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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
sumGridwheresumGrid[x][y] = sum(grid[i][j]whereiis inrange(x, N)andjinrange(y, N)So, each element in this new
sumGridis 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 innerforloops — making it much faster :)