Not able to think of the approach

#1

i am unable to get the appraoch of this problem … how can we apply dp here ?? what should be the thought process in making the solution?? the hint video is very confusing … and didn’t find it helpful plz help on this

#2

Hey @Shobhit_kumar read this carefully, and try to understand this,
Let’s solve the problem for rectangle W×H (W≥H). Of course, we can cover all rectangle with itself for cost W. To get something smaller than W we have to leave at least one column uncovered — otherwise we pay at least sum of w over all rectangles which is at least W. This gives us an idea to use DP on rectangles to solve the problem: dp[x1][x2][y1][y2] is minimal cost to cover the rectangle [x1;x2)×[y1;y2). It is initialized by max(x2−x1,y2−y1), and we have to try not to cover every column/row. Of course, we have to check if it is all white from the beginning; to do that we will precalculate 2D prefix sums. Total complexity is O(5*n).
W and H are width and height of rectangle drawing canvas
For our question W = H = N because its a square
If you use eraser at each valid position complexity will be 0(n^4)

Your solution can have at max 0(n^3)
So pre-calculate number of black cells in each k size segment in rows and cols
When you place an erase somewhere, row will become white if black cells in row == black cells in k size segment from that position
Similarly you can solve for columns

So n^2 cells and n time at each cells gives 0(n^3)

#3

Moreover if it’s still not understandable, refer this.

#4

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