Leetcode Best Time to Buy and Sell Stock problem solution

In this Leetcode Best Time to Buy and Sell Stock problem solution, we have given an array of prices where prices[i] is the price of a given stock on an 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.

Problem solution in Python.

class Solution(object):
    def maxProfit(self, prices):
        buy, profit = prices[0] if len(prices) else -1, 0
        for i in range(1, len(prices)):
            profit = max(profit, prices[i]-buy)
            buy = min(buy, prices[i])
        return profit

Problem solution in Java.

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

Problem solution in C++.

class Solution {
public:
    int maxProfit(vector<int>& prices) 
    {
        if(prices.size()==1 || prices.size()==0)
            return 0;
        int profit=0;
        int max=prices[prices.size()-1];
        for(int i=prices.size()-2;i>=0;--i)
        {
            if(max-prices[i]>=0 && max-prices[i]>profit)
            {
                profit=max-prices[i];
            }
            if(prices[i]>max)
                max=prices[i];
        }
        return profit;
    }
};

Problem solution in C.

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