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# Best Time to Buy and Sell Stock II LeetCode Solution

## Problem – 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] <= 104`

### Best Time to Buy and Sell Stock II LeetCode Solution in Java

``````public int maxProfit(int[] A) {
int n = A.length, lastBuy = -A, lastSold = 0;
if (n == 0) return 0;

for (int i = 1; i < n; i++) {
int curSold = Math.max(lastSold, lastBuy + A[i]);
lastSold = curSold;
}

return lastSold;
}
``````

### Best Time to Buy and Sell Stock II LeetCode Solution in C++

``````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;
}
};
``````

### Best Time to Buy and Sell Stock II LeetCode Solution in Python

``````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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