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Problem solution · Python

Count Pairs with Xor in a Range

Count Pairs with Xor in a Range: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Dynamic programming
Source
Kamyu LeetCode Solutions
Length
73 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

Dynamic programming

For Count Pairs with Xor in a Range, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 73 lines of Python from the credited upstream file count-pairs-with-xor-in-a-range.py.
  • The implementation visibly relies on sequence storage, hash lookup, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeCount Pairs with Xor in a Range · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(n) import collections  # dp solutionclass Solution(object):    def countPairs(self, nums, low, high):        """        :type nums: List[int]        :type low: int        :type high: int        :rtype: int        """        def count(nums, x):            result = 0            dp = collections.Counter(nums)            while x:                if x&1:                    result += sum(dp[(x^1)^k]*dp[k] for k in dp.iterkeys())//2  # current limit is xxxxx1*****, count xor pair with xxxxx0***** pattern                dp = collections.Counter({k>>1: dp[k]+dp[k^1] for k in dp.iterkeys()})                x >>= 1            return result            return count(nums, high+1)-count(nums, low)  # Time:  O(n)# Space: O(n)# trie solutionclass Trie(object):    def __init__(self):        self.__root = {}            def insert(self, num):        node = self.__root        for i in reversed(xrange(32)):            curr = (num>>i) & 1            if curr not in node:                node[curr] = {"_count":0}            node = node[curr]            node["_count"] += 1                    def query(self, num, limit):        node, result = self.__root, 0        for i in reversed(xrange(32)):            curr = (num>>i) & 1            bit = (limit>>i) & 1            if bit:                if curr in node:                    result += node[0^curr]["_count"]  # current limit is xxxxx1*****, count xor pair with xxxxx0***** pattern            if bit^curr not in node:                break            node = node[bit^curr]        return result  class Solution2(object):    def countPairs(self, nums, low, high):        """        :type nums: List[int]        :type low: int        :type high: int        :rtype: int        """        result = 0        trie = Trie()        for x in nums:            result += trie.query(x, high+1)-trie.query(x, low)            trie.insert(x)        return result

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