Search in Rotated Sorted Array II (With Duplicates)
Search in Rotated Sorted Array II (LeetCode 81) with duplicates: binary search that shrinks ambiguous ends, O(log n) average and O(n) worst case.
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Arrays
- Find the Largest Element in an Array
- Find the Second Largest Element in an Array
- Reverse an Array In Place
- Check if an Array Is Sorted
- Remove Duplicates from a Sorted Array
- Move Zeroes to the End of an Array
- Rotate an Array by K Positions
- Two Sum
- Maximum Subarray Sum (Kadane's Algorithm)
- Best Time to Buy and Sell Stock
- Sort an Array of 0s, 1s and 2s (Dutch National Flag)
- Find the Missing Number in an Array
- Find the Duplicate Number in an Array
- Union of Two Sorted Arrays
- Intersection of Two Sorted Arrays
- Prefix Sum Array and Range Sum Queries
- Majority Element (Boyer-Moore Voting Algorithm)
- Merge Overlapping Intervals
- Next Permutation
- Product of Array Except Self
- Subarray Sum Equals K
- Merge Two Sorted Arrays In Place
- Maximum Product Subarray
- 3Sum: Find All Triplets That Sum to Zero
- Trapping Rain Water
- Container With Most Water
- Rotate Image (Matrix 90 Degree Rotation)
- Spiral Matrix Traversal
- Set Matrix Zeroes
- Kth Largest Element in an Array (Quickselect)
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Strings
- Reverse a String In Place
- Check if a String Is a Palindrome
- Check if Two Strings Are Anagrams
- Count Frequency of Characters in a String
- First Non-Repeating Character in a String
- Reverse Words in a Sentence
- Check if a String Contains Only Digits
- Remove Duplicate Characters from a String
- Longest Substring Without Repeating Characters
- Group Anagrams
- Valid Parentheses
- String to Integer (atoi)
- Longest Common Prefix
- Implement strStr() - Naive Pattern Matching
- Run-Length Encoding of a String
- Check if a String Is a Rotation of Another
- Longest Palindromic Substring
- Minimum Window Substring
- Zigzag Conversion
- KMP Pattern Matching Algorithm
- Rabin-Karp Pattern Matching Algorithm
- Z-Algorithm for Pattern Matching
- Longest Palindromic Subsequence
- Edit Distance Between Two Strings
- Word Break Problem
- Decode Ways
- String Compression (In-Place)
- Count and Say
- Roman to Integer
- Integer to Roman
- Valid Palindrome II (At Most One Deletion)
- Longest Repeating Character Replacement
- Multiply Two Numbers Represented as Strings
- Why Strings Are Immutable (Interview Concept)
- Time Complexity of Common String Operations (Interview Co...
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Sliding Window
- Maximum Sum Subarray of Size K (Fixed Window)
- Average of All Subarrays of Size K
- Maximum Number of Vowels in a Substring of Length K
- First Negative Integer in Every Window of Size K
- K Radius Subarray Averages
- Max Consecutive Ones
- Max Consecutive Ones III (At Most K Flips)
- Longest Subarray With Sum At Most K
- Minimum Size Subarray Sum
- Fruit Into Baskets (Longest Subarray With At Most 2 Disti...
- Longest Subarray With At Most K Distinct Elements
- Subarray Product Less Than K
- Maximum Average Subarray I
- Sliding Window Maximum (Monotonic Deque)
- Find All Anagrams in a String (Sliding Window)
- Repeated DNA Sequences (Fixed Window Hashing)
- Longest Turbulent Subarray
- Maximum Erasure Value (Longest Unique Subarray Sum)
- Longest Continuous Increasing Subsequence
- Grumpy Bookstore Owner
- Maximum Points You Can Obtain from Cards
- Binary Subarrays With Sum (Exactly K Trick)
- Count Number of Nice Subarrays (Exactly K Odd Numbers)
- Continuous Subarray Sum (Multiple of K)
- Minimum Operations to Reduce X to Zero
- Maximum Frequency After K Increments
- Sliding Window Median
- Substring With Concatenation of All Words
- Count Complete Subarrays in an Array
- Number of Subarrays With Bounded Maximum
- Maximum Sum of Distinct Subarrays With Length K
- Shortest Subarray With Sum At Least K (With Negatives)
- Fixed-Size vs Variable-Size Sliding Window (Interview Con...
- When to Use Sliding Window vs Two Pointers (Interview Con...
