Відео

In problem D1, it says you start with some element X but what exactly is that X? We don't really take anything X as input so how do we exactly guess what x to take?

when will tle 12.0 be released?

sir you are best teacher a well as the guider to help in the journey of cp and dsa salute to you please make video on all the contest solved by your idea

Can anyone tell why it is giving WA- class Solution { public: using ll = long long; int MOD = 1e9 + 7; ll findexp(ll base, ll exponent) { ll result = 1; while (exponent > 0) { if (exponent % 2 == 1) { result = (result * base) % MOD; } base = (base * base) % MOD; exponent /= 2; } return result; } vector<int> getFinalState(vector<int>& nums, int k, int x) { if(x==1) return nums; int n = nums.size(); int maxi = *max_element(nums.begin(), nums.end()); priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq; for (int i = 0; i < n; i++) { pq.push({nums[i], i}); } while (k && pq.top().first * x <= maxi) { ll t = pq.top().first; int i = pq.top().second; pq.pop(); t*=x; pq.push({t, i}); k--; } ll remainingop = k / n; ll extras = k % n; ll z = findexp(x, remainingop); // Calculate the final states for the first n elements for (int i = 0; i < n; i++) { ll t = pq.top().first; int j = pq.top().second; pq.pop(); // Carefully multiply with z and apply the modulo in a controlled manner pq.push({(1LL*z * t)%MOD, j}); } // Handle the remaining extra operations individually for (int i = 0; i < extras; i++) { ll t = pq.top().first; int j = pq.top().second; pq.pop(); ll temp = (1LL * x) % MOD; temp = (temp * t) % MOD; pq.push({temp, j}); } // Extract the final state of the array while (!pq.empty()) { ll t = pq.top().first; int i = pq.top().second; pq.pop(); nums[i] = t%MOD; } return nums; } };

Loved D!!

What is the codeforces handle of Abhinav?

4th question ♥♥

Can I start cp with Java also?

Great video !

But what's the time complexity for D1? HOW N*L WORKS?

22:32 🤣🤣🤣😁

Awesome solution for D1 🔥Thanks Harsh Sir

good explanation viraj 👏👏

what is ans for D1 for below testcase. 2 0 1 0 2 0 3 3 50 1 2 3 1 2 3 4 0 1 2 3 And also ans to this.

Great, Harsh bhai.

Super 🎉

Thanks @VraijChandra

Great Explanation for D1

I think c explaining without the intuition is not worthy

I couldn't even solve A 😢

Beautifully Explained D1........ very simple explanation like Abhinav

Thanks Bro. Amazing

GUESSFORCES

Can anyone please tell me the rating of D1

VERY GOOD EXPLANATION OF ALL THE FOUR PROBLEMS

how are you so patient harsh.. loved every bit of it..

bro i hate myself for missing that (possibly zero), saddest that has happened to me in this contest

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3 explanation was 🔥

Whew! I wounder how he made it to the observation.

Best tutor at tle.

got 3rd problem very clearly

How is the initial state achieved? (1LL*multiplier*pq.top().first)%mod)<mx you are doing suppose we have array [2,1,3,5,6] and k=5 and m=4 my min heap will have 1(at top) 1*4=4 you push it into the pq & mx=6 1,(2,3,5,6,4) , you pop 1 2(3,5,6,4) ,multiply 2*4=8 & mx=6 it breaks here, but all elements are not multiplied by multiplier at least once Please explain, at least dry run the code that you write with a given example!

