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I like how you approached the problem — breaking it down into smaller cases really helps expose edge conditions before scaling up. It reminds me of planning a smooth ride before a long trip, where early preparation saves time later. Out of curiosity, are you finding more difficulty with handling time limits or with managing tricky edge cases in your solution?

The key to solving the Johnland problem is to realize that the shortest paths between cities will always lie within a minimum spanning tree (MST), since the edge weights are distinct powers of two. Once you construct the MST, each edge can be seen as a “bridge” that splits the graph into two groups. The contribution of that edge to the total distance is its weight multiplied by the number of node pairs it separates, and by summing these contributions you get the overall result. Finally, the answer is expressed in binary format. Visit site For a deeper dive into this concept and related problem-solving strategies.

Hi there, Johnland can be tricky without a clear plan. It helps to first check the constraints, try small examples by hand, and then start with a simple version to confirm the logic before optimizing. Often techniques like prefix sums or two-pointers make a big difference. I think of it like how connects short trips — small steps build the bigger solution. Are you running into time limit errors, or is it more about wrong answers on edge cases?

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