🤖 leetcode 121. Best Time to Buy and Sell Stock | easy | greedy | javascript

🗒️ Problems

121. Best Time to Buy and Sell Stock - Leetcode

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

You want to maximize your profit by choosing a single day
to buy one stock and choosing a different day in the future
to sell that stock.

Return the maximum profit you can achieve from this transaction.
If you cannot achieve any profit, return 0.

Input: prices = [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.
Note that buying on day 2 and selling on day 1 is not allowed
because you must buy before you sell.

Constraints:

1 <= prices.length <= 10^5
0 <= prices[i] <= 10^4

🤔 First attempt

• Buy in i day, and then sell in j day.

var maxProfit = function (prices) {
  const N = prices.length;
  let max = -Infinity;

  for (let i = 0; i < N; i += 1) {
    const buy = prices[i];
    for (let j = i + 1; j < N; j += 1) {
      const sell = prices[j];
      if (buy > sell) continue;
      if (max < sell - buy) max = sell - buy;
    }
  }
  return max === -Infinity ? 0 : max;
};

🙅‍♂️ Failed time complexity : O(n^2)

• Input size is 10^5, so that the nested loop will cause TLE.

✨ Idea

• When does the maximum profit occur ?
  • Buy the lowest price.
  • Sell the highest price.
• Can you do this in O(n) ?

🍀 Update min price

    let minPrice = Infinity;

    for (const price of prices) {
        if (minPrice > price) {
            minPrice = price
        } else {
          ...
        }
    }

🍀 Update max profit

• I can't sell in the same day which I bought.
• If it's NOT a min price, update the profit.

  let maxProfit = -Infinity

  for (const price of prices) {
    if (minPrice > price) {
      ...
    } else {
      const profit = price - minPrice
      if (maxProfit < profit) maxProfit = profit
    }
  }

🔥 My Solution

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

  for (const price of prices) {
    if (minPrice > price) {
      minPrice = price;
    } else {
      const profit = price - minPrice;
      if (maxProfit < profit) maxProfit = profit;
    }
  }

  return maxProfit === -Infinity ? 0 : maxProfit;
};

🙆‍♂️ Time complexity: O(n)