For E the problem can be reduced to finding sum of Manhattan distances between every pair of points . Now it can be seen that in all possible combination of selected k squares every pair of points will occur C(n*m-2,k-2) times as we have to select k-2 points from n*m-2 points as we have already fixed 2 points. So the answer can be simply written as = C(n*m-2,k-2)*d where d denotes the sum of Manhattan distances between all pair of points which can be easily found out

F: First, notice that the b values don't matter, accumulate them in a separate variable and add them to the answer when outputting.

Finding the $$$x$$$ that minimizes $$$\sum |x - a_i|$$$ is equivalent to finding the median of the set of $$${ a_i }$$$. This is a classic problem, which can be solved with two sets (or multisets if the values can repeat). Actually computing the function also requires storing the sum of each set.

Can someone explain F Binary indexed tree solution, plz

The solutions have already been explained by other people here. :)

The main insight to solve E is to count the number of times each pair is counted, rather than iterating over all $$$K$$$ combinations.

The main insight to solve F is to notice that Bs are constant, and that the answer to minimise $$$f(x)$$$ is the median. We can evaluate the function $$$f(x)$$$ quickly by dividing all the A's into two halves $$$A_1$$$ and $$$A_2$$$. Let $$$A_1$$$ contain all elements smaller than or equal to the median and let $$$A_2$$$ contain all elements larger than the median. If we just know the sum of $$$A_1$$$ and $$$A_2$$$, then we can find out $$$f(x)$$$ in $$$O(1)$$$ time.

Here is my GitHub repository consisting of all my programs and explanations. As an experiment, I have started posting some 'rough' notes that I make while working my way through the problem as well. It is not intended to be the full and formal solution. Just some thoughts and key points.