ARRP - Editorial

#1

Div1, Div2
Practice

Author: Misha Chorniy
Tester: Lewin Gan

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 $S* = \sum_{j=1}^{i} A[j]$. Let's make some observation here now:
• All the values of $S*$ 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:

$O(NlogN)$

AUTHOR'S AND TESTER'S SOLUTIONS

#2

Can anyone explain this through an example?

#3

Let’s consider the first example: 1 4 2 3 5. The array of prefix sums(S) will look like 1 5 7 10 15. Let’s make fractions from it: 1/15 5/15 7/15 10/15 15/15. Let’s make them irreducible 1/15 1/3 7/15 2/3 1/1. Now consider only denominators which are <= N=5(only they will affect the answer), i.e. (3)1/3 (3)2/3 (1)1/1. Now, for every K=1…N find the number of denominators which are divisors of K. K=1(1), K=2(1), K=3(3), K=4(1), K=5(1). We’re interested only in those values of K for which this number is equal to K, i.e 1 and 3.

#4

Thanks for the explanation.