Q \le N. When done iterating over j, you can run a sieve-like algorithm for updating the multiples of every cnt[Q.i] which will cost us O(NlogN).

You can refer to the setter’s solution for implementation details of this approach.

## ALTERNATE SOLUTION:

Based on above observations, we will try to devise our solution. Iterate over all the divisors of S[N] that is \le N. For every such divisor K we will check if all of the multiples of \frac{S[N]}{K} are present in the prefix sum array or not. If all of the multiples are present in the prefix sum array then this K is good. When doing naively, for every K it will require O(K.logN) operations to check if every multiple of K is present in prefix sum array. But creating good test data which costs O(K.logN) for every K is nearly impossible. Hence, in practice, this approach will run a lot fast than expected. Tester’s solution uses this idea. You can refer to it for the implementation details.

## Time Complexity:

O(NlogN)