javascript - return the maximum accumulated profit

Question

I'm doing an exercise which i'm no being able to solve. I need to get the maximum accumulated profit by buying and selling bitcoins. I have a function(A,Y) which receive an A = array of different prices during time and a Y = fee Restrictions:

Note: If a bitcoin was bought at 0 and sold at 1, we would have ad a loss of A[1] - A[0] =7050 -7200 - Y = -200. So, that movement was not made.

Note2: You can only have 1 bitcoin at the time. To sell, you have to have bought first. To buy, you need to have nothing or sold before.

Note3: Movements need to be time consequents. You cannot buy at A[5] and sell at A[4]

Note4: If no profit cannot be made, it should return 0

complexity is O(N)

A = [7200,7050,7300,7500,7440,7200,7300,7280,7400] //expected result 550
Y = 50
A[3] - A[1] - Y = 7500 - 7050 - 50 = 400
A[8] - A[5] - Y = 7400 - 7200 - 50 = 150
result = 550 //maximum accumulated profit

This is what i have

function solution(A, Y) {

 if(A.length < 2) {
     return 0;
 }

 var minIndex = (A[0] > A[1]) ? 1 : 0;
 var minPrice = A[minIndex];

 var acum = 0;
 var i = minIndex + 1

 for (i; i< A.length-1; i++) {
    if( (A[i] - minPrice - Y) > (A[i+1] - minPrice - Y  )) {
        acum += A[i] - minPrice - Y;
        i = i+1
    } else {
        acum += A[i + 1] - minPrice - Y;
        i = i+2
    }        
    minPrice = (A[i] > A[i+1]) ? A[i+1] : A[i];        
  }     
  return acum > 0 ? acum : 0;
}

Actually i'm getting 450 but it should be 550

Answer 1

It looks more complicated as it seems to be, because you need to check every single buying price with all possible selling price.

The result is a tree with this brute force approach.

This solution returns only the maximum profit with all buy/sell prices.

function maxima(array, fee) {

    function iter(prices, index, count) {
        var i = 0, profit = 0;
        if (index >= array.length) {
            if (!prices.length || prices.length % 2) {
                return;
            }
            if (prices.some((v, i, a) => i && (i % 2 ? a[i - 1] >= v : a[i - 1] < v))) {
                return;
            }
            while (i < prices.length) {
                profit += prices[i + 1] - prices[i] - fee;
                i += 2;
            }
            if (!result.length || result[0].profit < profit) {
                result = [{ profit, prices }];
            } else if (result[0].profit === profit) {
                result.push({ profit, prices });
            }
            return;
        }
        iter(prices.concat(array[index]), index + 1); // buy/sell
        iter(prices, index + 1);                      // no action
    }

    var result = [];
    iter([], 0, 0);
    return result;
}

console.log(maxima([7200, 7050, 7300, 7500, 7440, 7200, 7300, 7280, 7400], 50));

.as-console-wrapper { max-height: 100% !important; top: 0; }