maximize-subarrays-after-removing-one-conflicting-pair

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]