121. Best Time to Buy and Sell Stock.java

/*
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.
*/

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

My own solution:
public class Solution {
    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0){
            return 0;
        }
        
        int min = prices[0];
        int maxprofit = 0;
        //state: maxiunm transaction value
        int[] dp = new int[prices.length];
        dp[0] = 0;
        for(int i = 1; i < prices.length; i++){
            min = Math.min(min, prices[i]);
            dp[i] = Math.max(prices[i] - min, maxprofit);
            maxprofit = Math.max(maxprofit, dp[i]);
        }
        return maxprofit;
    }
}

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