#3480

Hard

LeetCode Problem Workspace

Maximize Subarrays After Removing One Conflicting Pair

Maximize the count of subarrays after removing one conflicting pair using array traversal and segment tree logic efficiently.

array segment tree enumeration prefix sum

Problem Statement

You are given an integer n representing an array nums with elements from 1 to n in order, and a list of conflictingPairs where each pair [a, b] indicates elements that cannot coexist in a subarray. Your task is to remove exactly one conflicting pair and calculate how many non-empty subarrays do not contain both elements of any remaining conflicting pair.

Return the maximum number of valid subarrays achievable after removing exactly one conflicting pair. Use efficient strategies such as prefix sums, enumeration, and segment tree updates to handle large n and multiple conflicting pairs within the constraints.

Examples

Example 1

Input: n = 4, conflictingPairs = [[2,3],[1,4]]

Output: 9

Example 2

Input: n = 5, conflictingPairs = [[1,2],[2,5],[3,5]]

Output: 12

Constraints

2 <= n <= 105
1 <= conflictingPairs.length <= 2 * n
conflictingPairs[i].length == 2
1 <= conflictingPairs[i][j] <= n
conflictingPairs[i][0] != conflictingPairs[i][1]

Solution Approach

Preprocess conflicting pairs

Map each number to the indices of pairs it participates in. Track the earliest ending index of a valid subarray starting from each position to quickly identify invalid segments.

Simulate removal of each pair

Iterate through each conflicting pair and temporarily ignore it. Update the segment tree or prefix structures to compute the longest valid subarrays for remaining pairs, keeping a running count of valid subarrays to determine the maximum.

Compute maximum subarrays

After evaluating each pair removal, combine results using prefix sums to efficiently count all non-empty valid subarrays. Return the highest count found, ensuring array and segment tree updates are applied correctly.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(n)

Time complexity is O(n) by processing each element and conflicting pair once using segment tree or prefix logic. Space complexity is O(n) to store mappings and segment structures for quick subarray evaluation.

What Interviewers Usually Probe

Check if candidates efficiently handle overlapping conflicting pairs using arrays and segment tree reasoning.
Look for correct calculation of longest valid subarrays starting from each index.
Assess whether the candidate correctly simulates removal of each conflicting pair for maximal subarrays.

Common Pitfalls or Variants

Common pitfalls

Forgetting to remove only one conflicting pair and recalculating subarrays for remaining pairs.
Inefficiently iterating over all subarrays instead of using segment tree or prefix sum optimizations.
Mishandling edge cases where conflicting pairs overlap at the start or end of the array.

Follow-up variants

Change the problem to allow removal of k conflicting pairs instead of one.
Compute maximum subarrays when conflicting pairs are dynamic and updated online.
Extend to arrays with repeated numbers instead of 1 to n sequential integers.

FAQ

What is the main strategy to solve Maximize Subarrays After Removing One Conflicting Pair?

Focus on removing one conflicting pair at a time and use segment tree or prefix sum logic to calculate the longest valid subarrays efficiently.

Can this problem be solved without segment trees?

Yes, with careful array traversal and prefix sums, but segment trees improve efficiency for larger n and overlapping conflicts.

How do conflicting pairs affect subarray counts?

Each conflicting pair prevents subarrays containing both elements, so identifying which pair to remove maximizes valid subarrays.

What is the time complexity for large n?

With proper preprocessing and segment tree updates, the problem can be solved in O(n) time for all pairs.

Is enumeration of subarrays necessary?

Direct enumeration is inefficient; instead, simulate pair removal and count valid subarrays using prefix sums or segment tree tracking.

terminal

Solution

Solution 1: Enumeration + Maintaining Minimum and Second Minimum Values

We store all conflicting pairs $(a, b)$ (assuming $a < b$) in a list $g$, where $g[a]$ represents the set of all numbers $b$ that conflict with $a$.

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class Solution:
    def maxSubarrays(self, n: int, conflictingPairs: List[List[int]]) -> int:
        g = [[] for _ in range(n + 1)]
        for a, b in conflictingPairs:
            if a > b:
                a, b = b, a
            g[a].append(b)
        cnt = [0] * (n + 2)
        ans = add = 0
        b1 = b2 = n + 1
        for a in range(n, 0, -1):
            for b in g[a]:
                if b < b1:
                    b2, b1 = b1, b
                elif b < b2:
                    b2 = b
            ans += b1 - a
            cnt[b1] += b2 - b1
            add = max(add, cnt[b1])
        ans += add
        return ans

Continue Practicing

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