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Problem solution · Python

Xor After Range Multiplication Queries I

Xor After Range Multiplication Queries I: a Python solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Hash-based lookup
Source
Kamyu LeetCode Solutions
Length
54 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

Hash-based lookup

For Xor After Range Multiplication Queries I, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 54 lines of Python from the credited upstream file xor-after-range-multiplication-queries-i.py.
  • The implementation visibly relies on sequence storage, hash lookup.
  • No explicit loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 codeXor After Range Multiplication Queries I · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(qlogm + (q + n) * sqrt(n))# Space: O(n * sqrt(n)) import collections  # sqrt decomposition, difference array, fast exponentiationclass Solution(object):    def xorAfterQueries(self, nums, queries):        """        :type nums: List[int]        :type queries: List[List[int]]        :rtype: int        """        MOD = 10**9+7        def inv(x):            return pow(x, MOD-2, MOD)         block_size = int(len(nums)**0.5)+1        diffs = collections.defaultdict(lambda: [1]*len(nums))        for l, r, k, v in queries:            if k <= block_size:                diffs[k][l] = (diffs[k][l]*v)%MOD                r += k-(r-l)%k                if r < len(nums):                    diffs[k][r] = (diffs[k][r]*inv(v))%MOD            else:                for i in xrange(l, r+1, k):                    nums[i] = (nums[i]*v)%MOD        for k, diff in diffs.iteritems():            for i in xrange(len(diff)):                if i-k >= 0:                    diff[i] = (diff[i]*diff[i-k])%MOD                nums[i] = (nums[i]*diff[i])%MOD        return reduce(lambda accu, x: accu^x, nums, 0)  # Time:  O(q * n)# Space: O(1)# simulationclass Solution2(object):    def xorAfterQueries(self, nums, queries):        """        :type nums: List[int]        :type queries: List[List[int]]        :rtype: int        """        MOD = 10**9+7         for l, r, k, v in queries:            for i in xrange(l, r+1, k):                nums[i] = (nums[i]*v)%MOD        return reduce(lambda accu, x: accu^x, nums, 0)

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