## Search sorted array free

This can be done in O(logN) time and O(1) space by using a slightly modified binary search. Since the elements in X are in increasing order, the elements in the new array Y will be in nondecreasing order. So a binary search for 0 in Y will give the answer. But creating Y will take O(N) space and O(N) time. We want to search for a given element in a circular sorted array in complexity not greater than O(log n). Example: Search for 13 in 5, 9, 13, 1, 3. . My idea was to convert the circular array into a regular sorted array then do a binary search on the resulting array, but my problem was the algorithm I came up was stupid that it takes O(n) in the worst case:**search sorted array** How can the answer be improved?

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i. e. , [0, 1, 2, 4, 5, 6, 7 might become [4, 5, 6, 7, 0, 1, 2). You are given a target value to search. If found in the array return its index, otherwise return 1. . You may assume no duplicate exists in the array. *search sorted array* Search in an almost sorted array. A simple solution is to linearly search the given key in given array. Time complexity of this solution is O(n). We can modify binary search to do it in O(Logn) time. The idea is to compare the key with middle 3 elements, if present then return the index. With a sorted array, however, you can use the binary search algorithm, which performs the following steps: Let a 0 and b 6 (or more generally, b is the upper bounds of the array, which is 6 in our example above) Let k (b a) 2. If A[k equals Rolla then