一个数组有2个元素。 一个元素是唯一的，另一个元素可以重复任意多次。 在<O（n）时间内找到不同的元素

Question

O(n) solution is straight forward. I can think of something in terms of binary search but that again will be O(n) in worst case as the input array need not be sorted. Is there any solution that runs in O(log n) time?

Answer 1

May be one of the solution can be based on divide and conquer techniques.

eg Array -> [2,2,2,2,3,2,2]

Take three variables `First`, `Mid` and `Last`
1. Find `First` first element     -> 2 in our case.
2. Find length of array   -> 7 
3. Find `Mid` element     -> 4
4. Sum till `Mid` element   -> 2+2+2+2=8 (calculate by Mid * First)
5. If sum from `First` and `Mid` is equal to (`Mid` * `First`) then set `First` as `Mid`+-1 else set `Last` as `Mid`+1

The above method will divide the array to half for each comparision in worst case scenario.