Okay, so if we do that, every player of one group will already have played with all the players of the other group and now should play against the players of his own group. Hence the problem is divided in $$$\left \lceil{\dfrac{n}{2}}\right \rceil$$$ and $$$\left \lfloor{\dfrac{n}{2}}\right \rfloor$$$ individual problems. Divide and Conquer Now I just need to merge, Since the matches are held in parallel, I guess the number of matches will be max of left and right, + 1.