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We need to add that 4th last term sumGrid[x+size][y+size] since we have subtracted it twice as part of terms 2 and 3. The solution now avoids two inner for loops — making it much faster :)
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A speed up I can suggest: pre-compute another grid, say
sumGrid
wheresumGrid[x][y] = sum(grid[i][j]
wherei
is inrange(x, N)
andj
inrange(y, N)
So, each element in this new
sumGrid
is the sum of all elements below and right to it, including itself.Then
sum(x,y,size)
is now just:As an example,
sum(3,2,5)
would look like this:We need to add that 4th last term
sumGrid[x+size][y+size]
since we have subtracted it twice as part of terms 2 and 3. The solution now avoids two innerfor
loops — making it much faster :)