Відео

Explanation is Fire

as i am a beginner ,,i should start with Div 3 or Div 4 section na??

Helpful

expl for 2nd was mind-blowing, just 2-3 mins into listening and got the complete idea

Informative Sir

how can I join this meet

Understood

sir If input is [0,5,9] and d=2 then if we check for mid=1 according to code it is possible that 1 can be the difference and its also true that we will get difference larger than 1 as well but sir in reality there cannot be two number chosen that can give difference=1 sir please explain this why we are marking it as possible value

Bhaiya, For Q4, Can you please explain why turn is not used for memoization ? How how it is redundant ?

waiting for ur video from morning bro

Join Whatsapp Channel: whatsapp.com/channel/0029Va55Idf1NCrU16kHOt3j

great explanation for D

In 1 to 7 only 25 substring is valid but according to your code pref [7]-pref[1] =28 Is it wrong or am I

thanks

I think the last question was similar to longest palindromic substring

sir aapki explanation itni acch hei !!! Genuine fan of ur explanation

class Solution { public: int findMultiplier(int health, int power){ int m = health/power; if(health%power > 0) m += 1; return m; } long long minDamage(int power, vector<int>& damage, vector<int>& health) { int n = damage.size(); vector<pair<double, int>> nums; long long total_damage = accumulate(damage.begin(), damage.end(), 0); long long min_damage_to_bob = 0; for(int i=0; i<n; i++){ int m = findMultiplier(health[i], power); double priority = double(m) / damage[i]; // this is efficiency and min priority is best // eg: ki m = 2 and damage[i] = 4 to humko m=2 round lene pdenge or humko calc krna hai ki // ek round me kitna damage kr rha hai (means damage per round) and humko min dam consider krna h // jo ek round me km damage kr rha hoga humko uske trf jana hoga nums.push_back({priority, i}); } sort(nums.begin(), nums.end()); for(auto it:nums){ int index = it.second; int multiplier = findMultiplier(health[index], power); min_damage_to_bob += multiplier * total_damage; total_damage -= damage[index]; } return min_damage_to_bob; } };

Madar... D question mai equation kaisi nikali... Wo tera baap samjayega...>?

How can i join you on meet while upsolving

Take it as a constructive criticism bro, but you need to explain & breakdown the solution as well as intuition behind the solution in a better fashion, just pasting the code an saying this is this, this is that especially for the ones where upsolving is required Q3s and Q4s.

Mast video Abhinav Bhai ❤❤ Khair mujhe nahi pta yeh sah kya hai par comment kar diya m😅😅😅😅

Q1 Biweekly code looks so unreadable, this is how I wrote it in contest ```class Solution { public: int generateKey(int num1, int num2, int num3) { int ans=0; int k=4; int p=0; while(k--){ int a=num1%10; int b=num2%10; int c=num3%10; num1/=10; num2/=10; num3/=10; int mn=min(min(a,b),c); ans+=mn*pow(10,p); p++; } return ans; } };``` stick to fundamentals.

Hii Sir, just puchna tha aapka koi course yaa phir sara youtube pe kyunki mera bhi logic hritu and kusharga jesa tha tle dia tha

Bhot acha explain kiya bhaiya apne Loved how you approached Biweekly Q3.

mastt hai

Bhai div 3 codeforces round 970 please 😢😢😢

great work bhaiya

👋👍👍

best for contest upsolving 💯💯

Nice

Waiting for it 😅

the trick to get past the TLE error is awesome

Amazing explanation

for making all possible variation of string we can use concept of generate permutation with a baseCase of -> if(swapCnt>=2 || index>=nums.size()){ store string } then it would work same but we have clear code.

3rd qs badiya samjhaya bro

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; } };

Hey can any one tell why this case only passed 570 test cases, q2 class Solution { public: int length(int num){ if(num == 0) return 1; int len = 0; while(num){ len++; num /= 10; } return len; } int reverseDigits(int num) { int rev_num = 0; while (num > 0) { rev_num = rev_num * 10 + num % 10; num = num / 10; } return rev_num; } bool check (int num1, int num2){ if(num1 == num2) return true; int temp1 = num1, temp2 = num2; int cnt = 0; if(length(num1) <=2 && length(num2) <=2 ) { if(num1 == reverseDigits(num2) || num2 == reverseDigits(num1) ) return true; else return false; } while(temp1 > 0 && temp2 > 0){ int a = temp1 % 10; int b = temp2 % 10; if(a != b) cnt++; temp1 /= 10; temp2 /= 10; } while (temp1 > 0 || temp2 > 0) { cnt++; temp1 /= 10; temp2 /= 10; } if(cnt != 2) return false; return true; } int countPairs(vector<int>& nums) { int cnt = 0; for(int i = 0;i<nums.size();++i){ for(int j = i+1; j<nums.size();++j){ // if(i == j) continue; if(check(nums[i], nums[j])) cnt++; } } return cnt; } };

