Contest: Division 1
Contest: Division 2

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




Implementation, Data-Structures.


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

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


  • 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.


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

We can iterate over every pair (i, j) of positions, check if A_i eq 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 eq 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 eq 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 :stuck_out_tongue:

Time Complexity

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


Setter’s solution
Tester’s solution
Editorialist’s solution

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


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:

ans = [“Poor Chef”,“Truly Happy”]
for ti in range(int(input())):
n = int(input())
ays = list(map(int,input().split()))
unq = set(ays)
aas = set(ays[a-1] for a in unq)
print( ans[len(aas) < len(unq)] )


why this solution is getting AC?
for (n>10000) I am not able to understand the approach…


Tester’s solution is giving wrong output for the following input

45 12 58 82 48 66 64 3 11 85 55 90
55 22 52 33 23 46 24 35 1

The expected output is

Poor Chef
Poor Chef

Solution’s output is

Poor Chef
Truly Happy


My O(n) approach

Using Hash tables


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.


Poor me, thought bruteforce was the only way. Thank you, kind sir Taranpreet


The Tester solved the problem without using Set. Can someone please explain his approach?


The Tester solved the problem without using Set. Can someone please explain his approach?


where is my code wrong i used set @vijju123


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*(k-1)/2=1 and add all such pairs over such special segments and get the final answer :slight_smile:
Nice problem, indeed :slight_smile:


@vijju123 please reply


Way to go kunnuuuuuu :smiley: :slight_smile:


His approach is wrong, he somehow got lucky with the larger TCs. Dont look at his code for correct approach.


No problem :slight_smile: