 # Best Time to Buy and Sell Stock – LeetCode – Solution

121. Best Time to Buy and Sell Stock – Solution

You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.

Example 1:

```Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (pr ice = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
```

Example 2:

```Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.
```

Constraints:

• `1 <= prices.length <= 105`
• `0 <= prices[i] <= 104`

## Solution

we need to find max(prices[j] - prices[i]), for every i and j such that j > i

### Approach 1: Brute Force

```public class Solution {
public int maxProfit(int prices[]) {
int n = prices.length;
int maxProfit = 0;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
maxProfit = Math.max(maxProfit, prices[j] - prices[i]);
}
}
return maxProfit;
}
}```

Time complexity: O(n2) Space complexity: O(1)

### Approach 2: Precalulate max array from right

```class Solution {
public int maxProfit(int[] prices) {
int n = prices.length;
int maxProfit = 0;
int[] maxArray = new int[n];
int maxTillNow = Integer.MIN_VALUE;

// calculate max array from right
for(int i=n-1; i>=0; i--){
maxTillNow = Math.max(maxTillNow,prices[i]);
maxArray[i] = maxTillNow;
}

for(int i=0; i<n-1; i++)
maxProfit = Math.max(maxProfit,maxArray[i+1]-prices[i]);
return maxProfit;
}
}```

Time complexity: O(n) Space complexity: O(n)

### Approach 3: One pass solution without extra space

```class Solution {

public int maxProfit(int[] prices) {
int n = prices.length;
int minSoFar = Integer.MAX_VALUE;
int maxProfit = 0;
for(int i=0;i< n;i++){
if(minSoFar > prices[i])
minSoFar = prices[i];
else
maxProfit = Math.max(prices[i] - minSoFar, maxProfit);
}
return maxProfit;
}
}```

Time complexity: O(n) Space complexity: O(1)