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Problem solution · Python

Maximum Total Subarray Value II

Maximum Total Subarray Value 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
144 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 Maximum Total Subarray Value 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

  • 144 lines of Python from the credited upstream file maximum-total-subarray-value-ii.py.
  • The implementation visibly relies on sequence storage, work queue.
  • 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 codeMaximum Total Subarray Value II · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O((n + k) * logn)# Space: O(n + k) import heapq  # heap, sort, two pointersclass Solution(object):    def maxTotalValue(self, nums, k):        """        :type nums: List[int]        :type k: int        :rtype: int        """        def nxt(left, right, i, j):            while not (left <= idxs[i] <= right):                i += 1            while not (left <= idxs[j] <= right):                j -= 1            return (i, j)                    idxs = range(len(nums))        idxs.sort(key=lambda x: (nums[x], x))        lookup = {(0, len(nums)-1):(0, len(idxs)-1)}        max_heap = [(-(nums[idxs[len(idxs)-1]]-nums[idxs[0]]), (0, len(idxs)-1))]        result = 0        while k:            v, (l, r) = heapq.heappop(max_heap)            i, j = lookup[(l, r)]            nl, nr = min(idxs[i], idxs[j]), max(idxs[i], idxs[j])            c = min((nl-l+1)*(r-nr+1), k)            k -= c            result += c*(-v)            if nl+1 <= r and (nl+1, r) not in lookup:                lookup[(nl+1, r)] = (ni, nj) = nxt(nl+1, r, i, j)                heapq.heappush(max_heap, (-(nums[idxs[nj]]-nums[idxs[ni]]), (nl+1, r)))            if l <= nr-1 and (l, nr-1) not in lookup:                lookup[(l, nr-1)] = (ni, nj) = nxt(l, nr-1, i, j)                heapq.heappush(max_heap, (-(nums[idxs[nj]]-nums[idxs[ni]]), (l, nr-1)))        return result  # Time:  O((n + k) * logn)# Space: O(nlogn)import heapq  # heap, rmq, sparse tableclass Solution2(object):    def maxTotalValue(self, nums, k):        """        :type nums: List[int]        :type k: int        :rtype: int        """        # RMQ - Sparse Table        # Template: https://github.com/kamyu104/GoogleCodeJam-Farewell-Rounds/blob/main/Round%20D/genetic_sequences2.py3        # Time:  ctor:  O(NlogN) * O(fn)        #        query: O(fn)        # Space: O(NlogN)        class SparseTable(object):            def __init__(self, arr, fn):                self.fn = fn                self.bit_length = [0]                n = len(arr)                k = n.bit_length()-1  # log2_floor(n)                for i in xrange(k+1):                    self.bit_length.extend(i+1 for _ in xrange(min(1<<i, (n+1)-len(self.bit_length))))                self.st = [[0]*n for _ in xrange(k+1)]                self.st[0] = arr[:]                for i in xrange(1, k+1):  # Time: O(NlogN) * O(fn)                    for j in xrange((n-(1<<i))+1):                        self.st[i][j] = fn(self.st[i-1][j], self.st[i-1][j+(1<<(i-1))])                    def query(self, L, R):  # Time: O(fn)                i = self.bit_length[R-L+1]-1  # log2_floor(R-L+1)                return self.fn(self.st[i][L], self.st[i][R-(1<<i)+1])                rmq_min = SparseTable(nums, min)        rmq_max = SparseTable(nums, max)        max_heap = [(-(rmq_max.query(i, len(nums)-1)-rmq_min.query(i, len(nums)-1)), (i, len(nums)-1)) for i in xrange(len(nums))]        heapq.heapify(max_heap)        result = 0        for _ in xrange(k):            v, (i, j) = heappop(max_heap)            result += -v            if i <= j-1:                heapq.heappush(max_heap, (-(rmq_max.query(i, j-1)-rmq_min.query(i, j-1)), (i, j-1)))        return result  # Time:  O((n + k) * logn)# Space: O(n)import heapq  # heap, segment treeclass Solution3(object):    def maxTotalValue(self, nums, k):        """        :type nums: List[int]        :type k: int        :rtype: int        """        class SegmentTree(object):            def __init__(self, N, build_fn, query_fn):                self.tree = [None]*(1<<((N-1).bit_length()+1))                self.base = len(self.tree)>>1                self.query_fn = query_fn                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 query(self, L, R):                if L > R:                    return None                L += self.base                R += self.base                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)             st_min = SegmentTree(len(nums), build_fn=lambda x: nums[x], query_fn=lambda x, y: y if x is None else x if y is None else min(x, y))        st_max = SegmentTree(len(nums), build_fn=lambda x: nums[x], query_fn=lambda x, y: y if x is None else x if y is None else max(x, y))        max_heap = [(-(st_max.query(i, len(nums)-1)-st_min.query(i, len(nums)-1)), (i, len(nums)-1)) for i in xrange(len(nums))]        heapq.heapify(max_heap)        result = 0        for _ in xrange(k):            v, (i, j) = heappop(max_heap)            result += -v            if i <= j-1:                heapq.heappush(max_heap, (-(st_max.query(i, j-1)-st_min.query(i, j-1)), (i, j-1)))        return result

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