Going to be honest here and say that this is my third foray into HackerRank (first one was a work thing, second one was a quiz for a potential job opportunity) and I have to say that these test write-ups for what is expected is REALLY difficult to suss out.

No, I'm not a college graduate, no I don't have extensive experience in high level math, but I'm pretty good at reading comprehension and the way this stuff is written makes my head hurt trying to figure out what is expected.

And I've been doing code reading and writing for a living for the past 25 years.

Perhaps you guys could offer a tutorial on how to read the test descriptions given that the use of this site for applicant filtering seems to have become an industry standard?

Just a thought.

Thanks! Mike Poz

I think the question is a bit ambigious: "Minimum units with which Priyanka could buy all of toys", I think "Minimum group of units with which Priyanka could buy all of toys" would explain the problem statement better, and in the given example it would be [(1,2,3),(10),(17)] -> 3 groups of units

I am sharing my solution below, feel free to leave comments please.

#!/usr/bin/python3

N = int(input())
W = set(map(int, input().split()))

i = 0
while len(W) > 0:
    i += 1
    m = min(W)
    W = W.difference(set(range(m, m + 5)))

print(i)

this passes all test cases.

int cmpfunc (const void * a, const void * b)

{ return ( (int)a - (int)b ); }

int main() {

int n,i,x,y,c=1;

int a[100000]={0};

scanf("%d",&n);

for(i=0;i<n;i++)

    scanf("%d",&a[i]);

qsort(a, n, sizeof(a[0]), cmpfunc);

x=a[0];

y=x+4;

for(i=0;i<n;i++)
    {
    if(a[i]>=x && a[i]<=y)
        continue ;
     else
         {
         c++;
         x=a[i];
         y=x+4;
     }
}
printf("%d",c);

return 0;

}

my code friends

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

int main() {
    int n;
    cin >> n;
    vector<int> v(n);
    for(int i = 0; i < n; ++i) cin >> v[i];
    sort(v.begin(), v.end());
    int cost = 0;
    for(int i = 0; i < n; i = upper_bound(v.begin(), v.end(), v[i]+4) - v.begin())
        cost++;
    cout << cost;
    return 0;
}

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    cin>>n;
    int a[n];
    for(int i=0;i<n;i++)
    cin>>a[i];
    int b=0;
    int total=0;
    sort(a,a+n);
    b=a[0];
    for(int i=1;i<n;i++)
    {
        if(a[i]-b>4)
        {
            total++;
            b=a[i];
        }
    }
    cout<<total+1;
    
}

This is a pretty short code passing all the test cases.

So I checked a unit case which I failed. Input : 0 2 9 10 13 16 17 17 18 19 Output : 3

This is not even possible taking into account all possibilities as below:

Take 0, Count=1,0+4=4 so FreeToys=(0,2), remaining ways=8...TotalWays=1+8=9 Take 2, Count=2,2+4=6 so FreeToys=(2), remaining ways=8...TotalWays=2+8=10 Take 9, Count=3,9+4=13 so FreeToys=(9,10,13), remaining ways=5...TotalWays=3+5=8 Take 10, Count=4,10+4=14 so FreeToys=(10,13), remaining ways=5...TotalWays=4+5=9 Take 13, Count=5,13+4=17 so FreeToys=(13,16,17,17), remaining ways=2...TotalWays=5+2=7 Take 16, Count=6,16+4=20 so FreeToys=(16,17,17,18,19), remaining ways=0...TotalWays=6+0=6 Take 17, Count=7,17+4=21 so FreeToys=(17,17,18,19), remaining ways=0...TotalWays=7+0=7 Take 17, Count=8,17+4=21 so FreeToys=(17,17,18,19), remaining ways=0...TotalWays=8+0=8 Take 18, Count=9,18+4=22 so FreeToys=(18,19), remaining ways=0...TotalWays=9+0=9 Take 19, Count=10,19+4=23 so FreeToys=(19), remaining ways=0...TotalWays=10+0=10

Final List of Possible ways = [9,10,8,9,7,6,7,8,9,10] Minimum way = 6

I could not understand the efficient approach explained in the editorial. Please help. Thanks.

Can there be a possibility of 2 toys having the same weight?