Importance of time problem

#1


I m getting TLE. Please suggest any changes or a different approach.

#2

hello @apoorvas1301

refer this approach->

  • The Idea is to Store the Calling order in Queue and the ideal order in an Array.
  • Then compare front element of the queue with the ideal order.
  • If it matches then increment the time and remove the element.
  • Else, Dequeue the element and enqueue it again.
  • The loop will run till the queue is consumed totally.

code implementation->

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

int importanceOfTime(queue<int> &q, const vector<int> &arr)
{
    //arr is the ideal order
    //q is the calling order
    int ans = 0;
    for (int i = 0; i < arr.size(); i++)
    {
        if (q.front() == arr[i])
        {
            ans++;
            q.pop();
        }
        else
        {
            while (q.front() != arr[i])
            {
                int var = q.front();
                q.push(var);
                q.pop();
                ans++;
            }
            ans++;
            q.pop();
        }
    }
    return ans;
}

int main()
{
    int n;
    cin >> n;
    vector<int> v(n);
    queue<int> q;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        q.push(x);
    }

    for (int i = 0; i < n; i++)
    {
        cin >> v[i];
    }

    cout << importanceOfTime(q, v);

    return 0;
}
#3

Thank you. Can you also point out the errors in my code?

#4

yeah , pls explain ur approach , i will try to debug it

#5

@apoorvas1301
check ur updated code here->

#include<iostream>
#include<deque>
using namespace std;
int timerequired(deque<int> &q1,deque<int> &q2,int n)
{
	int count=0;
	while(n--)
	{	
		while(q1.front()!=q2.front())
		{
			int x=q1.front();
			q1.pop_front();
			q1.push_back(x); // u were pushing it to q2
			count++;
		}
		// removed : count++;
		if(q1.front()==q2.front())
		{
			count++;
			q1.pop_front();
			q2.pop_front();
		}	
	}
	return count;
}
int main() {
	int n,a,b;
	cin>>n;
	int k=n;
	deque<int> q1,q2;
	while(n--)
	{
		cin>>a;
		q1.push_back(a);
	}
	n=k; // after this loop n will be zero so u need to again reinitilaise
	while(n--)
	{
		cin>>b;
		q2.push_back(b);
	}
	cout<<timerequired(q1,q2,k)<<endl;
	return 0;
}
#6

It worked .Thank you