LC-2944 Medium LeetCode

2944. Minimum Number of Coins for Fruits

Array Dynamic Programming Queue Heap (Priority Queue) Monotonic Queue

Read the full problem statement on LeetCode.

Difficulty: medium Acceptance: 46% Topics: Array, Dynamic Programming, Queue, Heap (Priority Queue), Monotonic Queue

Reading material

Arrays & Hashing 12 min 1-D Dynamic Programming 16 min Stacks & Queues 11 min

Reference solution (spoiler · python)

# Time:  O(n)
# Space: O(n)

import collections


# dp, mono deque
class Solution(object):
    def minimumCoins(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        dp = [float("inf")]*(len(prices)+1)
        dp[0] = 0
        dq = collections.deque()
        j = 0
        for i in xrange(len(prices)):
            while dq and dp[dq[-1]]+prices[dq[-1]] >= dp[i]+prices[i]:
                dq.pop()
            dq.append(i)
            while j+(j+1) < i:
                assert(len(dq) != 0)
                if dq[0] == j:
                    dq.popleft()
                j += 1
            dp[i+1] = dp[dq[0]]+prices[dq[0]]
        return dp[-1]


# Time:  O(nlogn)
# Space: O(n)
# dp, sorted list
from sortedcontainers import SortedList


class Solution2(object):
    def minimumCoins(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        dp = [float("inf")]*(len(prices)+1)
        dp[0] = 0
        sl = SortedList()
        j = 0
        for i in xrange(len(prices)):
            sl.add((dp[i]+prices[i], i))
            while j+(j+1) < i:
                sl.remove(((dp[j]+prices[j], j)))
                j += 1
            dp[i+1] = sl[0][0]
        return dp[-1]


# Time:  O(n^2)
# Space: O(n)
# dp
class Solution3(object):
    def minimumCoins(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        dp = [float("inf")]*(len(prices)+1)
        dp[0] = 0
        for i in xrange(len(prices)):
            for j in xrange(i//2, i+1):
                dp[i+1] = min(dp[i+1], dp[j]+prices[j])
        return dp[-1]

Solution from kamyu104/LeetCode-Solutions · MIT