# Flip Bits

Posted: 4 Jan, 2021
Difficulty: Easy

## PROBLEM STATEMENT

#### A flip operation is one in which you turn 1 into 0 and 0 into 1.

##### For example:
``````If you are given an array {1, 1, 0, 0, 1} then you will have to return the count of maximum one’s you can obtain by flipping anyone chosen sub-array at most once, so here you will clearly choose sub-array from the index 2 to 3 and then flip it's bits. So, the final array comes out to be {1, 1, 1, 1, 1} which contains five ones and so you will return 5.
``````
##### Input format :
``````The first line of input consists of a single integer T denoting the total number of the test case.

The first line of each test case contains an integer N, which represents the array's size.

The second line of each test case contains N space-separated integers representing the array elements accordingly.
``````
##### Output format :
``````For each test case, return a single integer representing the maximum number of 1's you can have in the array after at most one flip operation.
``````
##### Note:
``````You don’t have to print anything; it has already been taken care of. Just implement the given function.
``````
##### Constraints:
``````1 <= T = 100
1 <= N <= 10^4
0 <= ARR[i] <= 1

Time Limit: 1 sec
`````` Approach 1

The idea is to check for all the possible subarrays and inside each subarray, check for the highest value of the difference between the count for zeroes and ones for this. Let’s consider this highest difference to be MAX and initialize it to zero so formally MAX = count of zeroes in that subarray - count of ones in the same subarray.

• Initialize TOTALONES to zero, which will store the total count of ones in the array.
• Now by running two nested loops outer one starting from index I= 0 and the inner one starting from index I, and both running till the end of the array.
• Inside the loop: if you encounter one, update  TOTALONES by incrementing one to it.
• Initialize COUNT1 and COUNT0  to zero, which will store the count of 1 and 0, respectively.
• Consider all subarrays starting from I and find the difference between 1s and 0s.
• Update MAX on every iteration to store the answer.
• Finally, return the count of all the ones in the array by the sum TOTALONES + MAX.