最大単一販売利益質問する

Question

私はこの問題が大好きです。これは典型的な面接の質問であり、それについてどう考えるかによって、より良い解決策がどんどん得られます。これを O(n ² ) 時間よりも速く行うことは確かに可能です。ここでは、この問題について考えることができる 3 つの異なる方法を挙げました。

まず、分割統治法です。入力を半分に分割し、各サブ配列で問題を解決してから、2 つを組み合わせることで、この問題を解決できるかどうかを見てみましょう。実際にこれを行うことができ、効率的に実行できることがわかりました。直感的に理解できることは次のとおりです。1 日しかない場合、その日に購入して、利益なしで同じ日に売り戻すのが最善の選択肢です。それ以外の場合は、配列を 2 つに分割します。最適な答えは何かを考えると、次の 3 つの場所のいずれかにあるはずです。

正しい買い/売りペアは前半で完全に発生します。
正しい買い/売りペアは後半で完全に発生します。
正しい買い/売りペアは両方の半分にわたって発生します。つまり、前半で買い、後半で売ります。

前半と後半でアルゴリズムを再帰的に呼び出すことで、(1) と (2) の値を取得できます。オプション (3) の場合、最大の利益を得るには、前半の最低ポイントで購入し、後半の最高ポイントで売却します。入力に対して単純な線形スキャンを実行して 2 つの値を見つけるだけで、2 つの半分の最小値と最大値を見つけることができます。これにより、次の再帰を持つアルゴリズムが得られます。

T(1) <= O(1)
T(n) <= 2T(n / 2) + O(n)

使用方法マスター定理再帰を解くには、これは O(n lg n) 時間で実行され、再帰呼び出しに O(lg n) の空間を使用することがわかります。単純な O(n ² ) のソリューションを打ち負かしました。

でもちょっと待ってください。これよりずっと良い方法があります。再帰に O(n) 項がある唯一の理由は、各半分の最小値と最大値を見つけるために入力全体をスキャンしなければならなかったからです。すでに各半分を再帰的に探索しているので、再帰で各半分に格納されている最小値と最大値も返すようにすれば、もっと良い結果が得られるかもしれません。言い換えると、再帰では次の 3 つを返します。

利益を最大化するための購入と売却のタイミング。
範囲内の全体的な最小値。
範囲内の全体的な最大値。

これら最後の2つの値は、(1)を計算する再帰と同時に実行できる単純な再帰を使用して再帰的に計算できます。

単一要素範囲の最大値と最小値は、その要素のみです。
複数要素範囲の最大値と最小値は、入力を半分に分割し、各半分の最大値と最小値を見つけて、それぞれの最大値と最小値を取得することで見つけることができます。

このアプローチを用いると、再帰関係は

T(1) <= O(1)
T(n) <= 2T(n / 2) + O(1)

ここでマスター定理を使用すると、実行時間は O(n)、スペースは O(lg n) となり、元のソリューションよりもさらに優れたものになります。

しかし、ちょっと待ってください。これよりももっと良い方法があります。動的プログラミングを使用してこの問題を解くことを考えてみましょう。この問題を次のように考えます。最初の k 個の要素を見て、問題の答えがわかったとします。最初の (k+1) 番目の要素に関する知識と最初の解決策を組み合わせて、最初の (k+1) 個の要素の問題を解決できるでしょうか。できるとしたら、最初の要素、次に最初の 2 個、最初の 3 個というように、最初の n 個の要素について計算するまで、問題を解くことで優れたアルゴリズムを実現できます。

