Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:
[4,5,6,7,0,1,4] if it was rotated 4 times.[0,1,4,4,5,6,7] if it was rotated 7 times.Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array.
You must decrease the overall operation steps as much as possible.
Example 1:
Input: nums = [1,3,5] Output: 1
Example 2:
Input: nums = [2,2,2,0,1] Output: 0
Constraints:
n == nums.length1 <= n <= 5000-5000 <= nums[i] <= 5000nums is sorted and rotated between 1 and n times.Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?
The key idea is to apply a modified binary search on the rotated sorted array. In a rotated array, the minimum element lies in the portion where the order breaks. Normally, comparing nums[mid] with nums[right] helps determine whether the minimum lies in the left or right half.
However, this variation introduces duplicate values, which can make the decision ambiguous. If nums[mid] > nums[right], the minimum must be in the right half. If nums[mid] < nums[right], it lies in the left half including mid. When nums[mid] == nums[right], duplicates prevent a clear choice, so the search space is cautiously reduced by shrinking the boundary.
This approach keeps the efficiency of binary search in most cases while safely handling duplicates. The average complexity remains logarithmic, but in the worst case with many duplicates, the search may degrade toward linear time.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Modified Binary Search | O(log n) average, O(n) worst case (due to duplicates) | O(1) |
NeetCode
This approach utilizes hash maps (dictionaries in Python) to achieve optimal time and space complexity for operations such as insertion, deletion, and lookup. By storing elements as keys in a hash map, we benefit from average-case O(1) time complexity for these operations.
Time Complexity: O(1) on average for search, insert, and delete due to the hash map.
Space Complexity: O(n), where n is the number of elements stored.
1class MySet:
2 def __init__(self):
3 self._data = {}
4
5 def insert(self, key):
Python's dictionaries serve as a direct means to emulate hash maps, where presence in the dictionary indicates inclusion in the set.
This approach leverages sorted arrays to perform efficient binary searches. Operations are optimized for scenarios requiring sorted data, such as when frequent minimum/maximum queries are performed.
Time Complexity: O(n log n) for insertion (due to sorting), O(log n) for search (binary search).
Space Complexity: O(n).
1
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Yes, variations of rotated array problems are frequently asked in FAANG and other top tech company interviews. They test understanding of binary search, edge cases, and handling duplicates efficiently.
Duplicates can make it impossible to determine which half of the array is sorted by just comparing middle and boundary elements. In such cases, the algorithm must shrink the search space gradually, which can lead to linear time in the worst case.
The optimal approach uses a modified binary search. By comparing the middle element with the right boundary, we can determine which side contains the minimum. When duplicates appear, the boundary is reduced carefully to maintain correctness.
The problem primarily uses arrays along with the binary search technique. No additional data structures are required, allowing the solution to run with constant extra space.
Python's bisect module provides efficient insertion (keeping the list sorted) and binary search capabilities, making it ideal for this approach.