Best Time to Buy and Sell Stock II LeetCode Solution

Best Time to Buy and Sell Stock II You are given an integer array prices where prices[i] is the price of a given stock on the ith 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: 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: 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 * 104
• 0 <= prices[i] <= 10

Best Time to Buy and Sell Stock II Solutions

✅Time: O(n)
✅Space: O(1)

C++

class Solution {
 public:
  int maxProfit(vector<int>& prices) {
    int sell = 0;
    int hold = INT_MIN;

    for (const int price : prices) {
      sell = max(sell, hold + price);
      hold = max(hold, sell - price);
    }

    return sell;
  }
};

Java

class Solution {
  public int maxProfit(int[] prices) {
    int sell = 0;
    int hold = Integer.MIN_VALUE;

    for (final int price : prices) {
      sell = Math.max(sell, hold + price);
      hold = Math.max(hold, sell - price);
    }

    return sell;
  }
}

Python

  class Solution:
  def maxProfit(self, prices: List[int]) -> int:
    sell = 0
    hold = -math.inf

    for price in prices:
      sell = max(sell, hold + price)
      hold = max(hold, sell - price)

    return sell

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