Best Time To Buy And Sell Stock Ii

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

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

                                            
class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        total_profit = 0
        for i in range(1, len(prices)):
            if prices[i] > prices[i - 1]:
                total_profit += prices[i] - prices[i - 1]
        return total_profit

                                            
int maxProfit(int* prices, int pricesSize) {
    int profit = 0;
    
    for (int i = 1; i < pricesSize; i++) {
        if (prices[i] > prices[i - 1]) {
            profit += prices[i] - prices[i - 1];
        }
    }
    
    return profit;
}

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

                                            
/**
 * @param {number[]} prices
 * @return {number}
 */
const maxProfit = function(prices) {
    let totalProfit = 0;
    
    for (let i = 1; i < prices.length; i++) {
        if (prices[i] > prices[i - 1]) {
            totalProfit += prices[i] - prices[i - 1];
        }
    }
    
    return totalProfit;
};

                                            
function maxProfit(prices: number[]): number {
    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;
};

                                            
class Solution {
    /**
     * @param Integer[] $prices
     * @return Integer
     */
    function maxProfit($prices) {
        $profit = 0;
        for ($i = 1; $i < count($prices); $i++) {
            if ($prices[$i] > $prices[$i - 1]) {
                $profit += $prices[$i] - $prices[$i - 1];
            }
        }
        return $profit;
    }
}

                                            
class Solution {
    func maxProfit(_ prices: [Int]) -> Int {
        var maxProfit = 0
        
        for i in 1..<prices.count {
            if prices[i] > prices[i - 1] {
                maxProfit += prices[i] - prices[i - 1]
            }
        }
        
        return maxProfit
    }
}

                                            
class Solution {
    fun maxProfit(prices: IntArray): Int {
        var totalProfit = 0
        for (i in 1 until prices.size) {
            if (prices[i] > prices[i - 1]) {
                totalProfit += prices[i] - prices[i - 1]
            }
        }
        return totalProfit
    }
}

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

                                            
func maxProfit(prices []int) int {
    maxProfit := 0
    
    for i := 1; i < len(prices); i++ {
        if prices[i] > prices[i-1] {
            maxProfit += prices[i] - prices[i-1]
        }
    }
    
    return maxProfit
}

                                            
# @param {Integer[]} prices
# @return {Integer}
def max_profit(prices)
    max_profit = 0
    (1...prices.length).each do |i|
        max_profit += [0, prices[i] - prices[i - 1]].max
    end
    max_profit
end

                                            
object Solution {
    def maxProfit(prices: Array[Int]): Int = {
        var totalProfit = 0
        for (i <- 1 until prices.length) {
            if (prices(i) > prices(i - 1)) {
                totalProfit += prices(i) - prices(i - 1)
            }
        }
        totalProfit
    }
}

                                            
impl Solution {
    pub fn max_profit(prices: Vec<i32>) -> i32 {
        let mut max_profit = 0;
        
        for i in 1..prices.len() {
            if prices[i] > prices[i - 1] {
                max_profit += prices[i] - prices[i - 1];
            }
        }
        
        max_profit
    }
}

To solve this coding challenge, we need to maximize the profit from stock trades given a list of daily stock prices. The key insight for this problem is to recognize that we can achieve the maximum profit by summing up all positive differences between consecutive days. This approach is greedy and efficient because it captures all increments as profit opportunities on each "uphill" day.

Explanation

Understanding the Problem:

Input/Output:

We receive an array
prices

where
prices[i]

represents the stock price on the i-th day.
The output should be the maximum profit we can achieve by buying and selling the stock on specific days.

Constraints:

We can hold at most one share of the stock at any time.
We can buy and sell on the same day.
The array length is between 1 and 30,000, and each price is between 0 and 10,000.

Approach:

Initial Observations:

If the list is strictly decreasing, no profit can be made.
If the list is strictly increasing, the maximum profit is the difference between the last and the first day.
For mixed prices, every upward movement (where
prices[i] > prices[i-1]

) represents a profit opportunity.

Strategy:

Traverse the list of prices.
For each pair of consecutive days, if the price of the second day is higher than the price of the first day, buy on the first day and sell on the second day.
Sum all these positive differences to get the total profit.

Algorithm:

Initialize a total_profit variable to 0.
Loop through the
prices

array starting from the second element.
For each element, if it is greater than the previous one, add the difference to the total_profit.
Return the total_profit after the loop finishes.

Detailed Steps in Pseudocode

Initialization:

Initialize
total_profit

to 0, which will hold our total accumulated profit.

Iterating through Prices:

Loop through the prices array starting from the second price (i.e., from index 1).

Checking for Profit Opportunity:

For each day, compare the price with the previous day’s price.
If today's price is higher, compute the difference and add this difference to the
total_profit

.

Returning the Result:

After finishing the loop, return the
total_profit

which contains the sum of all positive differences.

Detailed Pseudocode with Comments

                                            
# Function definition to calculate maximum profit
Function maxProfit(prices):
    # Initialize total profit to 0
    total_profit = 0
    
    # Loop through the list of prices from the second element to the last
    For i from 1 to length of prices - 1:
        # If today's price is greater than yesterday's price
        If prices[i] > prices[i - 1]:
            # Calculate the profit of buying the stock yesterday and selling today
            profit = prices[i] - prices[i - 1]
            # Add this profit to the total profit
            total_profit = total_profit + profit
    
    # Return the total profit after the loop
    Return total_profit

Summary of Pseudocode:

Initialize the
total_profit

to 0.
Iterate through the prices from the second element.
On each iteration, if the current day's price is higher than the previous day's, add the profit to
total_profit

.
Return the accumulated
total_profit

after the loop ends.

The pseudocode provides a systematic approach to the problem, focusing on capturing profit from all price increases and ensuring we are summing all positive differences. This captures the essence of the problem effectively and leads to an optimal solution.