Codeforces 1189B. Number Circle Codeforces Round #572 (Div. 2)

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Let's suppose that array is sorted. First of all, if an≥an−1+an−2$a_n\ge a_{n-1}+a_{n-2}$, than the answer is — NO (because otherwise an$a_n$ is not smaller than sum of the neighbors). We claim, that in all other cases answer is — YES. One of the possible constructions (if the array is already sorted) is:

an−2,an,an−1,an−4,an−5,…,a1$a_{n-2},a_n,a_{n-1},a_{n-4},a_{n-5},\dots ,a_1$

It's easy to see, that all numbers except an$a_n$ will have at least one neighbor which is not smaller than itself. Complexity O(nlog(n))$O(nlog(n))$.

A solution in c++