121. Best Time to Buy and Sell Stock

Description

Say you have an array for which the i^th 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

DP, time O(n), space O(n)

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        if (n < 2) {
            return 0;
        }
        
        int[] max = new int[n];
        max[n - 1] = Integer.MIN_VALUE;
        for (int i = n - 2; i >= 0; --i) {
            max[i] = Math.max(max[i + 1], prices[i + 1]);
        }
        
        int maxProfit = 0;
        for (int i = 0; i < n - 1; ++i) {
            maxProfit = Math.max(maxProfit, max[i] - prices[i]);
        }
        
        return maxProfit;
    }
}

DP, time O(n), space O(1)

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        if (n < 2) {
            return 0;
        }
        
        int maxProfit = 0;
        int maxSell = prices[n - 1];
        
        for (int i = n - 2; i >= 0; --i) {
            maxSell = Math.max(maxSell, prices[i + 1]);
            maxProfit = Math.max(maxProfit, maxSell - prices[i]);
        }
        
        return maxProfit;
    }
}