Find Target In Rotated Sorted Array

Finding a target value in a rotated sorted array is a classic algorithmic challenge that extends the efficiency of binary search to dynamically restructured data. Unlike a standard sorted array where elements increase monotonically, a rotated sorted array undergoes a circular shift, splitting the sequence into two sorted subarrays. This structure preserves the overall sorted property in segments, allowing a modified binary search to locate any target in logarithmic time. Mastering this technique not only strengthens problem-solving skills for coding interviews but also deepens understanding of how traditional algorithms can adapt to non-trivial input arrangements.

Understanding the Rotated Sorted Array

A rotated sorted array originates from a normally sorted array that has been shifted left or right some number of times. But for example, the array [0, 1, 2, 4, 5, 6, 7] rotated at index 3 becomes [4, 5, 6, 7, 0, 1, 2]. The key observation is that at least one half of the array, when divided at the midpoint, will always remain sorted. This property forms the foundation of the modified search strategy Most people skip this — try not to..

The pivot element—the largest value in the array, or the point where the rotation occurs—divides the array into two regions: the left portion and the right portion, both internally sorted. Identifying which side contains the target, and whether the target lies within the sorted half, determines the next step of the search. This decision-making process is what distinguishes the rotated array search from its conventional counterpart Still holds up..

The Modified Binary Search Approach

The algorithm for finding a target in a rotated sorted array operates on the same divide-and-conquer principle as standard binary search, but with conditional logic to handle the rotation. At each iteration, the midpoint is calculated, and the relationship between the midpoint value, the leftmost value, and the target is evaluated The details matter here..

If the midpoint element equals the target, the search terminates successfully. Otherwise, the algorithm checks whether the left half (from low to mid) is sorted by comparing arr[low] and arr[mid]. If the left half is sorted and the target lies within its range (arr[low] ≤ target < arr[mid]), the search continues in the left half; otherwise, it shifts to the right half. Day to day, the reverse logic applies if the right half is sorted instead. This conditional branching ensures that each step eliminates half of the remaining elements, maintaining the O(log n) time complexity Easy to understand, harder to ignore..

Step-by-Step Procedure

To implement the search, follow these structured steps:

1. Initialize two pointers: low at index 0 and high at the last index of the array.
2. While low is less than or equal to high:
   • Calculate mid as low + (high - low) / 2 to avoid overflow.
   • If arr[mid] equals the target, return mid as the found index.
   • Determine which half is sorted:
     • If arr[low] ≤ arr[mid], the left half is sorted.
       • Check if the target lies between arr[low] and `arr[mid]