i guess if you can explain in Hindi only then your explanation will be better then what it is now

any java people who can provide the solution of 3 question, I tried to implement this approach but its failing on 519 case ``` java class Solution { public int[] getFinalState(int[] nums, int k, int multiplier) { if(multiplier==1)return nums; PriorityQueue<int[]> minHeap=new PriorityQueue<>((a, b)-> { if(a[0]!=b[0])return (a[0]-b[0] < 0 ? -1:1); return (a[1] - b[1]); }); int mod=1_000_000_007; int n=nums.length; int max=Integer.MIN_VALUE; for(int i=0;i<n;i++){ minHeap.offer(new int[]{nums[i], i}); max=Math.max(max, nums[i]); } while(k > 0 && (1L*multiplier*minHeap.peek()[0] <= max)){ int[] pair=minHeap.poll(); k--; minHeap.offer(new int[]{(int)((1L*pair[0]*multiplier)%mod), pair[1]}); } long pow=modPower(multiplier, k/n, mod); while(!minHeap.isEmpty()){ int[] pair=minHeap.poll(); int val=pair[0]; int index=pair[1]; if(k%n!=0){ k--; nums[index]=(int)((1L*multiplier%mod*pow%mod*val%mod)%mod); } else{ nums[index]=(int)((1L*pow%mod*val%mod)%mod); } } return nums; } public long modPower(int base, int exp, int mod){ long result=1; while(exp > 0){ if((exp&1)==1){ result=(result*base)%mod; } base=(base*base)%mod; exp>>=1; } return result; } }```

excellent explanation with quality code

(3&4) both are easy after seeing your video sir, I don't know why i am not thing of that...😢😢

Last question ka solution aisa lga jaise tle eliminator se copy paste kr diya ho 😂. But loved it

very good explanation bro,specially 3rd question

bhai 3rd question itna dhaansu smjhaya, maza aa gya thank you for the amazing explanation

class Solution { public: long long powermod(long long x, long long y, long long p){ long long res = 1; if (x == 0){return 0;} while (y > 0){ if (y & 1) res = (res * x) % p; y = y >> 1; x = (x * x) % p; } return res; } void print (priority_queue<long long, vector<long long >, greater<long long>> heap){ for (int i=0;i<5;i++) {cout<<heap.top()<<" "; heap.pop();}cout<<endl; } vector<int> getFinalState(vector<int>& nums, int k, int multiplier) { priority_queue<long long, vector<long long>, greater<long long>> heap; for (auto p : nums) { heap.push(p); } long long mx = *max_element(nums.begin(),nums.end()); while (k && 1LL*multiplier*heap.top()<=mx){ auto it = find(nums.begin(), nums.end(), heap.top()); *it = (heap.top() * multiplier)%1000000007 ; long long top = heap.top(); heap.pop(); heap.push((top * multiplier)%1000000007 ); k--; } long long turn = k/nums.size(); long long remain = k % nums.size(); cout<<turn<<" "<<remain<<endl; while(!heap.empty()){ heap.pop(); } for (long long i = 0; i < nums.size(); i++) { // auto it = find(nums.begin(), nums.end(), heap.top()); // *it = (heap.top() * (powermod(multiplier,turn,1000000007))) % 1000000007; // long long top = heap.top(); // heap.pop(); // heap.push((top * (powermod(multiplier,turn,1000000007)))%1000000007); nums[i] = (nums[i]%1000000007* (powermod(multiplier,turn,1000000007)))%1000000007; heap.push(nums[i]); } if (remain>0){ for (long long i = 0; i < remain; i++) { long long top = heap.top(); auto it = find(nums.begin(), nums.end(), top); *it = (top * multiplier) % 1000000007; heap.pop(); heap.push((top * multiplier) % 1000000007); }} // [66307295,441787703,589039035,322281864] // 900900704 // 641725 return nums; } }; nums = [889458628,338743558,875422936,684907163,233489834] k = 246181588 multiplier = 313380 stucked at test case 522 i had little bit diff sol please chek it

amazing

My question is for the 4th one. for every digit we are checking after 0th, 1st, 2nd swap they are previously present in the map or not and before that we are storing them in set(all unique ones). But here we are checking 0th swap on i with 0th,1st , 2nd swap on j element with the help of map. But what about 1st swap on i with 0th,1st,2nd(not 2nd because we are swaping ith element one time) swap on j....and like that 1|0 combination and 2|0 combination and 1|1 combination. I hope my questions are clear to you. How are you considering those conditions

Nice 😊

Great explanation of Q3. Watched many videos, but was unable to understand. Finally understood from ur video.

in first while loop in 3rd question you are not poping the top element why this will not give tle??