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 1point3acres0 <= L <= R < n . 1point3acres.com/bbsSample Input: 8 1 0 0 1 0 0 1 0 . 1point3acres.com/bbsSample 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

 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; 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    }34    for (int i=globalR+1; i<A.length; i++) {35         if (A[i] == 1) 36             OnesUnflip ++;37    }38    return global + OnesUnflip;39 }