Best Time to Buy and Sell Stock II

Here, We see Best Time to Buy and Sell Stock II problem Solution. This Leetcode problem done in many programming language like C++, Java, JavaScript, Python etc.

LeetCode Problem :-

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.```

Java Solution ->

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```class Solution {
public int maxProfit(int[] prices) {
int maxprofit = 0;
for (int i = 1; i < prices.length; i++) {
if (prices[i] > prices[i - 1])
maxprofit += prices[i] - prices[i - 1];
}
return maxprofit;
}
}
```Code language: Java (java)```

JavaScript Solution ->

``````var maxProfit = function(prices) {
let profits = [0]; // base condition

for(let i = 1; i < prices.length; i++) {
profits[i] = profits[i - 1] + Math.max(0, prices[i] - prices[i - 1])
}

return profits.pop()
};
```Code language: JavaScript (javascript)```

Python Solution ->

``````class Solution:
def maxProfit(self, prices: List[int]) -> int:
profit = 0
for i in range(1, len(prices)):
if prices[i] > prices[i - 1]:
profit += prices[i] - prices[i - 1]
return profit
```Code language: Python (python)```

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