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ArraysMedium

Stock buy sell

Problem link:

Problem Statement

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.

Brute Force

Intuition

Understand the problem and develop a solution approach.

Initialize two variables: min_price and max_profit.

-> min_price = minimum price in the array. -> max_profit = 0.

Iterate through the array, and for each price:

-> Update min_price to the minimum price seen so far. -> Update max_profit to the maximum profit seen so far, or the current price minus min_price, whichever is greater.

Return max_profit.

Approach

Initialize two variables: min_price and max_profit.

-> min_price = minimum price in the array. -> max_profit = 0.

Iterate through the array, and for each price:

-> Update min_price to the minimum price seen so far. -> Update max_profit to the maximum profit seen so far, or the current price minus min_price, whichever is greater.

Return max_profit.

Code (Java)

int maxProfit(ArrayList<int> &prices)
{
    int minprice = prices[0];
    int ans = 0;
    for (int i = 1; i < prices.size(); i++)
    {
        ans = max(ans, prices[i] - minprice);
        minprice = min(minprice, prices[i]);
    }
    return ans;
}
Time: O(N)Space: O(0)

Complexity Analysis

Time complexity: O(N) Space complexity: O(0)

Quick Revision (Brute Force)

  • Initialize two variables: min_price and max_profit.
  • min_price = minimum price in the array.
  • max_profit = 0.
  • Iterate through the array, and for each price:
  • Update min_price to the minimum price seen so far.

Quick Revision (Optimal)

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