Best time to buy and sell stock 2

//PEAK_VALLEY APPROACH

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        
        int n = prices.size();
        int diff = 0;
        for(int i=1;i<n;++i)
        {
            if(prices[i] > prices[i-1])
                diff += prices[i]-prices[i-1];
        }
        return diff;
    }
};




//LOOKUP_TABLE

class Solution {
    int find_maxProfit(bool buy,int pos,vector<int> prices,int n,vector<int> &dp_buy,vector<int> &dp_sell)
    {
        if(pos>=n || (buy==true && pos==n-1))
            return 0;
        else if(buy && dp_buy[pos]>0)
            return dp_buy[pos];
        else if(!buy && dp_sell[pos]>0)
            return dp_sell[pos];
        
        int profit = 0;
        //We have 3 options: Buy,Sell,Skip (current share)
        int skip = find_maxProfit(buy,pos+1,prices,n,dp_buy,dp_sell);  //Skip case
        if(buy)
        {
            profit = -prices[pos] + find_maxProfit(false,pos+1,prices,n,dp_buy,dp_sell);
            dp_buy[pos] = max(skip,profit);
            return dp_buy[pos];
        }
        else
        {
            profit = prices[pos] + find_maxProfit(true,pos+1,prices,n,dp_buy,dp_sell);
            dp_sell[pos] = max(skip,profit);
            return dp_sell[pos];
        }
        
        //Return by taking max of all 3 cases
        return max(profit,skip);
    }
public:
    int maxProfit(vector<int>& prices) {
        
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        
        int n = prices.size();
        if(n==0 || n==1)
            return 0;
        else if(n==2)
            return max(0,prices[1]-prices[0]);
        
        vector<int> dp_buy(n,0);
        vector<int> dp_sell(n,0);
        //vector<int> dp(n,0);
        
        return find_maxProfit(true,0,prices,n,dp_buy,dp_sell);
    }
};

Can you explain the intuition behind this lookup table it's hard to understand the recursion here and why did you take two vectors dp_buy,dp_sell and what are you doing with them

yes ,pls explain dp solution

@Pawanupadhyay10 see my approach but it will give tle in last test case

class Solution {
int dp[30005][2];
int find_maxProfit(bool buy,int pos,vector prices,int n)
{
if(pos>=n || (buy==true && pos==n-1))
return 0;
if(dp[pos][buy]!=-1)
return dp[pos][buy];
int profit = 0;
int skip = find_maxProfit(buy,pos+1,prices,n); //Skip case
if(buy)
{
profit = -prices[pos] + find_maxProfit(false,pos+1,prices,n);
}
else
{
profit = prices[pos] + find_maxProfit(true,pos+1,prices,n);
}
return dp[pos][buy]= max(profit,skip);
}
public:
int maxProfit(vector& prices) {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n = prices.size();
    if(n==0 || n==1)
        return 0;
    else if(n==2)
        return max(0,prices[1]-prices[0]);
    memset(dp,-1,sizeof(dp));
    return find_maxProfit(true,0,prices,n);
}

};