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GMEDIAN - Editorial (Unofficial)


Setter:Teja Vardhan Reddy
Editorialist: Utkarsh Pandey


Observation, Binomial Coefficients.

Given a sequence A1,A2,…,AN. Find the number of subsequences in which the median of the subsequence is present in that subsequence, modulo 1000000007.

Since we have to sort each subsequence to find its median, we will sort our original sequence (i.e. A1,A2,…,AN) at first itself. The ordering won't change.
For example, the good (desired) subsequences in the sequence [9,5,7,6,8,1] are same as that in the sequence [1,5,6,7,8,9].

The subsequences having odd length will definitely qualify to be one of our good subsequence (as in the subsequences with odd length, the element at the middle position is the median of the subsequence itself).

The subsequences with even length will qualify as good subsequence if and only if the middle two elements are same. It is so because the array is already sorted. View the content below in case you didn't understand this statement.

View Content

So our answer to this problem is : (Number of subsequences having odd length) + (Number of good subsequences having even length with both the middle elements equal)

First Subtask:
It is clearly written in this subtask: A is a permutation of integers 1 through N. Also, the number of elements in the sequence A is N which means each element from 1 to N is present once and only once. This tells us that there is no repetition of any number. So, no subsequence having even length will be counted.

So, answer to this subtask = Number of subsequences having odd length
= Number of ways of selecting 1 out of n elements + Number of ways of selecting 3 out of n elements + Number of ways of selecting 5 out of n elements + ............. + Number of ways of selecting n or (n-1) out of n elements

$=nC1 + nC3 + nC5 + ..... + nCn$ (if 'n' is odd) or $nC1 + nC3 + nC5 + ..... + nC(n-1)$ (if 'n' is even)
where nCx denotes the number of ways of choosing x elements out of n elements

= 2^(n-1) See the proof below if you didn't understand this completely Proof:

View Content

Second and Third Subtask:
The tricky part of this question is when the elements in the sequence starts repeating. In this case, we have to find the number of subsequences having even length. To find the subsequences of even length to be a good subsequence, we traverse through each element which is repeated at least once. For this element to be our median element in one of the subsequences of even length, we must have at least k elements at its index's left and we must have at least k elements at the other element index's right (Note that our array was sorted), where k is a non-negative integer.
An example is provided in the section below in case you didn't understand the lines written above:

View Content

Now, Calculating nCx in each iteration, we may have difficulties passing the time limit provided in the problem. To reduce the time complexity to pass the given time limit, we have to observe the given fact:

Let the repeating elements be found at index i and j Suppose, there are m elements at the left of the index i and n elements at the right of index j.

So, number of good subsequences: 1 (good subsequence of length 2) + $mC1 * nC1$ (good subsequences of length 4) + $mC2 * nC2$ (good subsequences of length 6) + ........... + $mCmin(m,n) * nC(min(m,n))$.

By Vandermonde's identity we can sum up this value as (m+n)C(min(m,n)), details of which can be found here :Vandermonde's Identity.

Bonus: Calculating queries of the type nCr % 1000000007 can be solved by precalculating $nCr % 1000000007$ upto n = whatever you need :p . Details of this can be found here: nCr % p in O(1)

Time Complexity = O($N^2$) (Enough to pass all the subtasks)

Editorialist's Solution:

asked 12 Nov '18, 15:48

utkarsh911's gravatar image

accept rate: 9%

edited 13 Nov '18, 19:21

Please add UnOfficial in the title.

(12 Nov '18, 16:52) aryanc4036★

@aryanc403 Done :)

(12 Nov '18, 17:49) utkarsh9115★

Using a proof of Vandermonde's identity to illustrate how to pick all the balanced enclosing sequences around a selected middle pair:

Suppose the chosen two central elements are at positions in the sorted array with $s$ elements before the first one and $t$ elements after the second. Then consider any selection of the same number of elements from each side, say $j$ elements from each side.

