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121. Best Time to Buy and Sell Stock 🔗

Date: Jun 03, 2024

Tags: two-pointers

Description

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.

Example 1:

      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.
      

Example 2:

      Input: prices = [7,6,4,3,1]
      Output: 0
      Explanation: In this case, no transactions are done and the max profit = 0.
      

Constraints:

  • 1 <= prices.length <= 105
  • 0 <= prices[i] <= 104

Optimal

function maxProfit(prices: number[]): number {
  let l = 0;
  let r = 1;
  let diff = -Infinity;

  while (r < prices.length) {
    const x = prices[l];
    const y = prices[r];

    if (y < x) {
      l = r;
    } else {
      diff = Math.max(y - x, diff);
    }

    r++;
  }

  return diff < 0 ? 0 : diff;
}

export { maxProfit };

Brute-Force

function maxProfit(prices: number[]): number {
  let diff = -Infinity;

  for (let i = 0; i < prices.length; i++) {
    for (let j = i + 1; j < prices.length; j++) {
      diff = Math.max(prices[j] - prices[i], diff);
    }
  }

  return diff < 0 ? 0 : diff;
}

export { maxProfit };