LeetCode 121 - Best Time to Buy and Sell Stock

• Difficulty: Easy.
• Related Topics: Array, Dynamic Programming.
• Link: leetcode

Problem

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Solution (Best)

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function (prices) {
  let maxProfit = 0;
  let minPrice = Number.MAX_SAFE_INTEGER;

  if (prices.length < 2) return maxProfit;

  for (let buyDay = 0; buyDay < prices.length; buyDay++) {
    if (prices[buyDay] < minPrice) {
      minPrice = prices[buyDay];
    } else if (prices[buyDay] - minPrice > maxProfit) {
      maxProfit = prices[buyDay] - minPrice;
    }
  }

  return maxProfit;
};

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).

Solution 2 (exceed time limit)

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function (prices) {
  let maxProfit = 0;

  if (prices.length < 2) return maxProfit;

  for (let buyDay = 0; buyDay < prices.length; buyDay++) {
    for (let sellDay = buyDay + 1; sellDay < prices.length; sellDay++) {
      if (prices[sellDay] - prices[buyDay] > maxProfit) {
        maxProfit = prices[sellDay] - prices[buyDay];
      }
    }
  }

  return maxProfit;
};

Complexity:

• Time complexity : O(n²).
• Space complexity : O(1).

Explain:

• 當天數小於 2 天時，直接 return 0
• solution 2 當陣列元素多的時候會超過時間限制，效能不好