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Problem solution · Python

Maximize Count of Distinct Primes After Split

Maximize Count of Distinct Primes After Split: 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
131 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 Maximize Count of Distinct Primes After Split, 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

  • 131 lines of Python from the credited upstream file maximize-count-of-distinct-primes-after-split.py.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 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 codeMaximize Count of Distinct Primes After Split · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(r + nlogn + qlogn), r = max(nums)# Space: O(r + n) from sortedcontainers import SortedList  # number theory, sorted list, segment treedef linear_sieve_of_eratosthenes(n):  # Time: O(n), Space: O(n)    primes = []    spf = [-1]*(n+1)  # the smallest prime factor    for i in xrange(2, n+1):        if spf[i] == -1:            spf[i] = i            primes.append(i)        for p in primes:            if i*p > n or p > spf[i]:                break            spf[i*p] = p    return spf  MAX_N = 10**5SPF = linear_sieve_of_eratosthenes(MAX_N)class Solution(object):    def maximumCount(self, nums, queries):        """        :type nums: List[int]        :type queries: List[List[int]]        :rtype: List[int]        """        class SegmentTree(object):            def __init__(self, N,                         build_fn=lambda x: 0,                         query_fn=lambda x, y: y if x is None else x if y is None else max(x, y),                         update_fn=lambda x, y: y if x is None else x+y):                self.tree = [None]*(1<<((N-1).bit_length()+1))                self.base = len(self.tree)>>1                self.lazy = [None]*self.base                self.query_fn = query_fn                self.update_fn = update_fn                if build_fn is not None:                    for i in xrange(self.base, self.base+N):                        self.tree[i] = build_fn(i-self.base)                    for i in reversed(xrange(1, self.base)):                        self.tree[i] = query_fn(self.tree[i<<1], self.tree[(i<<1)+1])             def __apply(self, x, val):                self.tree[x] = self.update_fn(self.tree[x], val)                if x < self.base:                    self.lazy[x] = self.update_fn(self.lazy[x], val)             def __push(self, x):                for h in reversed(xrange(1, x.bit_length())):                    y = x>>h                    if self.lazy[y] is not None:                        self.__apply(y<<1, self.lazy[y])                        self.__apply((y<<1)+1, self.lazy[y])                        self.lazy[y] = None             def update(self, L, R, h):  # Time: O(logN), Space: O(N)                def pull(x):                    while x > 1:                        x >>= 1                        self.tree[x] = self.query_fn(self.tree[x<<1], self.tree[(x<<1)+1])                        if self.lazy[x] is not None:                            self.tree[x] = self.update_fn(self.tree[x], self.lazy[x])                 L += self.base                R += self.base                # self.__push(L)  # enable if range assignment                # self.__push(R)  # enable if range assignment                L0, R0 = L, R                while L <= R:                    if L & 1:  # is right child                        self.__apply(L, h)                        L += 1                    if R & 1 == 0:  # is left child                        self.__apply(R, h)                        R -= 1                    L >>= 1                    R >>= 1                pull(L0)                pull(R0)             def query(self, L, R):                if L > R:                    return None                L += self.base                R += self.base                self.__push(L)                self.__push(R)                left = right = None                while L <= R:                    if L & 1:                        left = self.query_fn(left, self.tree[L])                        L += 1                    if R & 1 == 0:                        right = self.query_fn(self.tree[R], right)                        R -= 1                    L >>= 1                    R >>= 1                return self.query_fn(left, right)         def add(i, d):            x = nums[i]            if SPF[x] != x:                return            if d == 1:                lookup[x].add(i)            if len(lookup[x]) == 1:                st.update(0, len(nums)-2, d)            elif i == lookup[x][0]:                st.update(i, lookup[x][1]-1, d)            elif i == lookup[x][-1]:                st.update(lookup[x][-2], i-1, d)            if d == -1:                lookup[x].remove(i)         lookup = collections.defaultdict(SortedList)        st = SegmentTree(len(nums)-1)        for i in xrange(len(nums)):            add(i, +1)        result = [0]*len(queries)        for i, (idx, x) in enumerate(queries):            if nums[idx] != x:                add(idx, -1)                nums[idx] = x                add(idx, +1)            result[i] = st.tree[1]  # st.query(0, len(nums)-2)        return result

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