### haribhatt34's blog

By haribhatt34, history, 8 months ago, ,

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?

• 0

By haribhatt34, history, 14 months ago, ,

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.

• 0

By haribhatt34, history, 17 months ago, ,

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