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Leetcode Best Time to Buy and Sell Stock II problem solution

YASH PAL, 31 July 202419 January 2026

In this Leetcode Best Time to Buy and Sell Stock II problem solution, we have given an array of prices where prices[i] is the price of a given stock on an ith day. Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

Leetcode Best Time to Buy and Sell Stock II problem solution

Leetcode Best Time to Buy and Sell Stock II problem solution in Python.

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

Best Time to Buy and Sell Stock II problem solution in Java.

public int maxProfit(int[] prices) {
        int max_profit = 0, i = 0, current_profit;
        while(i < prices.length) {
            current_profit = 0;
            while (i+1 < prices.length && prices[i] > prices[i+1])
                i++;
            if(i >= prices.length)
                return max_profit;
            current_profit = current_profit - prices[i];
            while (i+1 < prices.length && prices[i] < prices[i+1])
                i++;
            if(i >= prices.length)
                return max_profit;
            current_profit = current_profit + prices[i];
            max_profit = max_profit + current_profit;
            i++;
        }
        return max_profit;
    }

Problem solution in C++.

int maxProfit(vector<int> &prices) {
    int ret = 0;
    for (size_t p = 1; p < prices.size(); ++p) 
      ret += max(prices[p] - prices[p - 1], 0);    
    return ret;
}

Problem solution in C.

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

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