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This is actually a math problem. Assume t is in the cycle with size 3*2^k, then k = floor(log2(t/3+1)). So, the problem is O(1).
Strange Counter
You are viewing a single comment's thread. Return to all comments →
This is actually a math problem. Assume t is in the cycle with size 3*2^k, then k = floor(log2(t/3+1)). So, the problem is O(1).