2813. Maximum Elegance of a K-Length Subsequence

# Time:  O(nlogk)
# Space: O(k)

import heapq
from sortedcontainers import SortedList


# heap, sorted list, greedy
class Solution(object):
    def findMaximumElegance(self, items, k):
        """
        :type items: List[List[int]]
        :type k: int
        :rtype: int
        """
        curr = 0
        lookup = set()
        stk = []
        for p, c in heapq.nlargest(k, items):
            if c in lookup:
                stk.append(p)
            curr += p
            lookup.add(c)
        sl = SortedList()
        lookup2 = {}
        for p, c in items:
            if c in lookup:
                continue
            if c in lookup2:
                if lookup2[c] >= p:
                    continue
                sl.remove((lookup2[c], c))
            sl.add((p, c))
            lookup2[c] = p
            if len(sl) > len(stk):
                del lookup2[sl[0][1]]
                del sl[0]
        result = curr+len(lookup)**2
        for p, c in reversed(sl):
            curr += p-stk.pop()
            lookup.add(c)
            result = max(result, curr+len(lookup)**2) 
        return result


# Time:  O(n + klogk)
# Space: O(n)
import random
import collections

# quick select, sort, greedy
class Solution2(object):
    def findMaximumElegance(self, items, k):
        """
        :type items: List[List[int]]
        :type k: int
        :rtype: int
        """
        def nth_element(nums, n, left=0, compare=lambda a, b: a < b):
            def tri_partition(nums, left, right, target, compare):
                mid = left
                while mid <= right:
                    if nums[mid] == target:
                        mid += 1
                    elif compare(nums[mid], target):
                        nums[left], nums[mid] = nums[mid], nums[left]
                        left += 1
                        mid += 1
                    else:
                        nums[mid], nums[right] = nums[right], nums[mid]
                        right -= 1
                return left, right
            
            right = len(nums)-1
            while left <= right:
                pivot_idx = random.randint(left, right)
                pivot_left, pivot_right = tri_partition(nums, left, right, nums[pivot_idx], compare)
                if pivot_left <= n <= pivot_right:
                    return
                elif pivot_left > n:
                    right = pivot_left-1
                else:  # pivot_right < n.
                    left = pivot_right+1
    
        def nlargest(k, nums):
            nth_element(nums, k-1, compare=lambda a, b: a > b)
            return sorted(nums[:k], reverse=True)
    
        curr = 0
        lookup = set()
        stk = []
        for p, c in nlargest(k, items):
            if c in lookup:
                stk.append(p)
            curr += p
            lookup.add(c)
        lookup2 = collections.defaultdict(int)
        for p, c in items:
            if c in lookup:
                continue
            lookup2[c] = max(lookup2[c], p)
        l = len(lookup)
        result = curr+l**2
        for p in nlargest(min(len(stk), len(lookup2)), lookup2.values()):
            curr += p-stk.pop()
            l += 1
            result = max(result, curr+l**2) 
        return result


# Time:  O(nlogn)
# Space: O(k)
# sort, greedy
class Solution3(object):
    def findMaximumElegance(self, items, k):
        """
        :type items: List[List[int]]
        :type k: int
        :rtype: int
        """
        items.sort(reverse=True)
        result = curr = 0
        lookup = set()
        stk = []
        for i in xrange(k):
            if items[i][1] in lookup:
                stk.append(items[i][0])
            curr += items[i][0]
            lookup.add(items[i][1])
        result = curr+len(lookup)**2
        for i in xrange(k, len(items)):
            if items[i][1] in lookup:
                continue
            if not stk:
                break
            curr += items[i][0]-stk.pop()
            lookup.add(items[i][1])
            result = max(result, curr+len(lookup)**2)
        return result

Solution from kamyu104/LeetCode-Solutions · MIT