Difficult to understand, need some more effective explanation

I feel brute force for question 2 could have been done in a simpler way, reducing its space complexity ``` class Solution { public: bool swapCheck(int a,int b){ string x=to_string(a); string y=to_string(b); int count=0; for(int i=0;i<y.length()-1;i++){ for(int j=i+1;j<y.length();j++){ string temp=y; swap(temp[i],temp[j]); if(stoi(temp)==stoi(x)){ return true; } } } return false; } int countPairs(vector<int>& nums) { int count=0; for(int i=0;i<nums.size()-1;i++){ for(int j=i+1;j<nums.size();j++){ if(nums[i]==nums[j]){ count++; cout<<nums[i]<<","<<nums[j]<<" "; continue; } if(swapCheck(nums[i],nums[j])||swapCheck(nums[j],nums[i])){ count++; cout<<nums[i]<<","<<nums[j]<<" "; } } } return count; } }; ``` During the swap itself, just check if you its a possible same string

In B , we can simplify the process : CHECK THIS OUT class Solution { public: bool swapit(int i, int j, vector<int> &nums) { string str1 = to_string(nums[i]); string str2 = to_string(nums[j]); while(str1.size()>str2.size()) str2 = '0' + str2; while(str1.size()<str2.size()) str1 = '0' + str1; if(str1 == str2) return true; unordered_set<string> set; for(int i=0;i<str2.size();i++){ for(int j=i+1;j<str2.size();j++){ swap(str2[i],str2[j]); set.insert(str2); swap(str2[i],str2[j]); } } if(set.find(str1)!=set.end()) return true; return false; } int countPairs(vector<int>& nums) { int n = nums.size(); int ans = 0; for(int i=0;i<n;i++){ for(int j=i+1;j<n;j++){ if(swapit(i,j,nums)) ans++; } } return ans; } }; CHECK THIS OUT

bhai tu hi CF and CC contests ka videos bana...bahut acccha seekhatha hai :)

Sir, in the code from line 28-34 how are we ensuring that we are multiplying the element which has least index among all the similar numbers(if there are) ?

What's the time complexity for D problem??

@viraj please check my code , adding trailing zeroesa nd comparing class Solution { public: int countPairs(vector<int>& nums) { int n=nums.size(); int ans=0; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { string a=to_string(nums[i]); string b=to_string(nums[j]); if(a.size()>b.size()) { int ex=a.size()-b.size(); string z=""; for(int k=0;k<ex;k++) z+='0'; b=z+b; } else if(b.size()>a.size()) { int ex=b.size()-a.size(); string z=""; for(int k=0;k<ex;k++) z+='0'; a=z+a; } int ml=a.size(); int l=0; int c=0; unordered_map<char,int>m1,m2; while(l<ml ) { if(a[l]!=b[l]) c++; m1[a[l]]++; m2[b[l]]++; l++; } if((c==2 || c==0) && (m1==m2)) ans++; } } return ans; } };

good explanation of time complexity

this is how I solved problem C in O(N) #include<iostream> #include<vector> #include<algorithm> #include<cmath> #include<set> #include<map> #include<climits> #include<string> #include<queue> #define ll long long int using namespace std; int main(){ ios_base::sync_with_stdio(false); cin.tie(NULL); int t=1; cin>>t; while(t--){ ll n,k; cin>>n>>k; vector<ll>v(n); ll check=LLONG_MIN; for(int i=0;i<n;i++){ cin>>v[i]; check=max(check,v[i]); } //sort(v.begin(),v.end()); //ll check=v[n-1]; ll maxi=-1; for(int i=0;i<n;i++){ ll n=check-v[i]; ll q=n/k; if(q%2==1){ maxi=max(maxi,k-(n%k)); } } if(maxi==-1){ cout<<check<<endl; continue; } check=check+maxi; maxi=-1; for(int i=0;i<n;i++){ ll n=check-v[i]; ll q=n/k; if(q%2==1){ maxi=max(maxi,k-(n%k)); } } if(maxi==-1){ cout<<check<<endl; continue; } cout<<-1<<endl; } }

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Runtime error on test 8 problem c

Plz combine all levels

at 46:40 88%3 == 1

I thought of the same approach for D but could not implement it

In Problem D, the time complexity for doing this goes same that is log(n) to insert each element in the multiset. Then what's the reason behind for this approach working?