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121.best-time-to-buy-and-sell-stock.rb

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+# -*- coding: utf-8 -*-
+#
+# @lc app=leetcode id=121 lang=ruby
+#
+# [121] Best Time to Buy and Sell Stock
+#
+# https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
+#
+# Say you have an array for which the ith element is the price of a
+# given stock on day i.
+#
+# If you were only permitted to complete at most one transaction
+# (i.e., buy one and sell one share of the stock), design an algorithm
+# to find the maximum profit.
+#
+# Note that you cannot sell a stock before you buy one.
+#
+# Example 1:
+#
+# Input: [7,1,5,3,6,4]
+# Output: 5
+# Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6),
+# profit = 6-1 = 5. Not 7-1 = 6, as selling price needs to be larger
+# than buying price.
+#
+# Example 2:
+#
+# Input: [7,6,4,3,1]
+# Output: 0
+# Explanation: In this case, no transaction is done, i.e. max profit = 0.
+
+
+# We want prices[j]-prices[i](j>i) to be maximum:
+#
+#   1. Keep track of and update the minimum price up to the given
+#      point (min), there are possible larger profit to its right.
+#   2. Keep track of and update the current profit if it's larger.
+#   3. The current profit at index i is prices[i]-min.
+
+# @param {Integer[]} prices
+# @return {Integer}
+def max_profit_a(prices)
+  i = 1
+  max, min = 0, prices[0]
+  while i < prices.size
+    profit = prices[i] - min
+    if profit > max
+      max = profix
+    elsif profit < 0
+      min = prices[i]
+    end
+    i += 1
+  end
+  max
+end
+
+
+# You may not see this at first, but this problem is similar to
+# problem 53:
+#
+#   1. Build an array of prices diff, diff[i] = prices[i]-prices[i-1].
+#   2. The max profit is the maximum sum of diff's contiguous subarray.
+#      I give you 7 dollars, you spend 5, make 4, you left with 6,
+#      your profit is 4-5=-1=(7-5)-7+(7-5+4)-(7-5).
+
+def max_profit_b(prices)
+  diff = (1...prices.size).reduce([0]) do |a, i|
+    a << prices[i] - prices[i-1]
+  end
+  max_sub_array(diff)
+end
+
+def max_sub_array(nums)
+  ans = nums[0]
+  sum = 0
+  (0...nums.size).each do |i|
+    sum += nums[i]
+    ans = [ans, sum].max
+    sum = [sum, 0].max
+  end
+  ans
+end
+
+
+# You don't need to build the diff array since you already know all diff[i].
+
+def max_profit(prices)
+  ans = sum = 0
+  (1...prices.size).each do |i|
+    sum += prices[i]-prices[i-1]
+    ans = [ans, sum].max
+    sum = [sum, 0].max
+  end
+  ans
+end