# 121. Best Time to Buy and Sell Stock

Reference: LeetCode

Difficulty: Easy

My Post: Java Solution with Comments and Explanation (BF, Divide and Conquer, One-Pass)

## Problem

Say you have an array for which the ith element is the price of a given stock on day

`i`

.

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

**Constraint:** Note that you cannot sell a stock before you buy one.

**Example:**

1 | Input: [7,1,5,3,6,4] |

1 | Input: [7,6,4,3,1] |

## Analysis

### Brute-Force

For each day `i`

, compare its prices with the following price of each day `j`

, and update the maximum profit.

**Note:**

`j`

should starts from`0`

.`maxProfit`

should be initialized with $0$.

1 | public int maxProfit(int[] prices) { |

**Time:** $O(N^2)$**Space:** $O(1)$

### Divide and Conquer

We divide the problem into two subproblems `left`

and `right`

. Then we calculate:

- The best buy time
`B1`

and the best sell time`S1`

from`left`

. - The best buy time
`B2`

and the best sell time`S2`

from`right`

. - The time
`M1`

of minimum price and the time`X1`

of maximum price in`left`

. - The time
`M2`

of minimum price and the time`X2`

of maximum price in`right`

.

In the `combine`

step, we have three cases to consider:

**Case 1:**In`left`

, we buy at`B1`

and sell at`S1`

.**Case 2:**In`right`

, we buy at`B2`

and sell at`S2`

.**Case 3:**In`left`

and`right`

, we buy at`M1`

and sell at`X2`

.- We cannot use
`B1`

and`S2`

because`B1`

may not be the time of minimum price and`S2`

may not be the time of maximum price. Check out the illustration below.

- We cannot use
- Beside, we need to update the time of minimum price and maximum price.

Here is an illustration of the text above. Notice the `left`

is `P1`

and `right`

is `P2`

.

The following solution uses a helper class `BuySell`

to encapsulate variables passed across recursive levels.

1 | private class BuySell { |

Here is the main part of the solution. Notice that it is a good approach to separate the combining code from the recursive function. It makes code more readable.

1 | public int maxProfit(int[] prices) { |

**Time:** $O(N)$ since $T(N) = 2T(N) + O(1)$**Space:** $O(\log{N})$ because of the call stack.

### One-Pass

Notice that the maximum profit that can be made by selling on a specific day is **determined** by the `minimum`

of the stock prices over the previous days. Thus, we can loop over each price in `prices`

, and keep updating the minimum price and at the same update the maximum profit.

**Note:** The ordering of updating `maxProfit`

and `minPrice`

does not matter.

1 | public int maxProfit(int[] prices) { |

**Time:** $O(N)$**Space:** $O(1)$