HackerEarth Not in Range problem solution

YASH PAL, 31 July 202414 February 2026

In this HackerEarth Not in Range problem solution, You are given (10 to power 6) boxes that are numbered from 1 to (10 to power 6). The value of each box is equal to its number.

There are N ranges and every range consists of two integers L and R denoting that the value of box in the range [L,R] will turn out to be zero.

Find the sum of values of all boxes from 1 to (10 to power 6).

HackerEarth Not in Range problem solution

HackerEarth Not in Range problem solution.

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
#define lli long long int
#define br cout<<"n";
#define plli pair<lli,lli>
#define vlli vector<lli>
#define vplli vector<plli>
#define pqlli priority_queue <lli>
#define pqplli priority_queue <plli>
#define mplli map<lli,lli>
#define pb push_back
#define mp make_pair
#define fr(i,n) for(i=0;i<n;i++)
#define fr1(i,n) for(i=1;i<=n;i++)
#define arr(a,n) lli a[n]; fr(i,n){ cin>>a[i]; }
#define arr1(a,n) lli a[n+5]; fr1(i,n){ cin>>a[i]; }
#define print(a,n) fr(i,n) {cout<<a[i]<<" ";} br
#define print1(a,n) fr1(i,n) {cout<<a[i]<<" ";} br
#define printv(vec) for(lli qq=0;qq<vec.size();qq++) {cout<<vec[qq]<<" ";}
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define MAXN 300005
#define mset(a) memset(a,0,sizeof a);
#define MAXNN 1000005
#define mod 1000000007
using namespace std;
using namespace __gnu_pbds;
const lli inf = 1e12 + 7;
lli power(lli x,  lli y, lli p){ lli res = 1; x = x % p; while (y > 0) {if (y & 1)res = (res%p*x%p) % p;y = y>>1;x = (x%p * x%p) % p;}return res;}
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
 lli a[MAXNN];
int main()
{
    ios_base::sync_with_stdio(false);
    lli i,j,k,l,r,x,y,n,m,t;
    mset(a);
    cin>>n;
    while(n--)
    {
        cin>>x>>y;
        a[x]+=1;
        a[y+1]+=-1;
    }
for(i=2;i<=1000000;i++)
{
    a[i]+=a[i-1];
}
x=0;
for(i=1;i<=1000000;i++)
{
    if(a[i]==0)
        x+=i;
}
cout<<x<<'n';
    return 0;
}

Second solution

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define endl "n"
#define eps 0.00000001
LL pow(LL a,LL b,LL m){ a%=m;LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
LL upd[1000001];
int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    vector<pair<int,int> > v;
    LL total = ((1000000LL) * (1000001LL)) / 2;
    int n;
    assert(cin >> n);
    for(int i = 1; i <= n; i++) {
        LL l, r;
        assert(cin >> l >> r);
        upd[l] += 1;
        upd[r + 1] += -1;
    }
    for(int i = 1; i <= 1000000; i++) {
        upd[i] += (upd[i - 1]);
    }
    LL ans = 0;
    for(int i = 1; i <= 1000000; i++) {
        if(upd[i] == 0)
            ans += i;
    }
    cout << ans;
}

coding problems solutions HackerEarth HackerEarth

Leave a Reply