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Diffstat (limited to 'max_subarr_remove_conflict_pair.py')
-rw-r--r-- | max_subarr_remove_conflict_pair.py | 38 |
1 files changed, 38 insertions, 0 deletions
diff --git a/max_subarr_remove_conflict_pair.py b/max_subarr_remove_conflict_pair.py new file mode 100644 index 0000000..ad322b0 --- /dev/null +++ b/max_subarr_remove_conflict_pair.py @@ -0,0 +1,38 @@ +# 3480. 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 +pairs '[a, b]'. return the maximum number of subarrays possible after +removing exactly one conflicting pair. +""" + + +class Solution(object): + def maxSubarrays(self, n, conflictingPairs): + """ + :type n: int + :type conflictingPairs: List[List[int]] + :rtype: int + """ + right = [[] for _ in range(n + 1)] + for a, b in conflictingPairs: + right[max(a, b)].append(min(a, b)) + ans, left = 0, [0, 0] + s = [0] * (n + 1) + for r in range(1, n + 1): + for l in right[r]: + left = max(left, [l, left[0]], [left[0], l]) + ans += r - left[0] + s[left[0]] += left[0] - left[1] + return ans + max(s) + + +if __name__ == "__main__": + obj = Solution() + print(obj.maxSubarrays(n=4, conflictingPairs=[[2, 3], [1, 4]])) + print(obj.maxSubarrays(n=5, conflictingPairs=[[1, 2], [2, 5], [3, 5]])) |