Rotated Sorted Array Search InterviewBit Solution

Problem: Rotated Sorted Array Search

Problem Description:

Given an array of integers A of size N and an integer B.

array A 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 B to search. If found in the array, return its index, otherwise, return -1.

You may assume no duplicate exists in the array.

Note:-

Input Format

The first argument given is the integer array A. The second argument given is the integer B.

Output Format

Return index of B in array A, otherwise return -1

Constraints

1 <= N <= 1000000
 1 <= A[i] <= 10^9
 all elements in A are disitinct.

For Example

Input 1:
  A = [4, 5, 6, 7, 0, 1, 2, 3]
  B = 4
 Output 1:
  0
 Explanation 1:
  Target 4 is found at index 0 in A.

Input 2:
  A = [5, 17, 100, 3]
  B = 6
 Output 2:
  -1

Since before rotation, the array was sorted, this means that we can use binary search.

In a binary search algorithm, we have three indexes, low, mid and high.

The mid variable divides the array into two part, one is [low,mid] and another is [mid,high].

You can clearly see that, whatever the mid value we choose, there will be at least one part that will be sorted.

For example,

A = [4, 5, 0, 1, 2, 3]
 
 For low = 0, high = 5
 mid = 0,
  [A[low], A[mid]] = [4] // Sorted
 
 mid = 1,
  [A[low], A[mid]] = [4, 5] // Sorted
 
 mid = 2,
  [A[low], A[mid]] = [4, 5, 0] // Unsorted but,
  [A[mid], A[high]] = [0, 1, 2, 3] // Sorted
 
 Similarly you can check for other indexes.

Once you know which part is sorted then it becomes quite easy to find which side you have to shift. See below code to understand conditions a bit further.

Time Complexity: O(logN)

Space Complexity: O(1)

- Ignoring space taken by input array, we have used only constant space.

Code in C++