Видео

bhaiya please comple heaps playlist as soon as possible ....please please please big bro

Good Job dude. Keep it up

can you made a separate video for this type of competitive programming concept ( like 1LL , 1e9+7 , ......)

pls could u help me in java code .. some test cases are failing ... class Solution { int Maximize(int arr[]) { Arrays.sort(arr); long ans=0; for(int i=0;i<arr.length;i++){ ans+=(long)arr[i]*i; ans%=100000007; } return (int)ans; } }

i have a doubt why have you used 1LL?

Great video brother !

😊 we are so excited 🥳🥳

love ya ❤❤❤❤❤❤

Woohoo🥳🥳

Nice ❤

My solution is also the same but I used vector instead of unordered_map since the latter can take O(n) time complexity in the worst case scenario

Thank you Mask-man, you are back!!

👍👍👍

Thank you bro Helping me out in building consistency

GFG POTD 25/07/2024 Node *buildBSTFromInorder(const vector<int> &inorder, int start, int end) { if (start > end) { return nullptr; } int mid = (start + end) / 2; Node *node = new Node(inorder[mid]); node->left = buildBSTFromInorder(inorder, start, mid - 1); node->right = buildBSTFromInorder(inorder, mid + 1, end); return node; } class Solution { public: Node* sortedArrayToBST(vector<int>& nums) { // Code here Node *root = buildBSTFromInorder(nums, 0, nums.size() - 1); return root; } };

My solution is also same except I declared prev as class attribute and used same function. Code: class Solution { int prev = -1; public: bool isBST(Node* root) { if(!root) return true; bool left = isBST(root->left); if(!left) return false; if(root->data <= prev) return false; prev = root->data; bool right = isBST(root->right); return left && right; } };

Can we do this without taking extra space to store the inorder traversal of root1 and root2?

missed your videos!!!

Bro why 3rd test case fails when submitted??

It's good to see you back bro. Hope you are doing good. By the way here is my solution: long long int findMaxProduct(vector<int>& arr) { // Write your code here int cnt=0,maxNeg=INT_MIN; long long ans = 1,mod = 1e9+7; for(int it:arr){ if(it < 0){ cnt++; maxNeg = max(maxNeg,it); } if(it != 0) ans = (ans*it)%mod; } if(arr.size() > 1 && (cnt & 1)) ans = ans/maxNeg; return ans; }

you have linked in account ? if yes give me link

I have been waiting for many long days ... finally it started again ..

Bhai yr video q nai bana rhe ho itna din sa??

when will you upload graphs , heaps , dp etc ? your videos are great and much appreciated. Thank you big brother

Great, keep doing ❤

Brow all okay? Hope you are doing well..as you have not posted from last 3 days

thanks, one suggestion: please like use a better color to highlight the thing you're talking about, it was hard to know which 0 you were talking about in the matrix

Nice approach.

it is not that easy as it tag ...

u explain well but handwriting can be improved ☺☺

My other solution using regex: a,b,c,d = re.findall(r'[0-9]+',str) p,q = int(a),int(b) r,s = int(c),int(d) if p/q > r/s: return a+'/'+b elif r/s > p/q: return c+'/'+d return "equal"

My python code: def compareFrac (self, str): a,b = str.split(",") b = b.strip() p,q = map(int,a.split("/")) r,s = map(int,b.split("/")) if p/q > r/s: return a elif r/s > p/q: return b return "equal" I think the time complexity will remain the same as split() takes O(n)

Bro i did this without using any loop😂🤣 constant time complexity

If this the case that is Str="abbccc" X=1,y=2 X+y=3 odd Is this crt??

❤😊

sir jo aapne first brute force approach bataya hai uska TC O(n) hoga na ki O(n^2)?

bro formula for volume u calculated in here is different from which u implemented in the code , in the code u implemented volume=l^2 *(p/4)-(2*l); which i could n't find in here 5:40

Bhai how can we become like you😢

Here is my code which is easy to Understand: double l = (p - sqrt((p * p) - (24 * a))) / 12; double ans = (l * l) * ((p / 4) - (2 * l)); return ans;

Bc problem toh chodo muze basic maths nahi samaz raha ab 🥲

Bro pls help it is getting stuck at 1110 test case I dont know whats the problem... // User function Template for Java class Solution { double maxVolume(double perimeter, double area) { // code here double h=(perimeter/2-Math.sqrt(perimeter*perimeter/4-6*area))/6; double v=h*area/2-h*h*perimeter/4+h*h*h; return v; } }

ans at 110 coming 1131004.73 but my output comes out to be 1131004.67

Thanks for the amazing explanation

also tried to code based on given volume formula but it was failing for big test case your code logic seems different ...

why are you caculating twice perimeter - (sqrt(pow(perimeter, 2) - (24 * area)))) / 12, Please try to upload cleaner code

yeh math ka question tha ya coding ka