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.

Related issue

• 122. Best Time to Buy and Sell Stock II
• 123. Best Time to Buy and Sell Stock III
• 188. Best Time to Buy and Sell Stock IV

classic solution, will time limit exceeded. O(n^2)

public class Solution {
    public int maxProfit(int[] prices) {
        int res = 0;
        if(prices == null || prices.length <= 1) return res;
        int[] dp = new int[prices.length];

        dp[0] = 0;
        for(int i=1; i<dp.length; i++){
            for(int j=0; j<i; j++){
                dp[i] = Math.max(prices[i]-prices[j], dp[i]);
            }
            res = Math.max(dp[i], res);
        }

        return res;
    }
}

However, you only need to keep track of the current minimal value ever found so far, use current value minus the min value, you get a profit, if current value is even smaller than the current minimal value, this is potentially a buy-in time, which will be sold later.

O(n)

public class Solution {
    public int maxProfit(int[] prices) {
        int res = 0;
        if(prices == null || prices.length <=1 )return res;
        int min = prices[0];
        for(int i=1; i< prices.length; i++){
            if(prices[i] > min){
                res = Math.max(res, prices[i] - min);
            }else{
                min = prices[i];
            }
        }

        return res;

    }
}