A,B,C | Codeforces Round 945 (Div. 2) Editorials | Cat, Fox and Double Maximum and the Lonely Array 

Welcome

can please tell me the resourse for learning the segmented tree..

Segment trees + binary search 😂

@@SamsIt-hi9gf watch striver video on segment tree , has has covered in detail and really good. Might feel long because its almost around 1hr

Can you please share your code

@@AbhinavAwasthi#include #include #include #include #include #include #include #include #include #define ll int using namespace std; class SegTree{ vectorseg; public: SegTree(int n){ seg.resize(4*n+1); } void build(int ind,int low,int high,vector&a){ if(low==high){ seg[ind]=a[low]; return; } else{ int mid=(low+high)>>1; build(2*ind+1,low,mid,a); build(2*ind+2,mid+1,high,a); seg[ind]=seg[2*ind+1]|seg[2*ind+2]; } } int query(int ind,int low,int high,int l,int r){ if(r1; if(iT; while(T--){ int n; cin>>n; vectorarr(n,0); for(int i=0;i>arr[i]; } SegTree s1(n); s1.build(0,0,n-1,arr); int left = 1, right = n, ans = n; while (left

Agree

Thanks

Hey, actually nowadays I am busy with some classes Probably from next week I will participate

because frequecy(count) of the number at particular index does not matter , the only thing matters in or operation is the avilabilty of bit. either it is set or not

Frequency equal is not compulsory, but frequency should be either 0 or more than 1 in both

Sure bro, actually from last few weeks I am a little more busy, because of which making quick editorials Will improve this

@@AbhinavAwasthi thanku so much

Yeah, I missed it

Thanks

Don't feel so bro, I can understand this happens sometimes

Bro 🙌🏻🙌🏻

1st plays 2 games with 3rd and 1 game with 2nd= 3 games 2nd plays 3 games with 3rd= 3 games Total 6 games

I think someone already explained

@@devtyagi7177 got it

even if the freq is 1 the bit would be set, you don't check frequencies rather than if the bit is set or not using the arrays.

@@nayanpahuja1919 Can you please explain it with example??

Yeah, it was actually a complex one

Sorry, by mistake

1100 because TLE-eliminator cp sheet ki 1100 rating me aise enough questions hai practice ke liye

@@kumkumslab5811 kuch bhi nearly 1300 rating hogi B ki

Pupil to specialist maanlo

@@SamsIt-hi9gf bhai maine solve kar lia tha newbie hu ye lo approach bool areArraysEqual( vll &arr1,vll &arr2) { if (arr1.size() != arr2.size()) { return false; } for (ll i = 0; i < arr1.size(); ++i) { if ((arr1[i] >=1 && arr2[i]=1 && arr1[i] 2*n) return false; vll temp(63,0); for (ll i = 0; i < mid && i < n; i++) { ll xr = arr[i]; for(int i=0;i>i)&1==1){ temp[i]++; } } } ll i=mid; ll j=0; vll temp2=temp; while(ik)&1==1){ temp2[k]--; } } ll newEle=arr[i]; for(int k=0;k>k)&1==1){ temp2[k]++; } } bool ans=areArraysEqual(temp,temp2); if(ans==0)return 0; i++; } return true; } void solve(){ ll n; cin>>n; vectorarr(n); for(ll i=0;i>arr[i]; } if(n==1){ cout

Building an Aquarium ye questions try kro

if there is answer for window size k, the condition is also true for k+1 window size.

@@ayushjaiswal2020 how??

@@TanmayKhandelwal-z9v because every smaller segement reside inside the large segment. just use pen try to draw by own. it will be clear.

@@ayushjaiswal2020 Yes please how you got the logic that it will be true for k+1 and further. I dry run it and found it to be true. But like how ??

@@aisharawat9102 1) If there is something to find min/max in problem, try to think if binary search can happen. 2) So to check if search space is monotonic, I evaluated for k+1 window size.

Hey, can you please dry run?

@@AbhinavAwasthi i have solved it 👍