LeetCode "Best Time to Buy and Sell Stock II" Explained: Step-by-Step

Hello, coding enthusiasts! 👋 Today, we’re diving into the popular LeetCode problem, "Best Time to Buy and Sell Stock II". This one’s a classic, and trust me, once you get the hang of it, you'll feel like a Wall Street guru 📈. Let's walk through the logic and various implementations with a sprinkle of humor and, of course, some emojis to keep things lively!

The Problem

Imagine you have an array where each element represents the stock price on that day. Your goal is to maximize your profit by choosing when to buy and sell. The catch is, you can buy and sell as many times as you want, but you can only hold one share of the stock at a time.

The Intuition

Okay, so here's the trick: we’re not looking for the absolute best single transaction. Instead, we’re capturing all those small profitable transactions like a greedy little squirrel 🐿️.

The key insight is to sum up all the positive differences between consecutive days. If the price goes up the next day, we add that profit to our total. It's like this: if you see a chance to make a profit, you take it! No hesitation!

The C++ Solution

Let's start with the C++ solution. Here's how we can implement this greedy approach:

#include <vector>
using namespace std;

int maxProfit(vector<int>& arr) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int len = arr.size();
    int profit = 0;
    for(int i = 1; i < len; i++){
        if(arr[i] > arr[i-1]){
            profit += arr[i] - arr[i-1];
        }
    }
    return profit;
}

Explanation

Iterate through the array: We start from day 1 (because you need at least one day to buy the stock).
Check for profit: If the price today is higher than yesterday, we add the difference to our profit.
Return the total profit: By the end of the loop, we have the maximum profit possible from all the little transactions.

Simple, right? It's all about capturing every small opportunity, just like a true opportunist 📉➡️📈.

Other Implementations

To make sure everyone can join the fun, let’s see how we can translate this into Python, Java, and JavaScript.

Python

def maxProfit(prices):
    profit = 0
    for i in range(1, len(prices)):
        if prices[i] > prices[i - 1]:
            profit += prices[i] - prices[i - 1]
    return profit

Java

public class Solution {
    public int maxProfit(int[] prices) {
        int profit = 0;
        for(int i = 1; i < prices.length; i++) {
            if(prices[i] > prices[i - 1]) {
                profit += prices[i] - prices[i - 1];
            }
        }
        return profit;
    }
}

JavaScript

var maxProfit = function(prices) {
    let profit = 0;
    for (let i = 1; i < prices.length; i++) {
        if (prices[i] > prices[i - 1]) {
            profit += prices[i] - prices[i - 1];
        }
    }
    return profit;
};

Wrapping Up

There you have it, folks! The "Best Time to Buy and Sell Stock II" problem is all about seizing the day, one tiny profit at a time. It’s a great exercise in thinking greedily and optimizing for the best outcome with simple logic.

Remember, the stock market (and coding) is not about getting it perfect every single time but making the best out of each opportunity. Happy coding, and may your profits always be positive! 📈💰

Feel free to drop any questions or comments below. Let's crack these problems together, one laugh at a time! 😄

That's it for now. Stay tuned for more coding fun and insightful tutorials. Until next time, keep coding and keep smiling! 😄

LeetCode "Best Time to Buy and Sell Stock II" Explained: Step-by-Step Solutions

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