You don't know uwi? Check this

You (yaowan).

Great!

Yes, the goal of Educational round is rather to practice than to compete

Physics.

you don't need binary search dude

You can't use binary search for this problem because the objective function (the length of the substring in your case) is not monotonic. Consider the example: 00111100. You can make balanced strings of lengths 2, 4, 8 but not 6.

I also had wrong answer in test 24. I realized this after the contest finished :/. you can't do binary search. try this case 10 1110000011

If k is even, then no solution because you always make 1 (main call)  + 2·m calls where m is the number of times you pass from condition 1. If k is odd, then there exists a solution if k is less than the maximum number of calls.

You can solve the problem recursively by calling a function that takes the current subarray endpoints and k and the integer to starting filling the array with in this subarray (if the integer is x and the subarray size is L, then you will fill it with the integers x, x + 1, ..., x + L - 1. These integers are the subarray values. The main call will be f(0, n, k, 1).

If k = 0, then put all the numbers sorted in the subarray so you do not make any mergeSort() call. Otherwise, split the subarray according to the rules. The left part will take the largest subarray values and the right part will take the smallest values. In this case, we guarantee that we will make 2 mergeSort() calls. Then you call f() for the left part giving it k - 2 to be maximum number of mergeSort() calls it can use and it will return the remainder of k - 2 (the unused calls). Then you call f() for the right part giving it this remainder. Finally, you return what the right part returns back. For a solution to exist, the returned value of the main call must be equal to 0

Let's reverse string and build suffix array. Since we have sorted suffixes, string 'a' will be a LCP of some consecutive suffixes in suffix array, next we should get LCP array (lcp[i] = lcp of suffix[i] and suffix[i + 1] when i and i + 1 indexes in sorted order). If we represent LCP array as columns along axis, our task transformed to 'find rectangle with maximum area', and it can be solved with two stacks: for each lcp[i] find nearest left and right index j, that lcp[j] < lcp[i], after that we should relax the answer with lcp[i] * (rightJ — leftJ), because there are (rightJ — leftJ) strings, that have lcp[i]. So the last question is how to handle forbidden cells? we should merge all lcp (in lcp array) into one between each two consecutive non-forbidden cells.

PS. If you not familiar with suffix structures, I hope it can be solved with the same aproach with replacing suffix array by hashing+binary search for sorting suffixes.

You just need prefix sum to solve this question and realize the following : initially sum = 0

add 1 to sum if you encounter a '1' subtract 1 from sum if you encounter a '0'

Suppose you are at current index 'i' and your Current sum is S. You are encountering this sum S for the first time. you map i to S. i.e, map[S] = i... Now suppose for some later index J you encounter S you just update ans as max(ans, j — map[S]) because if you observe carefully you need the best possible sum to be 0.

if you have current sum as 0 at index 'i' then you keep updating the answer with 'i' + 1

Problem is fixed, hacks will be rejudged by the end of hack phase.

Good day to you,

Well hope it won't be wrong, since it is not "proved yet" :)

Firstly, sort all "heights" (yet still remember their original position to be able to reconstruct at the end). Now, it is pretty easy to assign a value to each position (it is the height on the position minus the height on previous position [potentionaly 0]).

So not lets consider how would "brute-force" solution work? Well kinda simple.. just try EVERY triple in the array (check whether the x*2>=y is fine) and then take MAXIMUM from these (what is maximum is written in the "rules" — note tht the rightmost is always best diploma :) ). Now as you can see, it will obviously TLE, anyway as you can also see, it is just O(N³) which is not bad.

So now we can "simply" focus on reducing of the third cycle. As the other two iterations are fixed, we know pretty well, where the segment (if any) for the third cycle might be. EXAMPLE: N==10, i=7,j=9, we know the length of first segment is 2 and of second is 1, so the length of third segment must be between 1 and 2 (which begins on position 5 or 6). We are interested in the biggest number there so we can use some "RMQ" structure, which could get it. One way is for-example Sparse Table which can get maximum in O(1) making the total complexity O(N²).

So in short: Iterate with two (the best) diplomas and find the third by RMQ.

Good Luck And have Georgeous Day ^_^

Hope it was a little bit undestandable and hope I didn't lead you wrong way ^_^