GOOGLE KICK START ROUND-A -Workout(SOLVED)

general
1

Can somebody help me find out a test case for which my code fails @ssjgz ?
PROBLEM: Kick Start - Google’s Coding Competitions
MY SOLUTION :

#include<bits/stdc++.h>
using namespace std;
typedef long long ll ;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<ll> vi;
typedef vector<pll> vpll;
typedef unordered_map <ll,ll> umap ; 
//#pragma GCC optimize "trapv"
#define loop(i,a,b) for(ll i=a ;i<b;i++)
#define For(i,n) for(int i=0;i<(ll)n;i++)
#define Rev(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,n) for(int i=1;i<=n;++i)
#define F first
#define S second
#define pb push_back
#define em emplace_back
#define all(v) (v).begin(),(v).end()
#define mems(x, y) memset(x, y, sizeof(x))
#define sz(v) (v).size()
#define mp(a,b) make_pair(a,b)
#define pf(n) cout<<n<<"\n"
#define pff(n) cout << n << " " ; 
long const M=1e9+7;
const long mxN =1e5+2 ;
const long mxNN =1e6+2 ;
ll a[mxN] ;
void solve(){
    ll n , k ; cin >> n >> k ;
    multiset<ll> v ; 
    For(i,n){
        cin >>a[i]  ;
        
    }
    for(int i =1 ; i <n;i++){
        v.insert(a[i]-a[i-1]) ;
    }
    auto it = v.end() ;
    it-- ;
    while(*it >=1  && k){
        ll val = *it ;
        v.erase(it) ;
        v.insert(val-val/2) ;
        v.insert(val/2) ;
        it = v.end() ;
        it-- ;
        k-- ;
    }
    it = v.end() ;
    it-- ;
    cout << *it << endl ;
    
    

    
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    //solve() ;
    ll t ; cin >> t ;
    For(i,t){
    cout <<"Case #"<< i+1 << ": ";
    solve() ;
    }
	return 0;
}
2 Likes
2

check out the analysis its already mention there that dividing by 2 is not optimal

For this test case, we cannot perform such direct splits because repeatedly splitting the maximum difference into halves is not optimal. For example, given a sequence [2, 12] and K = 2, splitting into halves will result in [2, 12] → [2, 7, 12] → [2, 7, 9, 12]. This way, the difficulty would be 5. However, if we perform [2, 12] → [2, 5, 12] → [2, 5, 8, 12], the difficulty would be 4.

4 Likes
3

Got it thanks :smiley:

4

your approach is lokoks right but it will give you a time limit
Because i used same approach and get 11 point but for k>1 && k<=10^5 TLE

5

Dont you think that he has done that only because he is maintaining difference so there will be 5 only not 7

6

Same approach but giving a TLE . Please help how to optimize it

#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<vector>
#include<set>
#include<stack>
#include<queue>
#define ll long long int

using namespace std;

main()
{
	ll t;
	cin>>t;
	ll x=1;
	while(t--)
	{
	    ll n,k;
	    cin>>n>>k;
	    ll a[n];
	    vector<ll> v;
	    for(ll i=0;i<n;i++)
	      {  cin>>a[i];
	         v.push_back(a[i]);
	        }
	    for(ll i=n-1;i>=0;i--)
	       v[i]=v[i]-v[i-1];
	      
	     v[0]=0;
	     
	     
	     sort(v.begin(),v.end());
	     while(k--)
	     {
	        ll x=v[v.size()-1];
	        v.pop_back();
	        ll x1=x/2;
	        ll x2=x-x1;
	        v.push_back(x1);
	        v.push_back(x2);
	        sort(v.begin(),v.end());
	     }
	     cout<<"Case #"<<x<<": "<<v[v.size()-1]<<endl;
	    x++;
	}
	
	}```
8

this wont give optimal ans
look at the example mentioned above

9

u can decrease time complexity by using multiset instead of sorting array everytime

10

But still i think it will give TLE . I saw many of doing binary search but i did not understand that approach so if you then please help

11

My code is giving optimal because already i get 11 point means for small input it is working . BUt for larger iit gives TLE

12

first set of test cases is only for k=1 which means u can only input one value which works for your algo but it didnt work when k>1 look at this example:

For this test case, we cannot perform such direct splits because repeatedly splitting the maximum difference into halves is not optimal. For example, given a sequence [2, 12] and K = 2, splitting into halves will result in [2, 12] → [2, 7, 12] → [2, 7, 9, 12]. This way, the difficulty would be 5. However, if we perform [2, 12] → [2, 5, 12] → [2, 5, 8, 12], the difficulty would be 4.

1 Like
14

Thanks.
What optimal algorithm have you used for the k>1 cases? Can you please share.

15

Bdw , even if you decreased the time complexity , it will give wrong answer. This approach is wrong.

16

What if I gave you the array and number D and asked to check if you can place K workouts anywhere you want and bring the max difference down to D or less ?

The same approach is being applied, we are searching for the lowest possible D by using binary search.

Suppose Mid= (L+R)/2 ,
now if we can get a max difference time of Mid or less ,then we search in the left half (R=Mid-1) looking for a lower value of D ,

If we can’t get a max difference time of Mid or Less, then obviously the answer is larger than Mid, so we search in the right half (L=MId+1) .