By reyad, history, 5 months ago,

To solve this problem first we need to do some observations:
let's denote
l_k = no. of elements less than k,
g_k = no. of elements greater than k,
e_k = no. of elements equal to k.

(1) let us first state, "when there is no solution for this problem":
if l_k >= e_k + g_k for all possible subarray where each subarray length is greater than 1, then its not possible to have a solution for this problem, cause k would never be able to be the median of the subarray.

(2) if we can't construct a solution for any of the subarrays of length w, then there would be no solution of any subarray of length w+1. And if there is a solution for w(where w > 1), then it is also possible to construct a solution for w+1 no matter what the consecutive elements are. [note: this is the most important observation]

(3) the elements can be represented as only two types a[i] < k and a[i] >= k. So, if we turn them into +(positive transition) and -(negative transition) operation then, there would be only two types of transitions. And we can easily represent them in diagrams like below:

//         _   _   _
// (i)   _| |_| |_| |_...
//
// or
//
// (ii)  _
//        |_       _
//          |_   _| |_...
//            |_|
//
// or
//               _
//             _| |_
//           _|     |_...
//         _|
// (iii) _|
//
// etc...



The graph shows us one thing clearly, "when it is possible to make e_k + g_k > l_k":
e_k + g_k > l_k is possible if the graph contains any of the following segment for once:

//        _   _
// (i)   | |_| | : +, -, +
//
//           _
//         _|
// (ii)  _|     : +, +, any type of transitions
//
//
//            _
// (iii)    _|   : -, + , +
//       |_|
//


In summary it shows that if there are at least two positive transitions between three consecutive transitions, then it is possible to fill the array with k if there is at least one k present in the array.

(4) Let's denote positive transitions as +1 and negative transitions as 0, then in each subarray of size 3 if there is at least one subarray that confirms two positive transitions(considering k is present at least once in the array), we would be able to fill the array. Example of such subarray(for transitions) would be like: (1, 0, 1) or (1, 1, 0) or (0, 1, 1) or (1, 1, 1) etc...

Solution:

let flag = false, seenK = false;
let np = 0; // counts no. of positive transitions in subarray size of 3
for i = 0...n-1:
seenK |= (a[i] == k) // check if we've been seen k at least once in the array
np += ((a[i] >= k) ? 1 : 0);
if i >= 3:
np -= ((a[i-3] >= k) ? 1 : 0);
if np >= 2:
flag = true;
if seenK and (flag || (n == 1 && np == 1)/*check for exceptional case when n == 1*/):
print "yes"
else:
print "no"


The solution link is given here.

• +3

 » 5 months ago, # |   0 Hey thanks for making such a great editorial. Can you also mention how to prove the second point?(If there is a solution for length l, there is a solution for length l+1).
•  » » 5 months ago, # ^ |   +3 Consider the subarray {3, 1, 4, 4}. It has no immediate solution for length $2$, but does for length $3$. Once you make the changes for this, you can proceed to get valid subarrays of length $4, 5, 6, ...$ At least that's what I think he meant to say.
•  » » 5 months ago, # ^ | ← Rev. 4 →   +3 let's say there's a solution for subarray of length l(l > 1).let's consider this subarray of length l as a[i...j](where, j-i+1 = l)now, if a[i...j] have a solution, then we can change all the elements of a[i..j] to ki.e. a[i] = a[i+1] = a[i+2] = .... = a[j] = k now, consider all the possible values for a[i-1] and a[j+1]there's four possibilities where a[i-1] and a[j+1] none of them are k(let's say i-1>=1 and j+1<=n, n means length of a):(i) a[i-1....j+1] = [k+x, k, k, ...., k+y] (where, x > 0 && y > 0)(ii) a[i-1....j+1] = [k-x, k, k, ...., k-y] (where, x > 0 && y > 0)(iii) a[i-1....j+1] = [k+x, k, k, ...., k-y] (where, x > 0 && y > 0)(iv) a[i-1....j+1] = [k-x, k, k, ...., k+y] (where, x > 0 && y > 0) I hope you get the scenario.Part-01: for a[i....j+1]Now, let's consider the subarray a[i....j+1], can you construct a solution? (remember l>1)let's try to find the median, which is (((j+1)-(i)+1) / 2) = (h / 2) th element [note: j-i+1 = l, and so, h = l + 1]for h == 3(a[i, j, j+1]), median would be 2nd element, which is k(whatever, a[j+1] holds, cause, after sort the possible sequence would be (k, k, k+x) or (k-x, k, k) and 2nd element would always be k)for h == 4(a[i, j-1, j, j+1]), median would be 2nd element, which is k(whatever a[j+1] holds)......so on...so no matter what a[j+1] holds the median always would be some index which holds k Part-02: for a[i-1...j]You can just apply the rules as I have done in part-01 and prove it yourself. I hope this helps you.
•  » » » 5 months ago, # ^ |   0 Thanks a lot!