Hard
Minimum Pair Removal to Sort Array II — Python
Full explanation · Time O(nlogn) · Space O(n)
# Time: O(nlogn)
# Space: O(n)
from sortedcontainers import SortedList
# simulation, doubly linked list, sorted list
class Solution(object):
def minimumPairRemoval(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
def add(i):
if 0 <= i < right[i] < len(nums):
sl.add([nums[i]+nums[right[i]], i])
if nums[i] > nums[right[i]]:
cnt[0] += 1
def remove(i):
if 0 <= i < right[i] < len(nums):
sl.remove([nums[i]+nums[right[i]], i])
if nums[i] > nums[right[i]]:
cnt[0] -= 1
left = range(-1, (len(nums)+1)-1)
right = range(1, len(nums)+1)
cnt = [sum(nums[i] > nums[i+1] for i in xrange(len(nums)-1))]
sl = SortedList([nums[i]+nums[i+1], i] for i in xrange(len(nums)-1))
result = 0
while cnt[0]:
_, i = sl[0]
remove(left[i])
remove(i)
remove(right[i])
nums[i] += nums[right[i]]
left[right[right[i]]] = i
right[i] = right[right[i]]
add(left[i])
add(i)
result += 1
return result