Now invert the selection in the "before" section, so a previously selected item becomes unselected and vice versa. This now gives $s{-}j$ elements on that side. Combining this with the "after" section, you have a total of $(s{-}j) + j = s$ elements from the total pool, no matter what $j$ is.

You can easily see that every balanced selection can be turned into a corresponding choice of $s$ elements in this way, and vice versa - every choice of $s$ elements can be turned into a distinct balanced enclosing choice. So the number of options for any one central pair is $\binom{s+t}{s}$.


a b c d e f g h i j k l m n

chosen central pair h & j ($s=7, t=4$)

a b c d e f g X - X k l m n

choose say three from each side:

. b c . e . . X - X k l m .

invert the "before"s:

a . . d . f g X - X k l m .

and we have a choice of exactly $7$ elements.

And any choice of $7$ elements from the $7+4=11$ available can produce a unique balanced set:

a . c d e f . X - X k l . .

invert the "before"s:

. b . . . . g X - X k l . .

giving $2$ each side.

If you have a run of identical elements (more than two), you can make the summation a little more efficient too.

My code.


answered 14 Nov '18, 02:38

joffan's gravatar image

accept rate: 13%

edited 14 Nov '18, 04:15

I used another binomial identity--the hockey stick identity--for an O(n^2) solution.

Sums of binomial coefficients where n is changing have a closed form. There doesn't exist a closed form where k is changing.

$$\sum_{i=0}^n \binom{i}{k} = \binom{n+1}{k+1}$$

$$\sum_{i=m}^n \binom{i}{k} = \binom{n+1}{k+1} - \binom{m}{k+1}$$


answered 15 Nov '18, 16:18

wmoise's gravatar image

accept rate: 9%

edited 15 Nov '18, 16:29

Vandermonde's Identity was the crux to get AC with 100 points. I submitted many solutions but got this identity after a bit google surfing work. Such a soft dismissal. xD


answered 12 Nov '18, 20:48

gaurav_iiitian's gravatar image

accept rate: 10%

This will ( ) help in calculating nCr % p.


answered 12 Nov '18, 21:36

cenation092's gravatar image

accept rate: 6%

I derived the vandermonde's identity to get this done and still couldnt do it because """Bonus: Calculating queries of the type nCr % 1000000007 can be solved by precalculating upto n = whatever you need :p . Details of this can be found here: nCr % p in O(1) """


answered 13 Nov '18, 06:36

someshsingh22's gravatar image

accept rate: 0%

If I was in hinting mood I would just say "Pascal's triangle".

More fully, because of the ability to derive binomial coefficients - the entries of Pascal's Triangle - by summing two earlier values, it's very quick to generate the values you need in this problem. And of course since we're just adding, we can take mods as we go.

(13 Nov '18, 10:31) joffan5★

(m+n)C(min(m,n)) is the same as ${m+n \choose m}={m+n \choose n}$ and it can be derived directly, without using Vandermonde's identity.

We need to get the number of all combinations from $m+n$ elements, $m$ of which are on the left side and $n$ - on the right side, that have equal numbers of elements on each side. Lets replace the left-side part of each combination with its complement - the set of elements on the left side that don't belong to the original combination. If an original combination had $k$ elements on each side the resulting combination will contain $(m-k)+k=m$ elements. This number does not depend on $k$ and is the same for all the resulting combinations. And if we take any combination of $m$ elements and perform the same operation on its left side, we will obtain a combination with equal number of elements on each side. So we have a one-to-one correspondence, and the number of original combinations is equal to the number of resulting combinations, which is ${m+n \choose m}$.

