Link to the problem

Link to my solution

My Solution -

#include <bits/stdc++.h>
using namespace std;

map<int, int> mp;

int findPower(int i, int k)
{
    int t = i, cnt = 0;
    while (t <= k)
    {
	if (mp.find(t) != mp.end())
	{
	    cnt += mp[t];
	    break;
	}
	t *= 2;
	++cnt;
    }
    return mp[i] = cnt;
}

int main()
{
    int n, k;
    cin >> n >> k;
    // power of 2
    vector<int> p2(31);
    p2[0] = 1;
    for (int i=1; i<p2.size(); ++i)	p2[i] = p2[i-1] * 2;

    long double b = 0.0;
    for (int i=n; i>=1; --i)
    {
	int p = findPower(i, k);
	p = p2[p];
	b += (long double)1/p;
    }
    cout << ((long double)1/n) * b << '\n';
    return 0;
}

I know my solution is unnecessarily long, I was trying to catch value, so to use them later. I don't know what I am doing wrong, moreover I am not getting correct decimal precision. For sample — 100000 5, the output should be — 0.999973749998, but my program is returning 0.9999687500, why could it be?

Problem link — https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/

Problem Description:

Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee.

You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)

Return the maximum profit you can make.

Example 1: Input: prices = [1, 3, 2, 8, 4, 9], fee = 2

Output: 8 Explanation: The maximum profit can be achieved by: Buying at prices[0] = 1 Selling at prices[3] = 8 Buying at prices[4] = 4 Selling at prices[5] = 9 The total profit is ((8 — 1) — 2) + ((9 — 4) — 2) = 8.

Note: 0 < prices.length <= 50000; 0 < prices[i] < 50000; 0 <= fee < 50000.

My Solution : I have come up with a backtrack solution, which is working fine with smaller inputs, but gives TLE with bigger inputs. I am unable to memoize it. Need Help !!!

What I am doing is, I am checking if the we have already purchased a share, If Yes, either we can sell the share now or later. If No, we either purchase it now or later.

class Solution
{
    public:
    
    int solve (vector<int> &prices, bool hasPurchased, int lastPurchasedVal, int i, int fee)
    {
        if (i == prices.size()) return 0;
        
       // if (dp[hasPurchased][i] != -1) return dp[hasPurchased][i];
        
        if (hasPurchased)
        {
            int sold = INT_MIN;
            int profit = prices[i] - lastPurchasedVal - fee;
            if (profit > 0)
                sold = profit + solve (prices, !hasPurchased, 0, i+1, fee);
            int notsold = solve(prices, hasPurchased, lastPurchasedVal, i+2, fee);
            return  max(sold, notsold);
        }
        else
        {
            int purchased = solve (prices, !hasPurchased, prices[i], i+1, fee);
            int notPurchased = solve(prices, hasPurchased, lastPurchasedVal, i+1, fee);
            return  max(purchased, notPurchased);
        }       
    }
    
    int maxProfit(vector<int>& prices, int fee) 
    {
        //int dp[2][50001];
        //memset(dp, -1, sizeof dp);
        return solve (prices, false, 0, 0, fee);
    }
};

Thank you.

I am having trouble understanding this question. Sorry the images are not quite clear, it's a snapshot from the PC. This question was asked previous year by a company which came to our college for recruitment.

The input example given are more like the multiple of 5. But I think there is more the question I am missing.

Thanks

UPDATE Don't know why images are not getting uploaded. Image 1, part 1 Image 1, part 2