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Solution to (1):
- united November 14, 2014Let Fk (1<=k<=N) be the first k elements in the list. Let sk be the sum of elements in Fk.
If there exists k such that sk%N = 0, then return elements in Fk.
Otherwise, every sk%N is between 1 and N-1. So, among s1%N, s2%N, ..., sN%N, there must exist i, j, si%N = sj%N. Return elements in Fj but not in Fi (suppose j > i).
This takes O(N) time. The same solution doesn't apply for question (2). But it seems a BFS approach should still use O(N) time.