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Best Time to Buy and Sell Stock II - Solution/C++

Posted on Edited on In Disqus:

122. Best Time to Buy and Sell Stock II

Question:

Say you have an array prices for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
      Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.        Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Constraints:
  • 1 <= prices.length <= 3 * 10 ^ 4
  • 0 <= prices[i] <= 10 ^ 4

    • Source code

    Version 1

    Idea:
    Here is the same as Best Time to Buy and Sell Stock.

    At 8 line, it always gets lowest price to i-th day.
    At 9 - 10 line, we know the value of profit (prices[i] - minPrice), it must be checked whether it's bigger than 0. Because it avoids buying day and selling day on the same day, and guarantee the positive profit we get.
    At 11 - 14 lin, we should not sell stock immediately even the profit is prositive at i-th day, maybe it will higher tomorrow (i-th + 1 day). So, we can know the selling day that is the best.
    At 17 line, reinitialize minPrice variable for the next best buying day.

    Time complexity: O(n)
    Space complexity: O(1)

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    class Solution {
    public:
        int maxProfit(vector<int>& prices) {
            const int n = prices.size();
            if(n <= 1) return 0;
            int minPrice = INT_MAX, sum = 0;
            for(int i = 0; i < n; i++){
                minPrice = min(minPrice, prices[i]);
                int profit = prices[i] - minPrice, j = i;
                if(profit > 0){
                    while(j < n && prices[j] >= prices[i]){
                        i = j;
                        j++;
                    }
                    sum += prices[i] - minPrice;
                    minPrice= INT_MAX;
                }
            }
            return sum;
        }
    };

    Version 2

    Idea:
    This strategy is very simple, if the buying day subtracted from the selling day is positive (profit), we directly add profit into maxProfit variable.

    Time complexity: O(n)
    Space complexity: O(1)

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    class Solution {
    public:
        int maxProfit(vector<int>& prices) {
            const int n = prices.size();
            if(n <= 1) return 0;
            int maxProfit = 0;
            for(int i = 1; i < n; i++){
                if(prices[i] > prices[i-1]){
                    maxProfit += prices[i]- prices[i-1];
                }
            }
            return maxProfit;
        }
    };