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  1. Prepare
  2. Algorithms
  3. Dynamic Programming
  4. Equal
  5. Discussions

Equal

  • redb17
     + 1 comment

    Best possible solution can be found from (min,min-1,min-2). I think we don't need to check for (min-3) and (min-4). I have submitted by calculating optimal from (min,min-1,min-2). Can you provide an example where min-3 or min-4 is more optimal ?

    My code :

    #include<bits/stdc++.h>
using namespace std;
int main()
{
    int t,n;
    cin>>t;
    while(t--)
    {
        cin>>n;
        int arr[n],i,op1=0,op2=0,op3=0,mn=10000,left;
        for(i=0;i<n;i++)
        {
            cin>>arr[i];
            if(mn>arr[i])
                mn=arr[i];
        }
        for(i=0;i<n;i++)
        {
            left=arr[i]-mn; // for (min)
            op1+=left/5;
            left%=5;
            op1+=left/2;
            left%=2;
            op1+=left/1;

            left=arr[i]-(mn-1); // for (min-1)
            op2+=left/5;
            left%=5;
            op2+=left/2;
            left%=2;
            op2+=left/1;

            left=arr[i]-(mn-2); // for (min-2)
            op3+=left/5;
            left%=5;
            op3+=left/2;
            left%=2;
            op3+=left/1;
        }
        op1=min(op1,op2); // optimal from min-1 and min-2

        cout<<min(op1,op3)<<endl; // Most Optimal
    }
}