121. Best Time to Buy and Sell Stock

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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

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

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

Solution:

We need a buy and a sell to lock a profit.

If we know the following:

max profit if we sell at day 0

max profit if we sell at day 1

...

max profit if we sell at day prices.length - 1

Then the maximum profit is the the max of all above values.

The max profit we sell at day i, is prices[i] - min, where min is the lowest stock price before day i that we make a buy. When we go through the input, we simply maintains a lowest prices such that at day i we have already known the price to buy.

We only go through the input array once. So the time complexity is O(n).

Code:

public class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) {
            return 0;
        }
        int lowestPrice = prices[0];
        int maxProfit = 0;
        for (int i = 0; i < prices.length; i++) {
            maxProfit = Math.max(maxProfit, prices[i] - lowestPrice);
            lowestPrice = Math.min(lowestPrice, prices[i]);
        }
        return maxProfit;
    }
}