SCHEDULE - Editorial

editorial
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PROBLEM LINK:

Practice
Contest

Author: Hasan Jaddouh
Testers: Sergey Kulik, Kamil Dębowski
Editorialist: Pawel Kacprzak

DIFFICULTY:

EASY

PREREQUISITES:

Binary search

PROBLEM:

For a binary string A, let’s define a block as a substring of A containing either only 1's or only 0's, so 111 and 0 are blocks, while 10 and 0010 are not. Moreover, let a major block be a block of A, such that no block in A is longer that it and L be the length of a major block of A. Notice that there can be many major blocks, but all of them have the same length.

Then the problem can be reformulated as follow: for a given binary string A of length N and integer K, we have to return the minimum L (the length of a major block) that we can get by performing at most K bit flips on A.

For example, if A = 11100 and K=1, the best we can do is to either flip the first or the second bit to get for respectively 01100 and 10100. In both these strings, L is 2, which is the best we can get in this case.

QUICK EXPLANATION:

First, check if getting L=1 is possible. If yes the problem is solved, otherwise, perform a binary search on the result checking if getting L=M is possible in linear time.

EXPLANATION:

First of all, let’s take a closer look at the constraints. It is said that K \leq 10^6, but since also N \leq 10^6 and there is no point of flipping a single bit more than once, we can rewrite the constraint as K \leq N.

Subtask 1

In the first subtask N \leq 10, so one can generate all possible strings we can get by performing at most K flips on A, and compute the smallest L among them. This is possible because there are at most 2^N such strings and computing L for a single string takes O(N) time. Thus the overall time complexity of solving all test cases will be O(T \cdot 2^N \cdot N), which is completely fine for these constraints and T \leq 11000.

Subtask 2

In the second subtask we have N \le 10^6, so the problem has to be solved in a clever way. The crucial observation is that for any L and any block of A with length M, \lceil M / (L+1) \rceil flips are necessary to convert that block into a substring without blocks of length greater than L, and we can perform exactly that many flips to do that.

For example, if we have a block of length 5, 00000, and L=2, we flip its bits with indices L+1, 2 \cdot (L+1), 3 \cdot (L+1) \ldots, so in this particular case, we flip only its third bit to get 00100. Similarly, if we have a block of length 6, 000000, and L=2, we flip its third and sixth bits to get 001001.

Thus it looks like we can perform a binary search for the answer and for a fixed M, check if we can transform all the blocks in A into block of length at most X using at most K flips by just iterating over all blocks in A and computing the sum of needed swaps for all of these blocks. Notice that binary search is correct here, because if it is not possible to use at most K swaps to get A with L=X, then it is also not possible to use at most K swaps to get A with L \lt X. This will result in O(N \cdot \log N) time complexity for a single test case, but since the sum of N across all test cases is at most 10^6, this will run fast enough. However, we are not finished yet, because there is a tricky case to handle.

Notice that when we divide A into blocks, the blocks alternate in such a way that after a block consisting of 0's there is a block consisting of 1's, then one consisting of 0's again and so on. The observation we made says that for a block of length M, \lceil M / (L+1) \rceil flips are necessary and sufficient to convert this block into blocks of sizes at most L. It is true for an isolated block, however, since in some cases we are forced to flip the first or the last bit of such block, that may ruin our solution because that bit can create another block with adjacent blocks.

Let’s take a closer look where this situation can happen, so in what cases we are really forced to flip either the first of the last bit of a block? As mentioned above, for a block of length M and a fixed L, we can flip bits with indices L+1, 2 \cdot (L+1), 3 \cdot (L+1) \ldots. It follows that we are flipping its last bit if M \mod (L+1) = 0. However, if this happens and L \gt 1, we can always flip bits with indices L, 2 \cdot L, 3 \cdot L, \ldots and we avoid flipping the first and the last bits, which caused the problem. But what happens when L = 1? Well, then we cannot use that trick, however, this case is easy to handle: at the beginning, we can just check if A can be transformed using at most K flips to get L=1. Notice that there are just 2 possible strings of given length with L=1, either a sequence of alternating 0's and 1's starting with 0 or the same but starting with 1. If this is the case we are done. Otherwise, we use binary search described above searching for the answer in a range [2, N]. For implementation details, please refer to multiple solutions linked below.

AUTHOR’S AND TESTER’S SOLUTIONS:

Setter’s solution can be found here.
First tester’s solution can be found here.
Second tester’s solution can be found here.
Editorialist’s solution can be found here.

