PROBLEM LINK:Practice Setter: Misha Chorniy DIFFICULTY:Simple PREREQUISITES:Implementation, DataStructures. 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
EXPLANATIONFirst 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 Subtask. 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 Challenge Problem Count the number of such pairs for a given array. Enjoy :P Time ComplexityTime complexity is $O(N*logN)$ per test case. Can be optimized to $O(N)$ too. AUTHOR'S AND TESTER'S SOLUTIONS:Setter's solution Feel free to Share your approach, If it differs. Suggestions are always welcomed. :)
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asked 09 Nov '18, 23:08

Using Hash tables answered 20 Nov '18, 19:10

If set of distinct values of $A_i$ is greater than the set of distinct values of $A_{A_i}$, Chef will be truly happy, because there must be two (or more) different values of $A_i$ somewhere mapping on the same $A_{A_i}$. Hence my quick "existence" approach:
answered 15 Nov '18, 08:58

why this solution is getting AC? https://www.codechef.com/viewsolution/21455834 for (n>10000) I am not able to understand the approach... answered 16 Nov '18, 01:56
@vijju123 please reply
(22 Nov '18, 21:14)
His approach is wrong, he somehow got lucky with the larger TCs. Dont look at his code for correct approach.
(22 Nov '18, 21:24)

Tester's solution is giving wrong output for the following input
The expected output is
Solution's output is
answered 17 Nov '18, 22:27

Hi , can some one please explain how so many different people have same approach exactly with same variables also ? in this link from 2nd solution to next page few other people also has exactly has same solutions with exact variables also. Looks like code chef has some loop hole so they are cheating others. i have verified for submissions which submitted during contest. answered 01 Dec '18, 20:59

Poor me, thought bruteforce was the only way. Thank you, kind sir Taranpreet answered 06 Dec '18, 20:39

The Tester solved the problem without using Set. Can someone please explain his approach? answered 20 Dec '18, 10:55

The Tester solved the problem without using Set. Can someone please explain his approach? answered 20 Dec '18, 10:55

where is my code wrong i used set @vijju123 https://www.codechef.com/viewsolution/21543167 answered 26 Dec '18, 10:13

Solution to challenge problem : 1)Sort the given array,and also maintain its earlier indices by its side. 2)Whenever we find a segment of numbers which are equal, for example, lets consider '2' here, in my example, 1 2 2 2 2 3>The sorted array with original indices maintained by their side. I check the count of their indices and see if, they occurred at all, for example, 2 2 2 2> 0 1 0 1 So, 1+1=2 and number of pairs in this segment are k*(k1)/2=1 and add all such pairs over such special segments and get the final answer :) Nice problem, indeed :) answered 21 Jan, 21:37