これをどうやって行うか考えてみましょう。要素が 1 つしかない場合、それが最適な買い/売りペアであることは既にわかっています。ここで、最初の k 要素の最適な答えがわかっていて、(k+1) 番目の要素を見るとします。この値によって最初の k 要素よりも優れたソリューションが作成される唯一の方法は、最初の k 要素の最小値と新しい要素の差が、これまでに計算した最大の差よりも大きい場合です。したがって、要素を調べていくときに、これまでに見た最小値と、最初の k 要素だけで得られる最大利益の 2 つの値を追跡するとします。最初は、これまでに見た最小値は最初の要素で、最大利益はゼロです。新しい要素を見ると、まず、これまでに見た最低価格で購入し、現在の価格で販売した場合にいくら儲かるかを計算して、最適利益を更新します。これがこれまでに計算した最適値よりも優れている場合は、最適ソリューションをこの新しい利益に更新します。次に、これまでに見た最小要素を、現在の最小要素と新しい要素の最小値に更新します。

各ステップでは O(1) の作業のみを実行し、n 個の要素のそれぞれを 1 回だけ訪問するため、完了には O(n) の時間がかかります。さらに、使用する補助ストレージは O(1) のみです。これがこれまでのところ最高の成果です。

例として、入力に対してこのアルゴリズムが実行される様子を示します。配列の各値の間にある数字は、その時点でアルゴリズムが保持している値に対応します。実際にこれらすべてを保存するわけではありません (O(n) のメモリが必要になります) が、アルゴリズムが進化する様子を確認するのは役に立ちます。

            5        10        4          6         7
min         5         5        4          4         4    
best      (5,5)     (5,10)   (5,10)     (5,10)    (5,10)

答え: (5, 10)

            5        10        4          6        12
min         5         5        4          4         4    
best      (5,5)     (5,10)   (5,10)     (5,10)    (4,12)

答え: (4, 12)

            1       2       3      4      5
min         1       1       1      1      1
best      (1,1)   (1,2)   (1,3)  (1,4)  (1,5)

答え: (1, 5)

今、もっと良い方法がありますか? 残念ながら、漸近的な意味ではそうではありません。O(n) 未満の時間を使用する場合、大きな入力のすべての数値を調べることができないため、最適な答えを見逃さないことを保証できません (調べなかった要素に「隠す」ことができます)。さらに、O(1) 未満のスペースは使用できません。big-O 表記に隠された定数因子にいくつかの最適化があるかもしれませんが、それ以外では、根本的に優れたオプションが見つかるとは期待できません。

全体として、これは次のアルゴリズムがあることを意味します。

単純: O(n ² ) 時間、O(1) 空間。
分割統治: O(n lg n) 時間、O(lg n) スペース。
最適化された分割統治: O(n) 時間、O(lg n) スペース。
動的プログラミング: O(n) 時間、O(1) 空間。

編集: 興味があれば、コードを書きましたこれら4つのアルゴリズムのPythonバージョンこれらを試してみて、相対的なパフォーマンスを判断できるようにします。コードは次のとおりです。


# Four different algorithms for solving the maximum single-sell profit problem,
# each of which have different time and space complexity.  This is one of my
# all-time favorite algorithms questions, since there are so many different
# answers that you can arrive at by thinking about the problem in slightly
# different ways.
#
# The maximum single-sell profit problem is defined as follows.  You are given
# an array of stock prices representing the value of some stock over time.
# Assuming that you are allowed to buy the stock exactly once and sell the
# stock exactly once, what is the maximum profit you can make?  For example,
# given the prices
#
#                        2, 7, 1, 8, 2, 8, 4, 5, 9, 0, 4, 5
#
# The maximum profit you can make is 8, by buying when the stock price is 1 and
# selling when the stock price is 9.  Note that while the greatest difference
# in the array is 9 (by subtracting 9 - 0), we cannot actually make a profit of
# 9 here because the stock price of 0 comes after the stock price of 9 (though
# if we wanted to lose a lot of money, buying high and selling low would be a
# great idea!)
#
# In the event that there's no profit to be made at all, we can always buy and
# sell on the same date.  For example, given these prices (which might
# represent a buggy-whip manufacturer:)
#
#                            9, 8, 7, 6, 5, 4, 3, 2, 1, 0
#
# The best profit we can make is 0 by buying and selling on the same day.
#
# Let's begin by writing the simplest and easiest algorithm we know of that
# can solve this problem - brute force.  We will just consider all O(n^2) pairs
# of values, and then pick the one with the highest net profit.  There are
# exactly n + (n - 1) + (n - 2) + ... + 1 = n(n + 1)/2 different pairs to pick
# from, so this algorithm will grow quadratically in the worst-case.  However,
# it uses only O(1) memory, which is a somewhat attractive feature.  Plus, if
# our first intuition for the problem gives a quadratic solution, we can be
# satisfied that if we don't come up with anything else, we can always have a
# polynomial-time solution.

