Question

A terrible new disease, HORSEVID, has begun to spread among horses worldwide. Stable manager Jimmy is trying to take as many precautions as possible to protect his herd from infection.
Stable manager Jimmy's barn is a long narrow building containing N stalls in a row (2≤N≤105). Some of these stalls are currently occupied by horses, and some are vacant. Having read about the importance of "social distancing", Stable manager Jimmy wants to maximize D, where D is the distance between the closest two occupied stalls. For example, if stalls 3 and 8 are the closest that are occupied, then D=5.

Two new horses recently joined Jimmy's herd and he needs to decide to which formerly-unoccupied stalls they should be assigned. Please determine how he can place his two new horses so that the resulting value of D is still as large as possible. Stable manager Jimmy cannot move any of his existing horses; he only wants to assign stalls to the new horses.

INPUT:

The first line of input contains N.

The next line contains a string of length N of 0s and 1s describing the sequence of stalls in the barn. 0s indicate empty stalls and 1s indicate occupied stalls. The string has at least two 0s, so there is at least enough room for two new horses.

OUTPUT:

Please print the largest value of D

(the closest distance between two occupied stalls) that Stable manager Jimmy can achieve after adding his two new horses in an optimal fashion.

SAMPLE INPUT:

14
10001001000010


SAMPLE OUTPUT:
2
In this example, Stable manager Jimmy could add horses to make the occupancy string look like 10x010010x0010, where x's indicate the new horses. In this case D=2. It is impossible to add the new horses to achieve any higher value of D.

Answer 1

Answer:

Explanation:

Well here you only have to add 2 new horses .... so simply what you can do is find 2 slots where the gap is maximum ... lets take the given example in detail to understand the same.... here 10001001000010 here you can observe that the difference between the first two filled slots is 3 then 2 then 4 then 1....... make it a separate array ...... now as i said that we just need to place two new horses so we just need two of the biggest numbers from this array and place the new horses at the center of the corresponding slots ... like here 3 and 4 are greatest ... so one horse is placed in between the gap corresponding to 3  while other at 4 ..... after making the new array answer can be find in linear time itself... I hope this makes sense ....

Now lets understand how to program it easily so firstly after scanning all stuff ..... we will make a gap array .... then just think now what we will do is find the greatest two numbers then again putting the horses and then find our answers .... so now an easy approach is that instead of making the string again and again searching for answer .... what we do is just do half of the two greatest ones and again find the greatest one in array....