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divide-an-array-into-subarrays-with-minimum-cost-ii.py
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divide-an-array-into-subarrays-with-minimum-cost-ii.py
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# Time: O(nlogd)
# Space: O(d)
import heapq
# sliding window, heap
class Solution(object):
def minimumCost(self, nums, k, dist):
"""
:type nums: List[int]
:type k: int
:type dist: int
:rtype: int
"""
def get_top(heap, total):
while abs(heap[0][1]) < i-(1+dist):
heapq.heappop(heap)
total[0] -= 1
return heap[0]
def lazy_delete(heap, total):
total[0] += 1
if total[0] <= len(heap)-total[0]:
return
heap[:] = [x for x in heap if abs(x[1]) > i-(1+dist)]
heapq.heapify(heap)
total[0] = 0
max_heap, min_heap = [], []
total1, total2 = [0], [0]
mn, curr = float("inf"), 0
for i in xrange(1, len(nums)):
heapq.heappush(max_heap, (-nums[i], i))
curr += nums[i]
if i > k-1:
x, idx = get_top(max_heap, total1)
heapq.heappop(max_heap)
curr -= -x
heapq.heappush(min_heap, (-x, -idx))
if i > 1+dist:
x, idx = get_top(min_heap, total2)
if (x, idx) <= (nums[i-(1+dist)], -(i-(1+dist))):
lazy_delete(min_heap, total2)
else:
lazy_delete(max_heap, total1)
heapq.heappop(min_heap)
curr -= nums[i-(1+dist)]-x
heapq.heappush(max_heap, (-x, -idx))
if i >= k-1:
mn = min(mn, curr)
return nums[0]+mn
# Time: O(nlogd)
# Space: O(d)
import heapq
import collections
# sliding window, heap, freq table
class Solution2(object):
def minimumCost(self, nums, k, dist):
"""
:type nums: List[int]
:type k: int
:type dist: int
:rtype: int
"""
def get_top(heap, cnt, total):
while heap[0] in cnt:
x = heapq.heappop(heap)
cnt[x] -= 1
if cnt[x] == 0:
del cnt[x]
total[0] -= 1
return heap[0]
def lazy_delete(heap, cnt, total, x):
cnt[x] += 1
total[0] += 1
if total[0] <= len(heap)-total[0]:
return
new_heap = []
for x in heap:
if x not in cnt:
new_heap.append(x)
continue
cnt[x] -= 1
if cnt[x] == 0:
del cnt[x]
total[0] = 0
heapq.heapify(new_heap)
heap[:] = new_heap
max_heap, min_heap = [], []
cnt1, cnt2 = collections.Counter(), collections.Counter()
total1, total2 = [0], [0]
mn, curr = float("inf"), 0
for i in xrange(1, len(nums)):
heapq.heappush(max_heap, -nums[i])
curr += nums[i]
if (len(max_heap)-total1[0]) > k-1:
x = get_top(max_heap, cnt1, total1)
curr -= -x
heapq.heappush(min_heap, -heapq.heappop(max_heap))
if (len(max_heap)-total1[0])+(len(min_heap)-total2[0]) > 1+dist:
x = get_top(min_heap, cnt2, total2)
if x <= nums[i-(1+dist)]:
lazy_delete(min_heap, cnt2, total2, nums[i-(1+dist)])
else:
lazy_delete(max_heap, cnt1, total1, -nums[i-(1+dist)])
heapq.heappop(min_heap)
curr -= nums[i-(1+dist)]-x
heapq.heappush(max_heap, -x)
if len(max_heap)-total1[0] == k-1:
mn = min(mn, curr)
return nums[0]+mn
# Time: O(nlogd)
# Space: O(d)
from sortedcontainers import SortedList
# sliding window, sorted list
class Solution3(object):
def minimumCost(self, nums, k, dist):
"""
:type nums: List[int]
:type k: int
:type dist: int
:rtype: int
"""
sl1, sl2 = SortedList(), SortedList()
mn, curr = float("inf"), 0
for i in xrange(1, len(nums)):
sl1.add(nums[i])
curr += nums[i]
if len(sl1) > k-1:
curr -= sl1[-1]
sl2.add(sl1.pop())
if len(sl1)+len(sl2) > 1+dist:
if sl2[0] <= nums[i-(1+dist)]:
sl2.remove(nums[i-(1+dist)])
else:
sl1.remove(nums[i-(1+dist)])
curr -= nums[i-(1+dist)]-sl2[0]
sl1.add(sl2.pop(0))
if len(sl1) == k-1:
mn = min(mn, curr)
return nums[0]+mn
# Time: O(nlogd)
# Space: O(d)
from sortedcontainers import SortedList
# sliding window, sorted list
class Solution4(object):
def minimumCost(self, nums, k, dist):
"""
:type nums: List[int]
:type k: int
:type dist: int
:rtype: int
"""
sl = SortedList(nums[1:1+(1+dist)])
mn = curr = sum(sl[:k-1])
for i in xrange(1+(1+dist), len(nums)):
sl.add(nums[i])
curr += min(nums[i]-sl[k-1], 0)
curr -= min(nums[i-(1+dist)]-sl[k-1], 0)
sl.remove(nums[i-(1+dist)])
mn = min(mn, curr)
return nums[0]+mn