Maximize Subarrays After Removing One Conflicting Pair

Problem Description

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.

Key Insights

• The problem involves counting subarrays while considering conflicting pairs.
• By removing one conflicting pair, the constraints on the subarrays change, potentially allowing more valid subarrays.
• The total number of subarrays can be calculated using the formula: n * (n + 1) / 2.
• The main challenge is to efficiently determine the impact of each conflicting pair on the subarray count.

Space and Time Complexity

Time Complexity: O(m * n), where m is the number of conflicting pairs and n is the length of the array. Space Complexity: O(m), used for storing conflicting pairs and the counts of valid subarrays.

Solution

The solution involves iterating over the conflicting pairs and calculating the number of valid subarrays for each configuration after removing one pair. To do this:

1. Calculate the total number of subarrays possible without any restrictions.
2. For each conflicting pair, determine how many subarrays contain both elements of that pair.
3. By removing one conflicting pair, recalculate the total number of valid subarrays.
4. Keep track of the maximum valid subarray count obtained by removing each pair.

Code Solutions