Summary of "BS-5. Search Element in Rotated Sorted Array II"

Summary of “BS-5. Search Element in Rotated Sorted Array II”

This video continues a binary search tutorial series, focusing on searching for an element in a rotated sorted array with duplicates (unlike the previous part which dealt with unique elements only). The main challenge addressed is how duplicates affect the binary search approach and how to modify the algorithm to handle them.

Main Ideas and Concepts

Problem Context

Search for a target element in a rotated sorted array that may contain duplicates.
Unlike part 1 (unique elements), here the array can have repeated values.
The goal is to return a boolean indicating whether the target exists, not the index.

Recap of Part 1 (Unique Elements)

Use binary search.
Identify which half (left or right) is sorted.
Check if the target lies in the sorted half.
Narrow down the search space accordingly.
This approach relies heavily on being able to identify the sorted half by comparing arr[mid] with arr[low] and arr[high].

Why the Previous Approach Fails with Duplicates

When duplicates are present, especially when arr[low] == arr[mid] == arr[high], it becomes impossible to determine which half is sorted just by comparing these elements.
Example: Array like [3,3,3,3,1,2] where low, mid, and high all point to the same value.
This ambiguity breaks the logic of confidently eliminating half the search space.

Key Insight and Solution for Duplicates

When arr[low] == arr[mid] == arr[high], we cannot decide the sorted half.
The solution is to shrink the search space linearly by incrementing low and decrementing high by one.
This trimming removes duplicates from the edges and may eventually allow identifying the sorted half.
Continue this process until the ambiguity is resolved or the target is found.

Algorithm Steps (Detailed)

Compare arr[mid] with the target.
- If equal, return true.
If arr[low], arr[mid], and arr[high] are equal:
- Increment low by 1.
- Decrement high by 1.
- Continue to the next iteration.
Else, determine which half is sorted:
- If left half is sorted:
  - Check if target lies within arr[low] to arr[mid].
    - If yes, set high = mid - 1.
    - Else, set low = mid + 1.
- Else (right half is sorted):
  - Check if target lies within arr[mid] to arr[high].
    - If yes, set low = mid + 1.
    - Else, set high = mid - 1.
Repeat until low exceeds high.
If not found, return false.

Time Complexity

Average case: O(log n), similar to normal binary search.
Worst case: O(n), especially when many duplicates cause repeated shrinking of the search space.

General Problem-Solving Advice

Solve the problem first assuming unique elements.
Identify failure cases with duplicates.
Modify the solution to handle those edge cases (like trimming duplicates).
This approach applies broadly to problems involving duplicates.

Methodology / Instructions to Implement the Solution

Initialize: low = 0, high = arr.length - 1
While low <= high:
- Calculate mid = low + (high - low) / 2
- If arr[mid] == target, return true
- If arr[low] == arr[mid] == arr[high]:
  - Increment low by 1
  - Decrement high by 1
  - Continue to next iteration
- Else if left half is sorted (arr[low] <= arr[mid]):
  - If arr[low] <= target < arr[mid], set high = mid - 1
  - Else, set low = mid + 1
- Else (right half is sorted):
  - If arr[mid] < target <= arr[high], set low = mid + 1
  - Else, set high = mid - 1
Return false if target not found

Speakers / Sources

Primary Speaker: The video creator/host (name not explicitly mentioned)
The content is part of the Striver’s A to Z DSA course playlist.

Additional Notes

The video references the availability of code implementations in C++, Java, Python, and JavaScript in the description.
The problem link is also provided in the description.
The speaker encourages viewers to like, subscribe, and follow on social media.

This summary captures the essence of the problem, why duplicates pose challenges, and the key modification to the binary search algorithm to handle those duplicates effectively.