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Problem solution · Python

Handling Sum Queries After Update

Handling Sum Queries After Update: 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
103 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 Handling Sum Queries After Update, 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

  • 103 lines of Python from the credited upstream file handling-sum-queries-after-update.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 codeHandling Sum Queries After Update · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(nlogn + m + qlogn)# Space: O(n) # segment treeclass Solution(object):    def handleQuery(self, nums1, nums2, queries):        """        :type nums1: List[int]        :type nums2: List[int]        :type queries: List[List[int]]        :rtype: List[int]        """        class SegmentTree(object):  # 0-based index            def __init__(self, N,                        build_fn=lambda _: 0,                        query_fn=lambda x, y: y if x is None else max(x, y),                        update_fn=lambda x, y: y if x is None else x+y):                self.base = N                self.H = (N-1).bit_length()                self.query_fn = query_fn                self.update_fn = update_fn                self.tree = [None]*(2*N)                self.lazy = [None]*N                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[2*i], self.tree[2*i+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 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])                 if L > R:                    return                L += self.base                R += self.base                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):  # Time: O(logN), Space: O(N)                def push(x):                    n = self.H                    while n:                        y = x >> n                        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                        n -= 1                 result = None                if L > R:                    return result                 L += self.base                R += self.base                push(L)                push(R)                while L <= R:                    if L & 1:  # is right child                        result = self.query_fn(result, self.tree[L])                        L += 1                    if R & 1 == 0:  # is left child                        result = self.query_fn(result, self.tree[R])                        R -= 1                    L >>= 1                    R >>= 1                return result         st = SegmentTree(len(nums1),                         build_fn=lambda i: (nums1[i], nums1[i]^1),                         query_fn=lambda x, y: y if x is None else (x[0]+y[0], x[1]+y[1]),                         update_fn=lambda x, y: y if x is None else (x[1], x[0]) if y == (1, 0) else x)        result = []        total = sum(nums2)        for t, a, b in queries:            if t == 1:                st.update(a, b, (1, 0))            elif t == 2:                total += st.query(0, len(nums1)-1)[0]*a            elif t == 3:                result.append(total)        return result

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