Understanding the Problem
We are given a rotated sorted array that may contain duplicate elements. Our goal is to determine if a target number exists in this array.
A rotated sorted array is one that has been shifted at some pivot point. For example, the array [1, 2, 3, 4, 5]
might be rotated to become [4, 5, 1, 2, 3]
. The twist here is that the array may also contain duplicate values, which makes the traditional binary search approach slightly more complex.
Step-by-Step Solution with Example
Step 1: Start with Binary Search
We initialize two pointers: start
at index 0, and end
at the last index of the array. The idea is to repeatedly check the middle element and decide whether to move left or right based on sorted properties.
Step 2: Check the Middle Element
Find the middle index using mid = Math.floor((start + end) / 2)
. If arr[mid] == target
, return true
.
Step 3: Handle Duplicates at Boundaries
If arr[start] == arr[mid] == arr[end]
, we can't determine which side is sorted. In such cases, we cautiously move the start
pointer forward and the end
pointer backward by one step to eliminate duplicates.
Step 4: Determine Which Half is Sorted
If arr[start] <= arr[mid]
, the left half is sorted.
- If
target
lies between arr[start]
and arr[mid]
, narrow the search to the left half: end = mid - 1
.
- Otherwise, search the right half:
start = mid + 1
.
Step 5: Right Half is Sorted
If the left half isn’t sorted, then the right half must be sorted. Check if the target
lies in the range arr[mid] to arr[end]
.
- If it does, shift to the right half:
start = mid + 1
.
- Else, move to the left half:
end = mid - 1
.
Step 6: Repeat Until Found or Exhausted
Repeat the above steps while start <= end
. If the loop ends without finding the target, return false
.
Edge Cases
- Empty Array: If the input array is empty, the result is immediately
false
.
- Single Element: Check if the single element matches the target.
- All Elements Same: Binary search might degrade to linear time
O(n)
since we cannot skip chunks based on sorted halves.
- No Rotation: The array may already be sorted with no rotation — our approach still works correctly.
Finally
This approach enhances binary search to work with rotated arrays and duplicates. Though average-case complexity is O(log n)
, it can degrade to O(n)
in the worst case due to duplicates. Still, it handles all key scenarios:
- Fully sorted or rotated arrays
- Presence of duplicates
- Tricky ambiguous cases with same start, mid, and end values
Always take time to understand the problem's structure and apply conditional logic carefully when standard binary search rules don't apply.
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