Skip to content
Programmingoneonone
Programmingoneonone
  • Engineering Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
    • 100+ C++ Programs
  • Solutions
    • HackerRank
      • Algorithms Solutions
      • C solutions
      • C++ solutions
      • Java solutions
      • Python solutions
    • Leetcode Solutions
    • HackerEarth Solutions
  • Work with US
Programmingoneonone
Programmingoneonone

HackerEarth Magic Potion problem solution

YASH PAL, 31 July 202415 February 2026
In this HackerEarth Magic Potion problem solution, Shruti loves to play with Array. She would always be busy doing some random operation with her array. Today she got to know about Magic Potion. A Magic Potion is a special power that allows you to remove one element from your array either from the start or the end. After spending some time on Magic Potion she decided to use it on her arrays.
 
Shruti has an array of size N. She calls an array a Good array if the sum of the array is exactly K. She wants to apply Magic Potion on her array so that she could get a good array. She wants to find the count of all the Good arrays that could be formed from the given initial array by applying Magic Potion on them any number of times. She is also interested in finding out the minimum Magic Potion required to form a single Good array. Since she is already quite busy she asks you for help.
 
 
HackerEarth Magic Potion problem solution

 

 

HackerEarth Magic Potion problem solution.

#include <bits/stdc++.h>
#define _ ios_base::sync_with_stdio(false);cin.tie(0);
using namespace std;
#define pb push_back
#define pob pop_back
#define pf push_front
#define pof pop_front
#define mp make_pair
#define all(a) a.begin(),a.end()
#define bitcnt(x) __builtin_popcountll(x)
#define mod 1000000007
#define PI 3.14159265
#define tot 300005
#define BLOCK 20000
#define MAXN 500005
typedef unsigned long long int uint64;
typedef long long int int64;

int arr[100005];
int64 k;
int n;
int64 sum[100005];

int64 getsum( int r , int l ){
return sum[r]-sum[l]+arr[l];
}

int upper(int idx,int l,int r,int64 req){
if(req<0)
return -1;
int lft=l,rht=r;
int ret=-1;
while(lft<=rht){
int mid=(lft+rht);
mid/=2;
if(getsum(mid,idx)==req){
ret=mid;
lft=mid+1;
}
if(getsum(mid,idx)>req)
rht=mid-1;
if(getsum(mid,idx)<req)
lft=mid+1;
}
return ret;
}

int lower(int idx,int l,int r,int64 req){
if(req<0)
return -1;
int lft=l,rht=r;
int ret=-1;
while(lft<=rht){
int mid=(lft+rht);
mid/=2;
if(getsum(mid,idx)==req){
ret = mid;
rht = mid-1;
}
if(getsum(mid,idx)>req)
rht=mid-1;
if(getsum(mid,idx)<req)
lft=mid+1;
}
return ret;
}

int main(){
freopen("input.txt","r",stdin);
freopen("output1.txt","w",stdout);
cin>>n>>k;
int cnt=0;
for(int i=1;i<=n;i++){
scanf("%d",&arr[i]);
assert(arr[i]>=0);
//assert(arr[i]<=10000000000);
sum[i]=sum[i-1]+arr[i];
cnt++;
}
assert(cnt==n);
int64 ans=0;
int minstep = n;
for(int i=1;i<=n;i++){
int idx1 = upper(i,i,n,k);
int idx2 = lower(i,i,n,k);
if( idx1 != -1 )
ans+=(idx1-idx2+1);
minstep = min( minstep , (i-1)+(n-idx1) );
}
printf("%lld %dn",ans,minstep);
return 0;
}
 
 
coding problems solutions HackerEarth HackerEarth

Post navigation

Previous post
Next post

Leave a Reply

Your email address will not be published. Required fields are marked *

Programmingoneonone

We at Programmingoneonone, also known as Programming101 is a learning hub of programming and other related stuff. We provide free learning tutorials/articles related to programming and other technical stuff to people who are eager to learn about it.

Pages

  • About US
  • Contact US
  • Privacy Policy

Practice

  • Java
  • C++
  • C

Follow US

  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2026 Programmingoneonone | WordPress Theme by SuperbThemes