HackerEarth The Enlightened Ones problem solution

In this HackerEarth The Enlightened Ones problem solution There are N temples in a straight line and K monks who want to spread their enlightening power to the entire road of temples. All the monks have an enlightenment value, which denotes the range of enlightenment which they can spread in both the directions. Since, they do not want to waste their efficiency on trivial things of the world, they want to keep their range minimum.

Also, when we say that the N temples are in a straight line, we mean that that all the temples lie on something like an X-axis in a graph.

Find the minimum enlightenment value such that all the temples can receive it.

HackerEarth The Enlightened One’s problem solution.

#include<bits/stdc++.h>


using namespace std;

#define rep(i,n) for(i=0;i<n;i++)
#define ll long long
#define elif else if
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define CLEAR(array, value) memset(ptr, value, sizeof(array));
#define si(a)     scanf("%d", &a)
#define sl(a)     scanf("%lld", &a)
#define pi(a)     printf("%d", a)
#define pl(a)     printf("%lld", a)
#define pn        printf("n")
#define SZ(x)  (int)((x).size())

vector< int>tmp;
int check(int num,int k)
{
 int i,j,prev=tmp[0]+num;
k--;
 for(i=1;i<tmp.size();i++)
 {
   if(prev+num>=tmp[i])
    continue;
   if(k==0)
    return 0;
   prev=tmp[i]+num;
   k--;
 }
 return 1;
}
int main()
{
  int n,i,j,k;
  cin>>n>>k;
  tmp.resize(n);
  rep(i,n)
  cin>>tmp[i];
  sort(tmp.begin(),tmp.end());
  
   int l=0,h=0,mid,p=-1;
    h=100000000;
  while(h>=l)
  {
      mid=(l+h)/2; 
      int vv=check(mid,k);
  //  cout<<l<<" "<<h<<" "<<mid<<" "<<vv<<"n";
    if( check(mid,k) && check(mid-1,k)==0)break;
      if(vv==0)
        {
          p=mid;
          l=mid+1;
        }
      else
      h=mid-1;
  }
  cout<<mid<<"n";
  return 0;
}

Second solution

#include<bits/stdc++.h>


using namespace std;

#define rep(i,n) for(i=0;i<n;i++)
#define ll long long
#define elif else if
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define CLEAR(array, value) memset(ptr, value, sizeof(array));
#define si(a)     scanf("%d", &a)
#define sl(a)     scanf("%lld", &a)
#define pi(a)     printf("%d", a)
#define pl(a)     printf("%lld", a)
#define pn        printf("n")
#define SZ(x)  (int)((x).size())

vector< int>tmp;
int check(int num,int k)
{
 int i,j,prev=tmp[0]+num;
k--;
 for(i=1;i<tmp.size();i++)
 {
   if(prev+num>=tmp[i])
    continue;
   if(k==0)
    return 0;
   prev=tmp[i]+num;
   k--;
 }
 return 1;
}
int main()
{
  int n,i,j,k;
  cin>>n>>k;
  tmp.resize(n);
  rep(i,n)
  cin>>tmp[i];
  sort(tmp.begin(),tmp.end());
  
   int l=0,h=0,mid,p=-1;
    h=100000000;
  while(h>=l)
  {
      mid=(l+h)/2; 
      int vv=check(mid,k);
  //  cout<<l<<" "<<h<<" "<<mid<<" "<<vv<<"n";
    if( check(mid,k) && check(mid-1,k)==0)break;
      if(vv==0)
        {
          p=mid;
          l=mid+1;
        }
      else
      h=mid-1;
  }
  cout<<mid<<"n";
  return 0;
}

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