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Problem solution · Python

Subarrays with Xor at Least K

Subarrays with Xor at Least K: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Direct simulation
Source
Kamyu LeetCode Solutions
Length
122 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

Direct simulation

For Subarrays with Xor at Least K, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 122 lines of Python from the credited upstream file subarrays-with-xor-at-least-k.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeSubarrays with Xor at Least K · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(nlogr), r = max(max(nums), k, 1)# Space: O(nlogr) # bitmasks, prefix sum, trieclass Solution(object):    def countXorSubarrays(self, nums, k):        """        :type nums: List[int]        :type k: int        :rtype: int        """        class Trie(object):            def __init__(self, bit_length):                self.__lefts = [-1]*(1+(1+len(nums))*bit_length)  # preallocate to speed up performance                self.__rights = [-1]*(1+(1+len(nums))*bit_length)                self.__cnts = [0]*(1+(1+len(nums))*bit_length)                self.__i = 0                self.__new_node()                self.__bit_length = bit_length                        def __new_node(self):                self.__i += 1                return self.__i-1             def add(self, num):                curr = 0                for i in reversed(xrange(self.__bit_length)):                    x = (num>>i)&1                    if x == 0:                        if self.__lefts[curr] == -1:                            self.__lefts[curr] = self.__new_node()                        curr = self.__lefts[curr]                    else:                        if self.__rights[curr] == -1:                            self.__rights[curr] = self.__new_node()                        curr = self.__rights[curr]                    self.__cnts[curr] += 1                                    def query(self, prefix, k):                result = curr = 0                for i in reversed(xrange(self.__bit_length)):                    t = (k>>i)&1                    x = (prefix>>i)&1                    if t == 0:                        tmp = self.__lefts[curr] if 1^x == 0 else self.__rights[curr]                        if tmp != -1:                            result += self.__cnts[tmp]                    curr = self.__lefts[curr] if t^x == 0 else self.__rights[curr]                    if curr == -1:                        break                else:                    result += self.__cnts[curr]                return result            result = prefix = 0        mx = max(max(nums), k, 1)        trie = Trie(mx.bit_length())        trie.add(prefix)        for x in nums:            prefix ^= x            result += trie.query(prefix, k)            trie.add(prefix)        return result  # Time:  O(nlogr), r = max(max(nums), k, 1)# Space: O(t)# bitmasks, prefix sum, trieclass Solution_TLE(object):    def countXorSubarrays(self, nums, k):        """        :type nums: List[int]        :type k: int        :rtype: int        """        class Trie(object):            def __init__(self, bit_length):                self.__nodes = []                self.__cnts = []                self.__new_node()                self.__bit_length = bit_length                        def __new_node(self):                self.__nodes.append([-1]*2)                self.__cnts.append(0)                return len(self.__nodes)-1             def add(self, num):                curr = 0                for i in reversed(xrange(self.__bit_length)):                    x = (num>>i)&1                    if self.__nodes[curr][x] == -1:                        self.__nodes[curr][x] = self.__new_node()                    curr = self.__nodes[curr][x]                    self.__cnts[curr] += 1                                    def query(self, prefix, k):                result = curr = 0                for i in reversed(xrange(self.__bit_length)):                    t = (k>>i)&1                    x = (prefix>>i)&1                    if t == 0:                        tmp = self.__nodes[curr][1^x]                        if tmp != -1:                            result += self.__cnts[tmp]                    curr = self.__nodes[curr][t^x]                    if curr == -1:                        break                else:                    result += self.__cnts[curr]                return result                result = prefix = 0        mx = max(max(nums), k, 1)        trie = Trie(mx.bit_length())        trie.add(prefix)        for x in nums:            prefix ^= x            result += trie.query(prefix, k)            trie.add(prefix)        return result

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