Skip to content

Best Time to Buy and Sell Stock II⚓︎

Link

Description⚓︎

You are given an integer array prices where prices[i] is the price of a given stock on the i-th day.

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

Example 1:

  • Input: prices = [7,1,5,3,6,4]
  • Output: 7
  • Explanation:
1
2
3
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.
Total profit is 4 + 3 = 7.

Example 2:

  • Input: prices = [1,2,3,4,5]
  • Output: 4
  • Explanation:
1
2
Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Total profit is 4.

Example 3:

  • Input: prices = [7,6,4,3,1]
  • Output: 0
  • Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

Constraints:

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

Solution⚓︎

Dynamic Programming Method⚓︎

Way 1⚓︎

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(2));
        dp[0][0] = -prices[0];
        dp[0][1] = 0;
        for (int i = 1; i < n; i++) {
            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i]);  // different from Leetcode 121
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i]);
        }
        return dp[n - 1][1];
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(n)\).

Way 2⚓︎

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(2, vector<int>(2));
        dp[0][0] = -prices[0];
        dp[0][1] = 0;
        for (int i = 1; i < n; i++) {
            dp[i % 2][0] = max(dp[(i - 1) % 2][0], dp[(i - 1) % 2][1] - prices[i]);  // different from Leetcode 121
            dp[i % 2][1] = max(dp[(i - 1) % 2][1], dp[(i - 1) % 2][0] + prices[i]);
        }
        return dp[(n - 1) % 2][1];
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(1)\).

Easy Solution (Greedy)⚓︎

See reference 1 and reference 2 (Chinese).

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0;
        for (int i = 0; i + 1 < prices.size(); i++) {
            res += max(0, prices[i + 1] - prices[i]);
        }
        return res;
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(1)\).