def BruteForceSingleSellProfit(arr):
    # Store the best possible profit we can make; initially this is 0.
    bestProfit = 0;

    # Iterate across all pairs and find the best out of all of them.  As a
    # minor optimization, we don't consider any pair consisting of a single
    # element twice, since we already know that we get profit 0 from this.
    for i in range(0, len(arr)):
        for j in range (i + 1, len(arr)):
            bestProfit = max(bestProfit, arr[j] - arr[i])

    return bestProfit

# This solution is extremely inelegant, and it seems like there just *has* to
# be a better solution.  In fact, there are many better solutions, and we'll
# see three of them.
#
# The first insight comes if we try to solve this problem by using a divide-
# and-conquer strategy.  Let's consider what happens if we split the array into
# two (roughly equal) halves.  If we do so, then there are three possible
# options about where the best buy and sell times are:
#
# 1. We should buy and sell purely in the left half of the array.
# 2. We should buy and sell purely in the right half of the array.
# 3. We should buy in the left half of the array and sell in the right half of
#    the array.
#
# (Note that we don't need to consider selling in the left half of the array
# and buying in the right half of the array, since the buy time must always
# come before the sell time)
#
# If we want to solve this problem recursively, then we can get values for (1)
# and (2) by recursively invoking the algorithm on the left and right
# subarrays.  But what about (3)?  Well, if we want to maximize our profit, we
# should be buying at the lowest possible cost in the left half of the array
# and selling at the highest possible cost in the right half of the array.
# This gives a very elegant algorithm for solving this problem:
#
#    If the array has size 0 or size 1, the maximum profit is 0.
#    Otherwise:
#       Split the array in half.
#       Compute the maximum single-sell profit in the left array, call it L.
#       Compute the maximum single-sell profit in the right array, call it R.
#       Find the minimum of the first half of the array, call it Min
#       Find the maximum of the second half of the array, call it Max
#       Return the maximum of L, R, and Max - Min.
#
# Let's consider the time and space complexity of this algorithm.  Our base
# case takes O(1) time, and in our recursive step we make two recursive calls,
# one on each half of the array, and then does O(n) work to scan the array
# elements to find the minimum and maximum values.  This gives the recurrence
#
#    T(1) = O(1)
#    T(n) = 2T(n / 2) + O(n)
#
# Using the Master Theorem, this recurrence solves to O(n log n), which is
# asymptotically faster than our original approach!  However, we do pay a
# (slight) cost in memory usage.  Because we need to maintain space for all of
# the stack frames we use.  Since on each recursive call we cut the array size
# in half, the maximum number of recursive calls we can make is O(log n), so
# this algorithm uses O(n log n) time and O(log n) memory.

def DivideAndConquerSingleSellProfit(arr):
    # Base case: If the array has zero or one elements in it, the maximum
    # profit is 0.
    if len(arr) <= 1:
        return 0;

    # Cut the array into two roughly equal pieces.
    left  = arr[ : len(arr) / 2]
    right = arr[len(arr) / 2 : ]

    # Find the values for buying and selling purely in the left or purely in
    # the right.
    leftBest  = DivideAndConquerSingleSellProfit(left)
    rightBest = DivideAndConquerSingleSellProfit(right)

