CPMagical CP Library

Problem solution · Python

Subarrays Distinct Element Sum of Squares I

Subarrays Distinct Element Sum of Squares I: 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
184 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 Subarrays Distinct Element Sum of Squares I, 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

  • 184 lines of Python from the credited upstream file subarrays-distinct-element-sum-of-squares-i.py.
  • The implementation visibly relies on sequence storage, hash lookup, ordered 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 codeSubarrays Distinct Element Sum of Squares I · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(nlogn)# Space: O(n) import collectionsfrom sortedcontainers import SortedList  # bit, fenwick tree, sorted list, mathclass Solution(object):    def sumCounts(self, nums):        """        :type nums: List[int]        :rtype: int        """        MOD = 10**9+7        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] = (self.__bit[i]+val) % MOD                    i += (i & -i)             def query(self, i):                i += 1  # Extra one for dummy node.                ret = 0                while i > 0:                    ret = (ret+self.__bit[i]) % MOD                    i -= (i & -i)                return ret         def update(accu, d):            i = sl.bisect_left(idxs[x][-1])            accu = (accu + d*(len(nums)*(2*len(sl)-1) - (2*i+1)*idxs[x][-1] - 2*(bit.query(len(nums)-1)-bit.query(idxs[x][-1])))) % MOD            bit.add(idxs[x][-1], d*idxs[x][-1])            return accu         idxs = collections.defaultdict(list)        for i in reversed(xrange(len(nums))):            idxs[nums[i]].append(i)        result = 0        sl = SortedList(idxs[x][-1] for x in idxs)        accu = (len(nums)*len(sl)**2) % MOD        for i, x in enumerate(sl):            accu = (accu-(2*i+1)*x) % MOD        bit = BIT(len(nums))        for x in sl:            bit.add(x, x)        for x in nums:            result = (result+accu) % MOD  # accu = sum(count(i, k) for k in range(i, len(nums)))            accu = update(accu, -1)            del sl[0]            idxs[x].pop()            if not idxs[x]:                continue            sl.add(idxs[x][-1])            accu = update(accu, +1)        assert(accu == 0)        return result  # Time:  O(nlogn)# Space: O(n)# dp, segment tree, mathclass Solution2(object):    def sumCounts(self, nums):        """        :type nums: List[int]        :rtype: int        """        MOD = 10**9+7        # Template:        # https://github.com/kamyu104/LeetCode-Solutions/blob/master/Python/longest-substring-of-one-repeating-character.py        class SegmentTree(object):            def __init__(self, N,                         build_fn=None,                         query_fn=lambda x, y: y if x is None else x if y is None else (x+y)%MOD,                         update_fn=lambda x, y: y if x is None else (x+y)%MOD):                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])                self.count = [1]*len(self.tree)  # added                for i in reversed(xrange(1, self.base)):  # added                    self.count[i] = self.count[i<<1] + self.count[(i<<1)+1]             def __apply(self, x, val):                self.tree[x] = self.update_fn(self.tree[x], val*self.count[x])  # modified                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]*self.count[x])  # modified                 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)         result = accu = 0        sl = {}        st = SegmentTree(len(nums))        for i in xrange(len(nums)):            j = sl[nums[i]] if nums[i] in sl else -1            # sum(count(k, i)^2 for k in range(i+1)) - sum(count(k, i-1)^2 for k in range(i))            # = sum(2*count(k, i-1)+1 for k in range(j+1, i+1))            # = (i-j) + sum(2*count(k, i-1) for k in range(j+1, i+1))            accu = (accu+((i-j)+2*max(st.query(j+1, i), 0)))%MOD            result = (result+accu)%MOD            st.update(j+1, i, 1)  # count(k, i) = count(k, i-1)+(1 if k >= j+1 else 0) for k in range(i+1)            sl[nums[i]] = i        return result  # Time:  O(n^2)# Space: O(n)# hash tableclass Solution3(object):    def sumCounts(self, nums):        """        :type nums: List[int]        :rtype: int        """        MOD = 10**9+7        result = 0        for i in xrange(len(nums)):            lookup = set()            for j in reversed(xrange(i+1)):                lookup.add(nums[j])                result = (result+len(lookup)**2) % MOD        return result

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