121. Best Time to Buy and Sell Stock LeetCode Solution

In this guide, you will get 121. Best Time to Buy and Sell Stock LeetCode Solution with the best time and space complexity. The solution to Best Time to Buy and Sell Stock problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Best Time to Buy and Sell Stock solution in C++
  4. Best Time to Buy and Sell Stock solution in Java
  5. Best Time to Buy and Sell Stock solution in Python
  6. Additional Resources
121. Best Time to Buy and Sell Stock LeetCode Solution image

Problem Statement of Best Time to Buy and Sell Stock

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

See also  924. Minimize Malware Spread LeetCode Solution

Complexity Analysis

  • Time Complexity: O(n)
  • Space Complexity: O(1)

121. Best Time to Buy and Sell Stock LeetCode Solution in C++

class Solution {
 public:
  int maxProfit(vector<int>& prices) {
    int sellOne = 0;
    int holdOne = INT_MIN;

    for (const int price : prices) {
      sellOne = max(sellOne, holdOne + price);
      holdOne = max(holdOne, -price);
    }

    return sellOne;
  }
};
/* code provided by PROGIEZ */

121. Best Time to Buy and Sell Stock LeetCode Solution in Java

class Solution {
  public int maxProfit(int[] prices) {
    int sellOne = 0;
    int holdOne = Integer.MIN_VALUE;

    for (final int price : prices) {
      sellOne = Math.max(sellOne, holdOne + price);
      holdOne = Math.max(holdOne, -price);
    }

    return sellOne;
  }
}
// code provided by PROGIEZ

121. Best Time to Buy and Sell Stock LeetCode Solution in Python

class Solution:
  def maxProfit(self, prices: list[int]) -> int:
    sellOne = 0
    holdOne = -math.inf

    for price in prices:
      sellOne = max(sellOne, holdOne + price)
      holdOne = max(holdOne, -price)

    return sellOne
 # code by PROGIEZ

Additional Resources

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