    # Compute the best profit for buying in the left and selling in the right.
    crossBest = max(right) - min(left)

    # Return the best of the three
    return max(leftBest, rightBest, crossBest)
    
# While the above algorithm for computing the maximum single-sell profit is
# better timewise than what we started with (O(n log n) versus O(n^2)), we can
# still improve the time performance.  In particular, recall our recurrence
# relation:
#
#    T(1) = O(1)
#    T(n) = 2T(n / 2) + O(n)
#
# Here, the O(n) term in the T(n) case comes from the work being done to find
# the maximum and minimum values in the right and left halves of the array,
# respectively.  If we could find these values faster than what we're doing
# right now, we could potentially decrease the function's runtime.
#
# The key observation here is that we can compute the minimum and maximum
# values of an array using a divide-and-conquer approach.  Specifically:
#
#    If the array has just one element, it is the minimum and maximum value.
#    Otherwise:
#       Split the array in half.
#       Find the minimum and maximum values from the left and right halves.
#       Return the minimum and maximum of these two values.
#
# Notice that our base case does only O(1) work, and our recursive case manages
# to do only O(1) work in addition to the recursive calls.  This gives us the
# recurrence relation
#
#    T(1) = O(1)
#    T(n) = 2T(n / 2) + O(1)
#
# Using the Master Theorem, this solves to O(n).
#
# How can we make use of this result?  Well, in our current divide-and-conquer
# solution, we split the array in half anyway to find the maximum profit we
# could make in the left and right subarrays.  Could we have those recursive
# calls also hand back the maximum and minimum values of the respective arrays?
# If so, we could rewrite our solution as follows:
#
#    If the array has size 1, the maximum profit is zero and the maximum and
#       minimum values are the single array element.
#    Otherwise:
#       Split the array in half.
#       Compute the maximum single-sell profit in the left array, call it L.
#       Compute the maximum single-sell profit in the right array, call it R.
#       Let Min be the minimum value in the left array, which we got from our
#           first recursive call.
#       Let Max be the maximum value in the right array, which we got from our
#           second recursive call.
#       Return the maximum of L, R, and Max - Min for the maximum single-sell
#           profit, and the appropriate maximum and minimum values found from
#           the recursive calls.
#
# The correctness proof for this algorithm works just as it did before, but now
# we never actually do a scan of the array at each step.  In fact, we do only
# O(1) work at each level.  This gives a new recurrence
#
#     T(1) = O(1)
#     T(n) = 2T(n / 2) + O(1)
#
# Which solves to O(n).  We're now using O(n) time and O(log n) memory, which
# is asymptotically faster than before!
#
# The code for this is given below:

def OptimizedDivideAndConquerSingleSellProfit(arr):
    # If the array is empty, the maximum profit is zero.
    if len(arr) == 0:
        return 0

    # This recursive helper function implements the above recurrence.  It
    # returns a triple of (max profit, min array value, max array value).  For
    # efficiency reasons, we always reuse the array and specify the bounds as
    # [lhs, rhs]
    def Recursion(arr, lhs, rhs):
        # If the array has just one element, we return that the profit is zero
        # but the minimum and maximum values are just that array value.
        if lhs == rhs:
            return (0, arr[lhs], arr[rhs])

        # Recursively compute the values for the first and latter half of the
        # array.  To do this, we need to split the array in half.  The line
        # below accomplishes this in a way that, if ported to other languages,
        # cannot result in an integer overflow.
        mid = lhs + (rhs - lhs) / 2
        
        # Perform the recursion.
        ( leftProfit,  leftMin,  leftMax) = Recursion(arr, lhs, mid)
        (rightProfit, rightMin, rightMax) = Recursion(arr, mid + 1, rhs)