- Recognizing a Sliding Window Problem (Interview Concept)
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Two Pointers
- Two Sum II - Input Array Is Sorted
- Valid Palindrome (Two Pointer Check)
- Reverse Vowels of a String
- Remove Element In Place
- Move Zeroes Using Two Pointers
- 3Sum Closest
- 4Sum
- Sort Array By Parity
- Squares of a Sorted Array
- Merge Two Sorted Arrays (Two Pointer)
- Intersection of Two Arrays II
- Backspace String Compare
- Trapping Rain Water (Two Pointer Approach)
- Container With Most Water (Two Pointer Approach)
- Partition Labels
- Sort Colors (Dutch National Flag Two Pointer)
- Remove Duplicates from Sorted Array II
- Linked List Cycle Detection (Fast and Slow Pointers)
- Middle of the Linked List
- Palindrome Linked List Check
- Boats to Save People
- Valid Triangle Number
- Minimum Number of Swaps to Make Sequences Increasing
- Sentence Similarity Using Two Pointers
- Shortest Distance to a Character
- Interleaving String Check
- DNA Complement Matching
- Is Subsequence Check (Two Pointer)
- Interview Concept: Same Direction vs Opposite Direction Two Pointers
- Interview Concept: Fast and Slow Pointer Technique Explained
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Searching & Binary Search
- Binary Search (Classic Iterative)
- Binary Search (Recursive)
- Search Insert Position
- First and Last Position of Element in Sorted Array
- Search in Rotated Sorted Array
- Search in Rotated Sorted Array II (With Duplicates)
- Find Minimum in Rotated Sorted Array
- Find Peak Element
- Sqrt(x) Using Binary Search
- Koko Eating Bananas
- Capacity To Ship Packages Within D Days
- Split Array Largest Sum
- Median of Two Sorted Arrays
- Search a 2D Matrix
- Search a 2D Matrix II
- Find K Closest Elements
- Kth Smallest Element in a Sorted Matrix
- Aggressive Cows (Binary Search on Answer)
- Allocate Minimum Number of Pages
- Single Element in a Sorted Array
- Find the Smallest Divisor Given a Threshold
- Binary Search on a Function (Ternary Search Alternative)
- Count of Smaller Numbers After Self
- Time Based Key-Value Store
- Guess Number Higher or Lower
- H-Index II
- Find Right Interval
- Random Pick with Weight
- Interview Concept: Lower Bound vs Upper Bound Binary Search
- Interview Concept: Binary Search on Answer Pattern Explained
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Sorting
- Bubble Sort Implementation
- Selection Sort Implementation
- Insertion Sort Implementation
- Merge Sort Implementation
- Quick Sort Implementation
- Heap Sort Implementation
- Counting Sort Implementation
- Radix Sort Implementation
- Bucket Sort Implementation
- Sort an Array (Built-in vs Custom Comparator)
- Merge Intervals Using Sorting
- Largest Number Formed from an Array
- Custom Sort String
- Sort Characters By Frequency
- Kth Largest Element (Sorting Approach)
- Relative Sort Array
- Wiggle Sort
- Wiggle Sort II
- H-Index (Sorting Approach)
- Meeting Rooms (Can Attend All Meetings)
Problem Statement
Same as Search in Rotated Sorted Array, but the array may contain duplicates. Return true if the target exists, otherwise false.
Example
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: trueApproach
Duplicates break the "which half is sorted" test when nums[left] == nums[mid] == nums[right]. In that single case, shrink both ends by one (left++, right--) and continue. Otherwise the logic is identical to the no-duplicate version.
Complexity
- Time: O(log n) average, O(n) worst case
- Space: O(1)
Solution Code
Python
def search(nums, target):
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return True
if nums[left] == nums[mid] == nums[right]:
left += 1
right -= 1
elif nums[left] <= nums[mid]:
if nums[left] <= target < nums[mid]:
right = mid - 1
else:
left = mid + 1
else:
if nums[mid] < target <= nums[right]:
left = mid + 1
else:
right = mid - 1
return FalseJava
boolean search(int[] nums, int target) {
int left = 0, right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return true;
if (nums[left] == nums[mid] && nums[mid] == nums[right]) {
left++; right--;
} else if (nums[left] <= nums[mid]) {
if (nums[left] <= target && target < nums[mid]) right = mid - 1;
else left = mid + 1;
} else {
if (nums[mid] < target && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return false;
}C++
bool search(vector<int>& nums, int target) {
int left = 0, right = (int)nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return true;
if (nums[left] == nums[mid] && nums[mid] == nums[right]) {
left++; right--;
} else if (nums[left] <= nums[mid]) {
if (nums[left] <= target && target < nums[mid]) right = mid - 1;
else left = mid + 1;
} else {
if (nums[mid] < target && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return false;
}Key Takeaways
- Duplicates can force O(n) in the worst case (e.g. all equal values).
- The only new rule: when both ends equal mid, shrink the window by one.
- Reference: LeetCode 81 (Search in Rotated Sorted Array II).