Maximize Subarrays After Removing One Conflicting Pair Python/Java solution

Python and Java implementations follow the same structure: group pairs by right endpoint, iterate r from 1..n, maintain the largest and second-largest left endpoints, accumulate base valid subarrays, and track the best improvement from removing a pair. The algorithm runs in O(n + m) time and uses O(n) auxiliary space.

What is the best approach for Maximize Subarrays After Removing One Conflicting Pair?

The optimal approach is enumeration while maintaining the largest and second-largest conflicting left endpoints encountered so far. For each right boundary r, the largest left value determines the earliest valid subarray start. Removing the pair that causes this largest restriction yields a gain of (max1 - max2). This produces an O(n + m) time solution.

Is Maximize Subarrays After Removing One Conflicting Pair asked at Google/Amazon/Meta?

Hard array and interval-constraint problems like this frequently appear in interviews at large tech companies such as Google, Amazon, and Meta. The problem tests reasoning about constraints, prefix tracking, and optimizing from quadratic enumeration to linear-time counting.

What data structure is used in Maximize Subarrays After Removing One Conflicting Pair?

The optimal solution mainly uses arrays or adjacency lists to group conflicts by their right endpoint and tracks two running maximum values. Conceptually it resembles prefix constraint tracking; heavier structures like segment trees are unnecessary for the linear-time solution.

What is the time complexity of Maximize Subarrays After Removing One Conflicting Pair?

The optimal algorithm runs in O(n + m) time, where n is the number of elements and m is the number of conflicting pairs. Each pair is processed once when grouped by its right endpoint, and the array is scanned once while maintaining two prefix constraints.

Description Hints 5 Solution Video 55:07

3480. Maximize Subarrays After Removing One Conflicting Pair

Q: Is Maximize Subarrays After Removing One Conflicting Pair easy or hard?

The problem is classified as Hard because it requires recognizing how conflicting pairs restrict subarray start positions and how removing one pair changes those constraints. The final implementation is linear, but deriving the max and second-max prefix logic requires careful reasoning.

Q: How to solve Maximize Subarrays After Removing One Conflicting Pair in O(n)?

Store conflicts grouped by their right endpoint and iterate through the array from left to right. Track the largest and second-largest left endpoints among all processed conflicts. For each position r, add r - max1 valid subarrays and record the improvement max1 - max2 if the dominant pair were removed. The final answer is the base count plus the maximum improvement.

Hard64.6% acceptance5 companies

You are given an integer n which represents an array nums containing the numbers from 1 to n in order. Additionally, you are given a 2D array conflictingPairs, where conflictingPairs[i] = [a, b] indicates that a and b form a conflicting pair.

Remove exactly one element from conflictingPairs. Afterward, count the number of non-empty subarrays of nums which do not contain both a and b for any remaining conflicting pair [a, b].

Return the maximum number of subarrays possible after removing exactly one conflicting pair.

Example 1:

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

Output: 9

Explanation:

Remove [2, 3] from conflictingPairs. Now, conflictingPairs = [[1, 4]].
There are 9 subarrays in nums where [1, 4] do not appear together. They are [1], [2], [3], [4], [1, 2], [2, 3], [3, 4], [1, 2, 3] and [2, 3, 4].
The maximum number of subarrays we can achieve after removing one element from conflictingPairs is 9.

Example 2:

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

Output: 12

Explanation:

Remove [1, 2] from conflictingPairs. Now, conflictingPairs = [[2, 5], [3, 5]].
There are 12 subarrays in nums where [2, 5] and [3, 5] do not appear together.
The maximum number of subarrays we can achieve after removing one element from conflictingPairs is 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]

Asked at 5 companies

Google Amazon Microsoft Bloomberg

Topics

Hints open one at a time

Hint 1

Let f[i] (where i = 1, 2, 3, ..., n) be the end index of the longest valid subarray (without any conflicting pair) starting at index i.

Hint 2

The answer is: sigma(f[i] - i + 1) for i in [1..n], which simplifies to: sigma(f[i]) - n * (n + 1) / 2 + n.

Hint 3

Focus on maintaining f[i].

Hint 4

If we have a conflicting pair (x, y) with x < y: 1. Sort the conflicting pairs by y values in non-increasing order. 2. Update each prefix of the f array accordingly.

Hint 5

Use a segment tree or another suitable data structure to maintain the range update and sum query efficiently.

55:07

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1class Solution:
2    def maxSubarrays(self, n: int, conflictingPairs: List[List[int]]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

4
[[2,3],[1,4]]

Input

5
[[1,2],[2,5],[3,5]]

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