## Key Idea

### Solution 1: DP

There exists 2 states, and those can be expressed as:

Hence, we can use dynamic programming approach to this problem, representing state transfers.

• holding[i] = max( holding[i-1] , notHolding[i-1] - prices[i])
• notHolding[i] = max( notHolding[i-1] , holding[i-1] + prices[i] - fee )

Initializing base case shouldn’t be difficult,

• holding[0] = -prices[0]
• notHolding[0] = 0

The ultimate answer is $$\max(holding[n-1], notHolding[n-1])$$.

• Time: $$O(N)$$
• Space: $$O(N)$$

Note
We can optimize space complexity into $$O(1)$$. Implementation below does that.

### Solution 2: DP & Greedy

Example test cases tell us many things.

From left to right, update those values with following definitions.

• i: current index
• curMin: minimum price so far
• curMaxProfit: maximum profit so far

Say i=4, then curMin becomes 1 and curMaxProfit = 8 - 1 - 2(fee) = 5.
Here we need to decide wether to buy price[4] or not and it is always optimal to buy it when:
$$curMaxProfit + P - prices[i] - fee \gt P - curMin - fee$$
$$curMaxProfit + \bcancel{P} - prices[i] - \cancel{fee} \gt \bcancel{P} - curMin - \cancel{fee}$$
$$curMaxProfit \gt prices[i] - curMin$$
where P denotes some huge price in the future.

Thus, we can pick prices to buy in a greedy way, by checking the above condition.
From left to right,

• if $$curMaxProfit \gt prices[i] - curMin$$, then ans += curMaxProfit. prices[i] is a new curMin now.
• else if prices[i] < curMin, update curMin value.
• else update curMaxProfit value

• Time: $$O(N)$$
• Space: $$O(1)$$

## Implementation

### Solution 1: DP

typedef vector<int> vi;

class Solution {
public:
int maxProfit(vi &prices, int fee) {
int n= prices.size(), holding= -prices[0], notHolding= 0;
for (int i=1; i < n; ++i) {
int prevHolding= holding;
holding= max(prevHolding, notHolding-prices[i]);
notHolding= max(notHolding, prevHolding+prices[i]-fee);
}
return max(holding, notHolding);
}
};


### Solution 2: DP & Greedy

typedef vector<int> vi;

class Solution {
public:
int maxProfit(vi &prices, int fee) {
int n= prices.size(), curMin= prices[0], curMaxProfit= 0, ans= 0;
for (int i=1; i < n; ++i) {
if (curMin + curMaxProfit > prices[i]) {
ans += curMaxProfit;
curMin= prices[i];
curMaxProfit= 0;
}
else if (prices[i] < curMin) curMin= prices[i];
else curMaxProfit= max(curMaxProfit, prices[i]-curMin-fee);
}
ans += curMaxProfit;
return ans;
}
};


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