Search in Rotated Sorted Array (with Duplicates) Optimal Binary Search

Problem Statement❯

You are given a rotated sorted array that may contain duplicate values, and a target number key. Your task is to determine whether the key exists in the array or not.

The array was originally sorted in ascending order but then rotated at some pivot point. Due to rotation and the presence of duplicates, the structure can be irregular.

Your goal: Return true if the key is found, else return false.

Examples❯

Input Array	Key	Output	Description
[2, 5, 6, 0, 0, 1, 2]	0	true	Key is present after rotation with duplicates
[2, 5, 6, 0, 0, 1, 2]	3	false	Key not present in the array
[1, 0, 1, 1, 1]	0	true	Key is present, array has many duplicates
[1, 1, 1, 1, 1]	2	false	All values are duplicates, key is not present
[1]	1	true	Single-element array with match
[1]	0	false	Single-element array without match
[]	5	false	Empty array, so key cannot be present

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Solution❯

To solve this problem, we need to search for a given number in a rotated sorted array that may include duplicate values. At first glance, the problem seems similar to regular binary search, but rotation and duplicates introduce a few complications.

What Makes This Problem Tricky?

Rotation breaks the normal order of sorting at some pivot point.
Duplicates make it harder to decide which half of the array is sorted.

How Do We Approach It?

We use a variation of binary search that works even when we encounter duplicates. Here's how we handle different situations:

Case 1: Middle element matches the key

If the middle element is equal to the key, we return true. That’s the best-case scenario.

Case 2: Duplicates at the boundaries

If the elements at the start, mid, and end are all equal, we can’t tell which side is sorted. In this case, we safely move the start pointer forward and end pointer backward to reduce the ambiguity.

Case 3: Left half is sorted

If the left half from start to mid is sorted, we check if the key lies in that range. If it does, we search in the left half; otherwise, we move to the right half.

Case 4: Right half is sorted

If the right half from mid to end is sorted, and if the key lies in that part, we shift our search to the right half. Otherwise, we go left.

What If the Array is Empty?

If the array has no elements, the key can’t be found, so we simply return false.

Final Outcome

Using this strategy ensures we consider all scenarios, including:

Rotated sorted arrays
Completely sorted arrays
Arrays with lots of duplicates
Edge cases like empty or single-element arrays

This method still works in logarithmic time in the average case but may degrade to O(n) when many duplicates exist.

Algorithm Steps❯

Start with start = 0 and end = n - 1.
While start ≤ end, compute mid = start + (end - start) / 2.
If arr[mid] == key, return true.
If arr[start] == arr[mid] == arr[end], increment start and decrement end.
Else, check if left half is sorted (i.e., arr[start] ≤ arr[mid]).
→ If yes, and arr[start] ≤ key < arr[mid], move end = mid - 1.
→ Else, move start = mid + 1.
If right half is sorted (i.e., arr[mid] ≤ arr[end]):
→ If arr[mid] < key ≤ arr[end], move start = mid + 1.
→ Else, move end = mid - 1.
Repeat until found or range becomes invalid.

Code

Java

Python

JavaScript

C++

Kotlin

Swift

Php

public class RotatedArraySearchWithDuplicates {
  public static boolean search(int[] arr, int key) {
    int start = 0, end = arr.length - 1;

    while (start <= end) {
      int mid = start + (end - start) / 2;
      if (arr[mid] == key) return true;

      // Handle duplicates: shrink window
      if (arr[start] == arr[mid] && arr[mid] == arr[end]) {
        start++;
        end--;
      }
      // Left half is sorted
      else if (arr[start] <= arr[mid]) {
        if (arr[start] <= key && key < arr[mid]) {
          end = mid - 1;
        } else {
          start = mid + 1;
        }
      }
      // Right half is sorted
      else {
        if (arr[mid] < key && key <= arr[end]) {
          start = mid + 1;
        } else {
          end = mid - 1;
        }
      }
    }
    return false;
  }

  public static void main(String[] args) {
    int[] arr = {2, 5, 6, 0, 0, 1, 2};
    int key = 0;
    System.out.println("Is Key Present: " + search(arr, key));
  }
}

Time Complexity❯

Case	Time Complexity	Explanation
Best Case	O(1)	Key is found at the middle index in the first check.
Average Case	O(log n)	Binary search generally divides the array in half at each step.
Worst Case	O(n)	When duplicates exist everywhere, we might need to linearly shrink the search range.

Space Complexity❯

O(1)

Explanation: Only a constant number of pointers are used for binary search, without extra space.

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