Let there be an array containing only two different elements. Lets consider them to be 0 and 1. You have to find total number of 1’s you could make from this array after performing some task. And the task is to select a sub-array and invert each of its element(Convert each of the element from 0 to 1 or 1 to 0 for every element in sub-array).

This problem certainly have a simple approach. Check for all subsets and count number of resulting 1’s in the array. That would result in an O(N^{3}) solution and the pseudocode would look something like this

for i = 0 to n
for j = i to n
for k = i to j
invert(elements)
count(number of 1s)
revert(elements)

That is very high complexity and surely is not expected. But the solution for above problem can be condensed to O(n) complexity. Yeah! That’s great!! But how?

First count number of 1’s in given array and store them(C). Now make another array such that for all 0’s in given array, corresponding indexes in new array have value 1 and for all 1’s in given array, corresponding indexes have value of -1. Now find Maximum sub-array sum for this new array. It gives number of 1’s that would be added by the sub-array to actual array after being inverted/flipped. It also removes all previously counted 1’s in that part of sub-array. How? all 0’s are 1’s now which will be added to MaxSubarray sum and all 1’s are -1’s now which will be subtracted from that sum.

Adding C and M would result in number of maximum 1’s that could be formed as per the question.

C = 0
M = 0
for i = 0 to n
if A[i] = 1
C = C+1
for i = 0 to n
if A[i] = 1
B[i] = -1
else if B[i] = 0
B[i] = 1
M = MaxSubarraySum(B)
print C + M

**Example :**
**Array :** 1 0 0 1 0 1 1 0 0 0 1 0 0 1 1 1 1 0 0 1
**Inverting :** -1 1 1 -1 1 -1 -1 1 1 1 -1 1 1 -1 -1 -1 -1 1 1 -1
Count of 1's = 10
Maximum Subarray sum = 4
Answer for given array = 10 + 4 = 14
**Final Array :** 1 0 0 1 0 1 1 1 1 1 0 1 1 1 1 1 1 0 0 1

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