You pick a number (for first and third block) and iterate on its positions. Irregardless of the 'number' of numbers, there are only N positions in total to pick, thus the complexity is O(N * 200). Of course you have to implement it in an appropriate way. Similar to how, when you DFS/BFS, you iterate on the children of a node, and you do it for every node! It seems O(N^2) but there are only N distinct nodes to visit in total.