12 Likes
2

I had an algorithm to solve it by storing segments in heap, and dividing longest segment by two and storing both parts again in heap. Will it work?
Is there a easy way to check if it is possible to make answer 1. If it does suggest.

3

@saurabh9456 … that won’t work… consider the string “111111111” and K=2 … the optimal answer is 3 and with ur algorithm the answer would be 4.

2 Likes
4

It’s nice to editorials too early.

2 Likes
5

This question can also be solved very easily by priority queue.
First lets calculate all the lengths of consecutive 1s and 0s.Lets now make a priority queue of pair of pairs.
priority_queue<pair<long,pair<long,long> > > pqueue;

Now we need to insert the (value of length,(some big number, initial value));
some big number can be 100000 for instance.
Now we just need to pop from the priority queue and decrement the big number and divide the initial value by the (big number- new big number+1) and insert in the priority queue (new value,(new big number,initial value)).
when pqueue.top().first==1
break the loop.

Solve for the answer 1 separately.
Here is the code snap:

while(k–)
{

        long currtop= pq.top().first;
         if(currtop<=2){
            break;
        }
        long ix=pq.top().second.second;
        long trueval= val[ix];
        long id= pq.top().second.first;
        id--;
        long im=100000-id;
        currtop= trueval/(im+1);
        pq.pop();
        pq.push(make_pair(currtop,make_pair(id,ix)));
    }

And here is the AC code: https://www.codechef.com/viewsolution/13057467

6 Likes
6

I am getting NZEC error in this code.
https://www.codechef.com/viewsolution/13067855
can anyone explain why i am getting this error?

7

My code is working in 3/9 case can someone help me with this .
code link-> https://code.hackerearth.com/ec26cap?key=8873a98d212090ee34b470913e133bfb

8

Alternative Solution: Maintain a set of all original segments.Think of ‘k’ as total resources you have to allot to these segments to minimise the maximum contigous same bits.One can calculate the closed form for maximum subsegment size of the original segment when ‘x’ resources are alloted to the segment.The first element always corresponds to the original segment having maximum sub-segment size.At each iteration we allot one more resources to the first segment and calculate its new position in the set.The procedure terminates when k=0 .
Complexity : O(n*logn)

1 Like
9

can you explain how and why it works?What is the use of big number

10

@ajaman13 … There seems to be no code out there.

11

nice editorial…

12

if k is greater than (n+1)/2 then will the answer be 1?

13

Yes @aminuteman . For any given sequence, for the answer to be 1, the minimum number of moves required will always be less than or equal to half the length of the sequence. So, if k is >(n+1)/2 , we will surely be left with one or two excess moves even in the worst case. It can also be put this way - For the answer to be 1, there must be n/2 ones and n/2 zeroes (considering n is even, when n is odd, there will be an extra 1 or 0). So, to change the sequence so that answer becomes 1, at max we require n/2 moves or (n+1)/2 moves. Hope you got it.

14

I need the explaination for the editorialist’s solution’s [[Binary Search]] Logic. Thanks in Advance.

15

it can also be solved by checking for each answer ( not more than the length of longest segment ) …
if we are checking for an answer (say target) ,we try to reduce all the lengths of segments greater than our target by dividing it into n parts which will require n-1 values of given ‘k’(allowed flips).

n = length of segment / (target + 1)

if all the segments can be divided requiring not more than remaining k flips…then our target is achieved and we have the answer, otherwise we will check for the next target(i.e target + 1).

here is the link to my code

https://www.codechef.com/viewsolution/13072300

16

my solution :

https://www.codechef.com/viewsolution/13088704

17

Wouldn’t an approach like the following work?

https://www.codechef.com/viewsolution/13050938

18

the binary search logic:

maxL = 1e6
minL = 1

m = (maxL + minL)/2

now calculate the number of bits need to change/flip to have m as the answer.
if the count is less than k less then the answer can be less than m(even lower than m, so maxl =m ) else m cannot be the answer so answer lies in between m and maxL(minl=m).

in the end minl is the answer

19

@padamchopra Can you explain your approach in words ?

20

it is simple greedy technique.The longest segment stays on the top of the priority queue, we pop it and divide it by the required factor.Now purpose of a big number is when the top and the second element of the queue are the same i.e suppose we have 4,4.Now we wish to pop that element first which has been divided by a lesser factor.Hence initially we select a big number in the priority queue.