        # Our result is the maximum possible profit, the minimum of the two
        # minima we've found (since the minimum of these two values gives the
        # minimum of the overall array), and the maximum of the two maxima.
        maxProfit = max(leftProfit, rightProfit, rightMax - leftMin)
        return (maxProfit, min(leftMin, rightMin), max(leftMax, rightMax))

    # Using our recursive helper function, compute the resulting value.
    profit, _, _ = Recursion(arr, 0, len(arr) - 1)
    return profit

# At this point we've traded our O(n^2)-time, O(1)-space solution for an O(n)-
# time, O(log n) space solution.  But can we do better than this?
#
# To find a better algorithm, we'll need to switch our line of reasoning.
# Rather than using divide-and-conquer, let's see what happens if we use
# dynamic programming.  In particular, let's think about the following problem.
# If we knew the maximum single-sell profit that we could get in just the first
# k array elements, could we use this information to determine what the
# maximum single-sell profit would be in the first k + 1 array elements?  If we
# could do this, we could use the following algorithm:
#
#   Find the maximum single-sell profit to be made in the first 1 elements.
#   For i = 2 to n:
#      Compute the maximum single-sell profit using the first i elements.
#
# How might we do this?  One intuition is as follows.  Suppose that we know the
# maximum single-sell profit of the first k elements.  If we look at k + 1
# elements, then either the maximum profit we could make by buying and selling
# within the first k elements (in which case nothing changes), or we're
# supposed to sell at the (k + 1)st price.  If we wanted to sell at this price
# for a maximum profit, then we would want to do so by buying at the lowest of
# the first k + 1 prices, then selling at the (k + 1)st price.
#
# To accomplish this, suppose that we keep track of the minimum value in the
# first k elements, along with the maximum profit we could make in the first
# k elements.  Upon seeing the (k + 1)st element, we update what the current
# minimum value is, then update what the maximum profit we can make is by
# seeing whether the difference between the (k + 1)st element and the new
# minimum value is.  Note that it doesn't matter what order we do this in; if
# the (k + 1)st element is the smallest element so far, there's no possible way
# that we could increase our profit by selling at that point.
#
# To finish up this algorithm, we should note that given just the first price,
# the maximum possible profit is 0.
#
# This gives the following simple and elegant algorithm for the maximum single-
# sell profit problem:
#
#   Let profit = 0.
#   Let min = arr[0]
#   For k = 1 to length(arr):
#       If arr[k] < min, set min = arr[k]
#       If profit < arr[k] - min, set profit = arr[k] - min
#
# This is short, sweet, and uses only O(n) time and O(1) memory.  The beauty of
# this solution is that we are quite naturally led there by thinking about how
# to update our answer to the problem in response to seeing some new element.
# In fact, we could consider implementing this algorithm as a streaming
# algorithm, where at each point in time we maintain the maximum possible
# profit and then update our answer every time new data becomes available.
#
# The final version of this algorithm is shown here:

def DynamicProgrammingSingleSellProfit(arr):
    # If the array is empty, we cannot make a profit.
    if len(arr) == 0:
        return 0

    # Otherwise, keep track of the best possible profit and the lowest value
    # seen so far.
    profit = 0
    cheapest = arr[0]

    # Iterate across the array, updating our answer as we go according to the
    # above pseudocode.
    for i in range(1, len(arr)):
        # Update the minimum value to be the lower of the existing minimum and
        # the new minimum.
        cheapest = min(cheapest, arr[i])

        # Update the maximum profit to be the larger of the old profit and the
        # profit made by buying at the lowest value and selling at the current
        # price.
        profit = max(profit, arr[i] - cheapest)

    return profit

# To summarize our algorithms, we have seen
#
# Naive:                        O(n ^ 2)   time, O(1)     space
# Divide-and-conquer:           O(n log n) time, O(log n) space
# Optimized divide-and-conquer: O(n)       time, O(log n) space
# Dynamic programming:          O(n)       time, O(1)     space

Answer 1