Profit From Multiple Stock Trades

medium array greedy

Problem

Given a daily price series, you can buy and sell as often as you like, holding at most one share at a time. What is the maximum total profit? The greedy answer: sum every positive day-over-day climb. Any longer winning streak from a low to a high decomposes exactly into those single-day gains, so capturing every up-day is optimal.

Inputprices = [7, 1, 5, 3, 6, 4]

Output7

Buy at 1, sell at 5 (+4). Buy at 3, sell at 6 (+3). Total 7 — equal to the sum of (5−1) + (6−3) = (5−1) is split into (1→2 etc). Equivalently, sum positive deltas: (5−1)+(3−5 skip)+(6−3)+(4−6 skip) = 4 + 3.

prices (comma-separated)

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

function maxProfit(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;
}

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;
    }
}

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

Time: O(n) Space: O(1)