Bro how to apply data analyst job for freshers ?? when openings be for data anayst role in amazon?? pls tell me bro

My is for 3 months in phonepay will I get PPO they said we can't say right now

JAVA CODE package String_StringBuilder_ArrayList.LoveBabbarDsaSheet; import java.io.*; import java.util.*; //Count and say problem public class q7 { public static void main(String[]args){ Scanner sc= new Scanner(System.in); String str= sc.nextLine(); System.out.println(CountandSay(str)); } public static String CountandSay(String str){ StringBuilder sb = new StringBuilder(); int count=1; for(int i=1; i<str.length();i++){ char curr = str.charAt(i); char prev = str.charAt(i-1); if(curr==prev){ count++; } else{ if(count>1){ sb.append(count); sb.append(prev); count =1; } else { sb.append(prev); count = 1; } } } sb.append(count); sb.append(str.charAt(str.length() - 1)); return sb.toString(); } }

JAVA code:- package String_StringBuilder_ArrayList.LoveBabbarDsaSheet; import java.util.*; //WAP to Check if Strings Are Rotations of Each Other public class q5 { public static void main(String[]args){ Scanner sc = new Scanner(System.in); System.out.println("Enter String 1: "); String str1 = sc.nextLine(); System.out.println("Enter String 2: "); String str2 = sc.nextLine(); System.out.println(rotationCheck(str1, str2)); } public static boolean rotationCheck(String str1, String str2){ if (str1.length() != str2.length()) { return false; } String concatenated = str1 + str1; if(concatenated.contains(str2)){ return true; } return false; } }

bhai muje tumhara content babber or straiver se bhi jaada accha lagta hai. seedhi baat , no bakwaas.

I think we can keep track of first non repeating character and update it accordingly. By this we can solve in O(n)

just learn the code and record it on youtube

corner cases is not covered,

i have written same code but it says time limit exceeded void inorder(TreeNode<int>* root, TreeNode<int>* &prev){ if(!root) return; inorder(root->left, prev); prev->right=root; prev->left=NULL; prev=root; inorder(root->right,prev); } TreeNode<int>* flatten(TreeNode<int>* root) { // Write your code here TreeNode<int>*temp=new TreeNode<int>(-1); TreeNode<int>*prev=temp; inorder(root,prev); prev->right=NULL; prev->left=NULL; TreeNode<int>*newRoot=temp->right; temp->left=NULL; temp->right=NULL; delete temp; return newRoot; }

Did you take interview preparation course or Competitive programming course of Coding blocks

bhai tum na gym gaye na tumne coffee pii hai, natak kyu ye samjhe nhi aaya bus

mast samjhaya h bhai!

nice explaination

I have a question

Thank you🙏😊

Sir can you share all the resources plz It will help me so much

I received assignments mail, please help me

yr koi tabulation bata hi ni r h

zeotap m apply kiya tha, unlogo neh assignment bheja Rule engine with ast bnane, can anybody help with this project ya kisine yeh project kiya ho phle ?

Correct code: class Solution { public: /*This function returns true if the tree contains a duplicate subtree of size 2 or more else returns false*/ string solver(Node *root, unordered_map<string, int> &mp) { // base case if (root == NULL) { return "$"; } // for the leaf node if (root->left == NULL && root->right == NULL) { // since the leaf nodes can be same but it would not be >= 2 so we are not adding it to the map return to_string(root->data) + "$"; } string str = ""; // if not a leaf node then get the string and go left and right str += to_string(root->data) + "$"; str += solver(root->left, mp); str += solver(root->right, mp); // finally while backtracking add the string with it's freq in the map mp[str]++; return str; } int dupSub(Node *root) { // creating a map for storing the string and the freq unordered_map<string, int> mp; solver(root, mp); for (auto it : mp) { if (it.second >= 2) { return 1; } } return 0; } };

Asa lgta hai ap he confuse hore hore ho 😢

Can u help me in my assignment

Given 80 hour workweeks that an hourly equivalent to 60000 pm for a 40 hour job 😂

bakwaaaas explanation

focus more on dry run than the code , that would make the learning more easy!!

Thanks vro ham bangali hau

optimized 8:51

Easy Approach Node* addOne(Node* head) { Node* newhead= reverse(head); Node* curr=newhead; while(curr!=NULL){ if(curr->data<9){ curr->data+=1; break; } curr->data=0; if(curr->next==nullptr){ Node* temp=new Node(1); curr->next=temp; break; } curr=curr->next; } return reverse(newhead); } };

You are a coding prodigy 🎉

NYC sir

Have you stored these codes in the githu??/

Great!

ans will be 0 only when the arr[i]==x then why we need to move for even one step??

best explanation on this question over all the youtube!thanks alot ,that explanation through table was too good.

you drink so little litter. Water costs me 70 for 5 liters in Bangalore, Drink more water

u are great.....

How can I gte that sheet

Finally getting a approach which is more intutive.

lovely............

At first I was confused, but I am clear after 2nd explanation

Best explanation

Bhai tum samjhate Aacha ho bs samjhaya mt kro

@CodeLibrary can you explain when we take array as 1, 2, 3, 2, 4 the output for this is coming 3 5 how

Criteria Kaya hai just pay ka

jhantu explanation

Thoda pani pe leta hu, bohot pyaas lagi hai🤡

Bhai mtt bna videos bht bekaar explanation h.

I watched it 2-3 times before I fully understood the approach. I'm tired.

sale "Shorted" nai "sorted" hota hai

I not able to understand your language but i understand your code and logic thank you😊