How do you solve Maximum Segment Sum After Removals on LeetCode?

Maximum Segment Sum After Removals (LeetCode #2382) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n α(n)) time complexity and O(n) space complexity. Key insight: Using Union-Find helps manage segments efficiently.

What is the brute force solution for Maximum Segment Sum After Removals?

The brute force approach for Maximum Segment Sum After Removals is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves removing each element specified in the removeQueries one by one and recalculating the maximum segment sum after each removal. This is straightforward but inefficient as it recalculates sums repeatedly. The optimal Optimal Solution approach improves this to O(n α(n)).

What algorithm or data structure is used to solve Maximum Segment Sum After Removals?

Maximum Segment Sum After Removals uses the following concepts: Array, Union-Find, Prefix Sum, Ordered Set. The recommended patterns to study are: Union-Find, Dynamic Programming.

What are the interview tips for Maximum Segment Sum After Removals?

Explain your thought process clearly while coding. Discuss trade-offs between brute force and optimal solutions. Practice similar problems to get comfortable with Union-Find.

What are common mistakes when solving Maximum Segment Sum After Removals?

Not considering edge cases where all elements are removed. Failing to properly merge segments after removals.

#2382

Maximum Segment Sum After Removals

Hard

ArrayUnion-FindPrefix SumOrdered SetUnion-FindDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n α(n))
Space	O(1)	O(n)

💡

Intuition

Time O(n α(n))Space O(n)

The optimal solution leverages a Union-Find (Disjoint Set Union) data structure to efficiently manage segments as elements are removed. This allows us to dynamically merge segments and compute their sums without recalculating everything from scratch.

⚙️

Algorithm

3 steps

1Step 1: Initialize a Union-Find structure to manage segments and an array to store segment sums.
2Step 2: For each removal index in removeQueries, mark the element as removed and check its neighboring elements to merge segments if they are still present.
3Step 3: Update the segment sums accordingly and track the maximum segment sum after each removal.

solution.py38 lines

1class UnionFind:
2    def __init__(self, n):
3        self.parent = list(range(n))
4        self.size = [1] * n
5        self.sum = [0] * n
6    def find(self, x):
7        if self.parent[x] != x:
8            self.parent[x] = self.find(self.parent[x])
9        return self.parent[x]
10    def union(self, x, y):
11        rootX = self.find(x)
12        rootY = self.find(y)
13        if rootX != rootY:
14            if self.size[rootX] < self.size[rootY]:
15                rootX, rootY = rootY, rootX
16            self.parent[rootY] = rootX
17            self.size[rootX] += self.size[rootY]
18            self.sum[rootX] += self.sum[rootY]
19    def add(self, idx, value):
20        self.sum[idx] = value
21
22
23def maxSegmentSum(nums, removeQueries):
24    n = len(nums)
25    uf = UnionFind(n)
26    for i in range(n):
27        uf.add(i, nums[i])
28    answer = []
29    max_sum = sum(nums)
30    for idx in removeQueries:
31        uf.add(idx, 0)
32        if idx > 0 and nums[idx - 1] != 0:
33            uf.union(idx, idx - 1)
34        if idx < n - 1 and nums[idx + 1] != 0:
35            uf.union(idx, idx + 1)
36        max_sum = max(max_sum, max(uf.sum))
37        answer.append(max_sum)
38    return answer

ℹ

Complexity note: The time complexity is efficient due to the union-find operations, which are nearly constant time due to path compression and union by size. The space complexity is O(n) because we store additional arrays for parent, size, and sum.

1Using Union-Find helps manage segments efficiently.
2Dynamic updates to segment sums reduce redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.