How do you solve Handling Sum Queries After Update on LeetCode?

Handling Sum Queries After Update (LeetCode #2569) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) per query time complexity and O(n) space complexity. Key insight: Using a segment tree can significantly reduce the time complexity for range updates and queries.

What is the time complexity of Handling Sum Queries After Update?

The optimal solution for Handling Sum Queries After Update runs in O(log n) per query time and uses O(n) space. The segment tree allows us to perform updates and queries in logarithmic time, making it efficient for large inputs.

What is the brute force solution for Handling Sum Queries After Update?

The brute force approach for Handling Sum Queries After Update is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach directly applies each query to the arrays, flipping values and updating sums as needed. This is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(log n) per query.

What algorithm or data structure is used to solve Handling Sum Queries After Update?

Handling Sum Queries After Update uses the following concepts: Array, Segment Tree. The recommended patterns to study are: Segment Tree, Lazy Propagation.

What are the interview tips for Handling Sum Queries After Update?

Always clarify the constraints and edge cases before jumping into coding. Discuss your thought process and potential optimizations with the interviewer. Practice implementing data structures like segment trees to become comfortable with their usage.

What are common mistakes when solving Handling Sum Queries After Update?

Not properly handling the lazy propagation in segment trees. Failing to account for the range updates correctly.

#2569

Handling Sum Queries After Update

Hard

ArraySegment TreeSegment TreeLazy Propagation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n) per query
Space	O(1)	O(n)

💡

Intuition

Time O(log n) per querySpace O(n)

Using a lazy segment tree allows us to efficiently handle updates and queries in logarithmic time. This is crucial for large inputs where brute force would be too slow.

⚙️

Algorithm

4 steps

1Step 1: Build a lazy segment tree to manage the flips and updates efficiently.
2Step 2: For type 1 queries, mark the range for flipping in the segment tree.
3Step 3: For type 2 queries, propagate the updates through the segment tree to adjust nums2.
4Step 4: For type 3 queries, calculate the sum of nums2 using the segment tree.

solution.py59 lines

1class SegmentTree:
2    def __init__(self, n):
3        self.n = n
4        self.tree = [0] * (4 * n)
5        self.lazy = [0] * (4 * n)
6
7    def update(self, l, r, val, node=1, start=0, end=None):
8        if end is None:
9            end = self.n - 1
10        if self.lazy[node] != 0:
11            self.tree[node] ^= self.lazy[node]
12            if start != end:
13                self.lazy[node * 2] ^= self.lazy[node]
14                self.lazy[node * 2 + 1] ^= self.lazy[node]
15            self.lazy[node] = 0
16        if start > r or end < l:
17            return
18        if start >= l and end <= r:
19            self.tree[node] ^= val
20            if start != end:
21                self.lazy[node * 2] ^= val
22                self.lazy[node * 2 + 1] ^= val
23            return
24        mid = (start + end) // 2
25        self.update(l, r, val, node * 2, start, mid)
26        self.update(l, r, val, node * 2 + 1, mid + 1, end)
27        self.tree[node] = self.tree[node * 2] + self.tree[node * 2 + 1]
28
29    def query(self, idx, node=1, start=0, end=None):
30        if end is None:
31            end = self.n - 1
32        if self.lazy[node] != 0:
33            self.tree[node] ^= self.lazy[node]
34            if start != end:
35                self.lazy[node * 2] ^= self.lazy[node]
36                self.lazy[node * 2 + 1] ^= self.lazy[node]
37            self.lazy[node] = 0
38        if start == end:
39            return self.tree[node]
40        mid = (start + end) // 2
41        if idx <= mid:
42            return self.query(idx, node * 2, start, mid)
43        else:
44            return self.query(idx, node * 2 + 1, mid + 1, end)
45
46def handleQueries(nums1, nums2, queries):
47    n = len(nums1)
48    seg_tree = SegmentTree(n)
49    results = []
50    for query in queries:
51        if query[0] == 1:
52            seg_tree.update(query[1], query[2], 1)
53        elif query[0] == 2:
54            p = query[1]
55            for i in range(n):
56                nums2[i] += seg_tree.query(i) * p
57        elif query[0] == 3:
58            results.append(sum(nums2))
59    return results

ℹ

Complexity note: The segment tree allows us to perform updates and queries in logarithmic time, making it efficient for large inputs.

1Using a segment tree can significantly reduce the time complexity for range updates and queries.
2Understanding lazy propagation in segment trees is crucial for optimizing updates.

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