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Problem solution · Python

Number of Integers with Popcount Depth Equal to K II

Number of Integers with Popcount Depth Equal to K II: a Python solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Segment tree or range structure
Source
Kamyu LeetCode Solutions
Length
70 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

Segment tree or range structure

For Number of Integers with Popcount Depth Equal to K II, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 70 lines of Python from the credited upstream file number-of-integers-with-popcount-depth-equal-to-k-ii.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

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 codeNumber of Integers with Popcount Depth Equal to K II · PythonPython
Use this to learn the idea, then write your own version.
# Time:  precompute: O((logr) * log(logr) + log*(r) * (logr)) = O((logr) * log(logr)), r = max(n)#        runtime:    O(nlogr + max_k * n + nlogn + qlogn)# Space: O(logr + max_k * n) # fenwick treedef popcount(x):    return bin(x).count('1')  def ceil_log2(x):    return (x-1).bit_length()  class BIT(object):  # 0-indexed.    def __init__(self, n):        self.__bit = [0]*(n+1)  # Extra one for dummy node.     def add(self, i, val):        i += 1  # Extra one for dummy node.        while i < len(self.__bit):            self.__bit[i] += val            i += (i & -i)     def query(self, i):        i += 1  # Extra one for dummy node.        ret = 0        while i > 0:            ret += self.__bit[i]            i -= (i & -i)        return ret  MAX_N = 10**15MAX_BIT_LEN = MAX_N.bit_length()D = [0]*(MAX_BIT_LEN+1)for i in xrange(2, MAX_BIT_LEN+1):    D[i] = D[popcount(i)]+1MAX_K = 0while MAX_N != 1:  # O(log*(MAX_N)) times    MAX_N = ceil_log2(MAX_N)    MAX_K += 1class Solution(object):    def popcountDepth(self, nums, queries):        """        :type nums: List[int]        :type queries: List[List[int]]        :rtype: List[int]        """        def count(x):            return D[popcount(x)]+1 if x != 1 else 0                bit = [BIT(len(nums)) for _ in xrange(MAX_K+1)]        for i in xrange(len(nums)):            bit[count(nums[i])].add(i, +1)        result = []        for q in queries:            if q[0] == 1:                _, l, r, k = q                assert(k < len(bit))                result.append(bit[k].query(r)-bit[k].query(l-1))            else:                _, i, x = q                old_d = count(nums[i])                new_d = count(x)                if new_d != old_d:                    bit[old_d].add(i, -1)                    bit[new_d].add(i, +1)                nums[i] = x        return result

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