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Problem solution · Python

Smallest Unique Subarray

Smallest Unique Subarray: a Python solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
41 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Smallest Unique Subarray, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 41 lines of Python from the credited upstream file smallest-unique-subarray.py.
  • The implementation visibly relies on sequence storage, hash lookup.
  • No explicit loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSmallest Unique Subarray · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(nlogn)# Space: O(n) import collections  # rolling hash, binary searchclass Solution(object):    def smallestUniqueSubarray(self, nums):        """        :type nums: List[int]        :rtype: int        """        MOD = 10**9+7        B = 29        def binary_search(left, right, check):            while left <= right:                mid = left+(right-left)//2                if check(mid):                    right = mid-1                else:                    left = mid+1            return left         def get_hash(l, r):            return (prefix[r+1]-prefix[l]*base[r-l+1])%MOD if l <= r else 0         def check(l):            cnt = collections.defaultdict(int)            for i in xrange(len(nums)-l+1):                cnt[get_hash(i, i+l-1)] += 1            return any(x == 1 for x in cnt.itervalues())            prefix = [0]*(len(nums)+1)        for i in xrange(len(prefix)-1):            prefix[i+1] = (prefix[i]*B+nums[i])%MOD        base = [1]*(len(nums)+1)        for i in xrange(len(base)-1):            base[i+1] = (base[i]*B)%MOD        return binary_search(1, len(nums)-1, check)

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