3480. Maximize Subarrays After Removing One Conflicting Pair
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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]fromconflictingPairs. Now,conflictingPairs = [[1, 4]]. - There are 9 subarrays in
numswhere[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
conflictingPairsis 9.
Example 2:
Input: n = 5, conflictingPairs = [[1,2],[2,5],[3,5]]
Output: 12
Explanation:
- Remove
[1, 2]fromconflictingPairs. Now,conflictingPairs = [[2, 5], [3, 5]]. - There are 12 subarrays in
numswhere[2, 5]and[3, 5]do not appear together. - The maximum number of subarrays we can achieve after removing one element from
conflictingPairsis 12.
Constraints:
2 <= n <= 1051 <= conflictingPairs.length <= 2 * nconflictingPairs[i].length == 21 <= conflictingPairs[i][j] <= nconflictingPairs[i][0] != conflictingPairs[i][1]
Hints
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.