Twitter OA prepare: Flipping a bit

You are given a binary array with N elements: d[0], d[1], ... d[N - 1]. 

You can perform AT MOST one move on the array: choose any two integers [L, R], and flip all the elements between (and including) the L-th and R-th bits. L and R represent the left-most and right-most index of the bits marking the boundaries of the segment which you have decided to flip.



What is the maximum number of '1'-bits (indicated by S) which you can obtain in the final bit-string? . more info on 1point3acres.com



'Flipping' a bit means, that a 0 is transformed to a 1 and a 1 is transformed to a 0 (0->1,1->0). 

Input Format 

An integer N 

Next line contains the N bits, separated by spaces: d[0] d[1] ... d[N - 1] 



Output: 

S 



Constraints: 

1 <= N <= 100000 

d can only be 0 or 1f -google 1point3acres

0 <= L <= R < n 

. 1point3acres.com/bbs

Sample Input: 

8 

1 0 0 1 0 0 1 0 . 1point3acres.com/bbs



Sample Output: 

6 



Explanation: 



We can get a maximum of 6 ones in the given binary array by performing either of the following operations: 

Flip [1, 5] ==> 1 1 1 0 1 1 1 0

分析：这道题无非就是在一个数组内，找一个区间，该区间  0的个数  与  1的个数  差值最大。如果我们把这个想成股票的话，0代表+1，1代表-1，那么这道题就转化成了Best Time to Buy and Sell Stock, 找0和1的个数差值最大就变成了找max profit。

因为需要找到这个区间，所以在Stock这道题的基础上还要做一定修改，记录区间边缘移动情况

 1 public int flipping(int[] A) {

 2    int local = 0;

 3    int global = 0;

 4    int localL = 0;

 5    int localR = 0;

 6    int globalL = 0;

 7    int globalR = 0;
      int OnesFlip = 0;

 8    int OnesUnFlip = 0;   //those # of ones outside the chosen range


 9    for (int i=0; i<A.length; i++) {

10         int diff = 0;

11         if (A[i] == 0) diff = 1;

12         else diff = -1;

13 

14         if (local + diff >= diff) {

15             local = local + diff;

16             localR = i;

17         }

18         else {

19             local = diff;

20             localL = i;

21             localR = i;

22         }

23 

24         if (global < local) {

25             global = local;

26             globalL = localL;

27             globalR = localR;

28         }

29    }

30    for (int i=0; i<globalL; i++) {

31         if (A[i] == 1) 

32             OnesUnflip ++;

33    }
　　　 for (int i=globalL; i<=globalR; i++) {
           if (A[i] == 1)
　　　　　　     OnesFlip ++;
      }

34    for (int i=globalR+1; i<A.length; i++) {

35         if (A[i] == 1) 

36             OnesUnflip ++;

37    }

38    return (global+OnesFlip) + OnesUnflip;

39 }