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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] <= 104
public int maxProfit(int[] A) {
int n = A.length, lastBuy = -A[0], lastSold = 0;
if (n == 0) return 0;
for (int i = 1; i < n; i++) {
int curBuy = Math.max(lastBuy, lastSold - A[i]);
int curSold = Math.max(lastSold, lastBuy + A[i]);
lastBuy = curBuy;
lastSold = curSold;
}
return lastSold;
}
class Solution {
public:
int maxProfit(vector<int>& P) {
int profit = 0;
for(int i = 1; i < size(P); i++)
if(P[i] > P[i-1]) // yesterday was valley, today is peak
profit += P[i] - P[i-1]; // buy yesterday, sell today
return profit;
}
};
class Solution:
def maxProfit(self, P: List[int]) -> int:
return sum(P[i]-P[i-1] if P[i] > P[i-1] else 0 for i in range(1, len(P)))
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