PROBLEM LINK:

Practice
Contest: Division 1
Contest: Division 2

Setter:Misha Chorniy
Tester:Zhong Ziqian
Editorialist:Taranpreet Singh

DIFFICULTY:

Simple

PREREQUISITES:

Implementation, Data-Structures.

PROBLEM:

Given an array $A$ of length $N$, Determine whether the array contains two positions $i$ and $j$, $i \neq j$ such that $A_i \neq A_j$ and $A_{A_i} == A_{A_j}$.

Print Truly Happy, if we can find such pair of positions, otherwise print Poof Chef.

SUPER QUICK EXPLANATION

• We can maintain a set for each distinct value, and for every position, x, insert $A_i$ in the set corresponding to value $A_{A_i}$.
• This way, We can select two positions satisfying the required criteria if any set has more than one distinct value.

EXPLANATION

First of all, let's see the brute force solution.

We can iterate over every pair $(i, j)$ of positions, check if $A_i \neq A_j$ and $A_{A_i} == A_j$ holds. If this holds for any pair, we can make the chef Truly Happy. But Sadly for us, this solution has Time Complexity $O(N^2)$ and thus, will time out for Last Sub-task.

Now, Focus on the condition for a valid Pair $(i, j)$, $A_i \neq A_j$ and $A_{A_i} == A_{A_j}$.

Inequality is usually harder to handle than equality, so, focusing on Equality first tells us the following.

For the required pair of positions, if it exists, it holds that $A_{A_i} == A_{A_j}$. This means, that we can consider all positions having the same value of $A_{A_i}$ together.

Now, For every distinct value of $A_{A_i}$, we have a number of values. The problem reduces to finding two distinct values $A_i$ and $A_j$ in the same set which has $A_i \neq A_j$.

We can see, the simplest way to do so is to make set for every distinct value $A_{A_i}$ and add to it, the values $A_i$. Now, Chef will be happy, if we can select 2 elements from any one set.

This implies that Final condition for Existence of required Pair is, If any set has more than one element, It is always possible to pick at least one pair and make Chef Truly Happy.

Alternative Implementation
We can also replace Array of Sets with a single map, or even a single array, Implementation of which is left as an exercise.

Challenge Problem

Count the number of such pairs for a given array. Enjoy :P

Time Complexity

Time complexity is $O(N*logN)$ per test case. Can be optimized to $O(N)$ too.

AUTHOR'S AND TESTER'S SOLUTIONS:

Setter's solution
Tester's solution
Editorialist's solution

Feel free to Share your approach, If it differs. Suggestions are always welcomed. :)