Update: joffan essentially describes the same.


answered 14 Nov '18, 06:08

eugalt's gravatar image

accept rate: 4%

edited 15 Nov '18, 00:17

You can use DP to find number of sequences (of all lengths) which start and end at a particular element.Then finding number of sequences which use that element for even and odd length becomes easy .


answered 13 Nov '18, 10:07

ashforever007's gravatar image

accept rate: 0%

Testcases are weak. I solved it in O(N^4)


answered 13 Nov '18, 10:39

mightymercado's gravatar image

accept rate: 11%

Lol I wrote a brute O(N ^ 3) for each test then optimized it by precalculating an array in O(1000 ^ 3), then each test only takes O(N ^ 2). Only takes 0.3 secs. (Of course, that other thing was just 1000 ^ 3 / 6, which obviously will pass)


answered 13 Nov '18, 17:25

ducpro's gravatar image

accept rate: 0%

Answer is hidden as author is suspended. Click here to view.

answered 13 Nov '18, 18:02

karangreat234's gravatar image

accept rate: 0%


The problem with your code arises in counting the number of good subsequences of even length. One of your failed test case will be
A=1 2 2 2 2 3
Output should be 54
Your code's output : 46
I suggest you to read the editorial completely. I've shown the same test case as an example to pass subtask 2

(13 Nov '18, 23:58) utkarsh9115★

Its late, but thanks!:)

(17 Jan, 10:01) karangreat2344★

My approach same as editorial , but precompte nCr using dp and i think it's quite easy. This is my code:


answered 13 Nov '18, 19:36

savaliya_vivek's gravatar image

accept rate: 0%

edited 13 Nov '18, 19:37

Whats wrong with this code

for(long int i=0; i<n-1; i++){
    ans += 1;
    ans += min(right, left);
        long int right_ = right -1;
        ans += 1;
        ans += min(right_, left);
    left += 1;
    right -= 1;
    ans %= MOD;

i am iterating in array and updating the ans. left and right are number of elements in the left and right respectively


answered 14 Nov '18, 10:29

rational_kunal's gravatar image

accept rate: 0%

Your code fails for sequence having multiple duplicates. Also, you can't just add the min(left,right) directly to your ans. I suggest you to read the example I've given in the editorial above

(14 Nov '18, 12:41) utkarsh9115★

What's wrong in this solution:


answered 15 Nov '18, 00:17

rahul1995's gravatar image

accept rate: 0%

Guys my code does the exact same thing. Please someone look into this why it's giving WA.


answered 15 Nov '18, 17:56

rahul1995's gravatar image

accept rate: 0%

Since we have to sort each subsequence to find its median, we will sort our original sequence (i.e. A1,A2,…,AN) at first itself. The ordering won't change. For example, the good (desired) subsequences in the sequence [9,5,7,6,8,1] are same as that in the sequence [1,5,6,7,8,9].

There is a logical error here. I assume that you mean that the number of good subsequences is independent of the ordering of the sequence. But consider [1, 2, 3, 2, 3] and its ordered counterpart [1, 2, 2, 3, 3]. In the former, [1, 2, 3, 2] and [1, 2, 2, 3] must be counted as two distinct good subsequences, while in the latter, [1, 2, 2, 3] is the only corresponding good subsequence.

I am surprised that this solution gets AC. Either the test cases were weak or I don't understand the problem correctly.


answered 23 Nov '18, 16:08

lllllllllllll's gravatar image

accept rate: 0%

In [1,2,2,3,3] -> [1,2,2,3] (with 3 of index 3) and [1,2,2,3] (with 3 of index 4), both are good subsequence

(25 Nov '18, 11:50) utkarsh9115★

I gather that this solution counts [1, 2, 2, 3] twice for the sequence [1, 2, 2, 3, 3] (in the step where we choose 1 number form the 2 numbers to the right of 2's, we use 2C1 which counts both the 3's individually). But, isn't this wrong? The problem becomes a lot more difficult if we try to take into account that many numbers can be non-distinct and then the Vandermonde's identity expression is invalidated!


answered 23 Nov '18, 16:22

lllllllllllll's gravatar image

accept rate: 0%

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question asked: 12 Nov '18, 15:48

question was seen: 3,436 times

last updated: 17 Jan, 10:01