[LeetCode] 122. Best Time to Buy and Sell Stock II

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Problem

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

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Input: [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.

Example 2:

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Input: [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.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

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Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Explanation

  1. Check if the array is empty, if it is, then return 0;

  2. Loop through the array, compare current element with its next element.

    • If the next element is greater, then we do noting, just wait for the next element to subtract the minimum element, if reach the end and the next element is still greater, then we sell the minimum using the next element and return the result.
    • If the next element is smaller, then we know we need to sell our stock with the current price, and then update the minimum to the next element.
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     [1, 3, 5, 7, 9, 2] => 9 - 1 = 8

 min is the first element 1, we wait for the 9 to subtract it, and update the min to 2.

Solution

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class Solution {
    public int maxProfit(int[] prices) {
        if (prices == null || prices.length == 0) return 0;
        int profit = 0, buy = prices[0];

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

        return profit;
    }
}