Please link submissions that fail this test-case
Your code gives No if input is just one element - 0.
4 Likes
Didn’t expected that this will be done in O(n^2) … wasted 2 hours over thinking about optimized solution …
RIP ratings
18 Likes
no bro, i just checked, total pages of submission are 83, and pages for python submission are 4, so, rest 79 pages are for c++ or java or any other language.
Answer was just a google search away…
Please take case of this…
Link 1: https://www.geeksforgeeks.org/count-distinct-bitwise-or-of-all-subarrays/
Link 2: - LeetCode
anyone can just find the total distinct OR values of that array by this code:
and simply check if ((n*(n+1))/2) is equal to the above result or not. If its equal print yes. else no.
check out the soultion:
https://www.codechef.com/viewsolution/35817176
10 Likes
I have do this with another logic & this also accepted.
#include <stdio.h>
#include<math.h>
typedef long long int ll;
int cmpfunc(const void a,const void b){
if((ll)a==(ll)b) return 0;
else if((ll)a>(ll)b) return 1;
else return -1;
}
int main()
{
int t,n,flag,l;
scanf("%d",&t);
while(t–)
{
scanf("%d",&n);
l=2*(n-1);
ll a[n],b[n],c[l],total=0,left=0,right=0;flag=1;
for(int i=0;i<n;i++)
{
scanf("%lld",&a[i]);
b[i]=a[i];
total|=a[i];
}
qsort(a,n,sizeof(ll),cmpfunc);
for(int i=1;i<n;i++)
{
if(a[i]==a[i-1])
{
printf(“NO\n”);
flag=0;
break;
}
}
if(flag)
{
for(int i=0;i<n-1;i++)c[(2i)]=(left|=b[i]);
for(int i=n-1;i>0;i–)c[(2(i-1))+1]=(right|=b[i]);
qsort(c,l,sizeof(ll),cmpfunc);
for(int i=0;i<l;i++)
{
if((i!=0)&(c[i]==c[i-1])){flag=0;break;}
if(c[i]==total){flag=0;break;}
}
if(!flag)printf(“NO\n”);
else printf(“YES\n”);
}
}
}
Weak Test Case for ORTHODOX
Input:
1
3
1 50 99
Output:
NO
I have seen Solutions where the Output is YES.
@admin
2 Likes
Bhai koi plz mujhe ye answer smjha dega, meko bilkul smjh nhi aaya ye answer.
Please bhaiyo.
Please format your code
Thank you very much.
After adding just one more if statement it got accepted.
Seriously , same. And then i just saw submissions rising like a missile. The gfg and leetcode makes sense now.
2 Likes
why the answer is no if N>62??
5 Likes
Yeah bro 
this was pure hard luck 
2 Likes
I don’t Understand I had done the same thing except this if (n >= 100) { cout << "NO\n";…I got Tle First then i had to rethink and this cost me 1.30 hours i could have been in top 1000 now i am in late 2000… that’s pathetic…
The complexity of this solution is O(n^2). So how is it getting accepted within one second? Is it due to weak test cases?
Someone please reply and clear this dilemma.
2 Likes
Along with the weak test cases, don’t you think time limit was a bit off? O(N^2) within one second, maybe they haven’t put the worst cases as the test case?
1 Like
Its n ^ 2 when n <= 62, otherwise it’s always a NO.
So for n <= 62, you can simply do brute force, won’t TLE.
1 Like
Many people who unable to solve ORTHODOX and after the contest they get to know, it’s already on google / net -
to setters -
21 Likes