0122. Best Time to Buy and Sell Stock II

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii

Description

You are given an integer array prices where prices[i] is the price of a given stock on the ith day.

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

Example 1:

**Input:** prices = [7,1,5,3,6,4]
**Output:** 7
**Explanation:** Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Total profit is 4 + 3 = 7.

Example 2:

**Input:** prices = [1,2,3,4,5]
**Output:** 4
**Explanation:** Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Total profit is 4.

Example 3:

**Input:** prices = [7,6,4,3,1]
**Output:** 0
**Explanation:** There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

Constraints:

• 1 <= prices.length <= 3 * 104

• 0 <= prices[i] <= 104

ac1: one pass

very smart solution

class Solution {
    public int maxProfit(int[] prices) {
        // edge case
        if (prices == null || prices.length <= 1) return 0;

        // walk through
        int profit = 0;
        for (int i = 0; i < prices.length - 1; i++) {
            if (prices[i+1] > prices[i]) {
                profit += prices[i+1] - prices[i];
            }
        }

        return profit;
    }
}
/*
[7, 1, 5, 3, 6, 4]
buy point, valley
sell point, peak

*/

ac2: find peak and valley

class Solution {
    public int maxProfit(int[] prices) {
        // edge case
        if (prices == null || prices.length <= 1) return 0;

        // walk through
        int profit = 0;
        for (int i = 0; i < prices.length - 1; i++) {
            while (i < prices.length - 1 && prices[i] >= prices[i+1]) i++;  // find buy point
            int buy = prices[i];
            while (i < prices.length - 1 && prices[i] <= prices[i+1]) i++;  // find sell point;
            int sell = prices[i];
            profit += sell - buy;
        }

        return profit;
    }
}
/*
[7, 1, 5, 3, 6, 4]
buy point, valley
sell point, peak

*/