LeetCode #121: Best Time to Buy and Sell Stock (C/C++).

easy

source: https://leetcode.com/problems/best-time-to-buy-and-sell-stock
C/C++ Solution to LeetCode problem 121. Best Time to Buy and Sell Stock.

Problem

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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.

Examples

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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

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

Solution

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This is a greedy problem. To solve it:

• Iterate throught every value.
  • If current value is lower than a previous lower, then we have a new buy price (lower).
  • If the profit of selling at current value (current - lower) is greater than the current profit, we have a new maximum.
• Return the max profit found.

class Solution {
public:
  int maxProfit(vector<int>& prices) {
    int lowestPrice = INT_MAX;
    int maxProfit = 0;
    int cProfit = 0;
    
    for (int i = 0; i < prices.size(); i++) {
      if (prices[i] < lowestPrice)
        lowestPrice = prices[i];
      
      cProfit = prices[i] - lowestPrice;
      if (cProfit > maxProfit)
        maxProfit = cProfit;
    }

    return maxProfit;
  }
};