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Hi there,
- roberto.langarica February 22, 2020Trying to get the worst case scenario (the most flips required) where the array has only one position with 1 and the rest filled with 0's.
In this scenario if iterate from position (0,0) to the (n-1)'th position (n-1,n-1) the flips required will be the addition of both the x and y coordinate and can be expressed as: 2(n-1).
So for a 3x3 matrix the flips are: 4
4x4: 6
....
9x9:16 and so on
In scenarios where the only 1 present is in other position rather than the last one then the flips required are still represented by the addition of both coordinate components (x,y) so no mather the order of the matrix if we iterate the matrix in order starting from (0,0) then the flips needed to fill the matrix with 1's are the addition of the coordinates x+y.
So if the 1 is in (1,1) then te amount of flips required are: 2
(2,2): 4
(3,1): 4
...
If there is more than one 1 then the flips needed could be calculated by searching the first 1 and then the last rule apply, the flips needed are described by the addition of the coordinate componentes x+y.