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Hello guys..
Here you can find all the important programming questions that i have solved.., if you get any doubt feel free to ask below in comment section :)

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// sieve of Atkin is a popular way to find all the prime numbers in range using the dp approach and is //better than the erronthesis approach // C++ program for implementation of Sieve of Atkin #include <bits/stdc++.h> using namespace std;   int SieveOfAtkin(int limit) {     // 2 and 3 are known to be prime     if (limit > 2)         cout << 2 << " ";     if (limit > 3)         cout << 3 << " ";       // Initialise the sieve array with false values     bool sieve[limit];     for (int i = 0; i < limit; i++)         sieve[i] = false;     for (int x = 1; x * x < limit; x++) {         for (int y = 1; y * y < limit; y++) {                           // Main part of Sieve of Atkin             int n = (4 * x * x) + (y * y);             if (n <= limit && (n % 12 == 1 || n % 12 == 5))                 sieve[n] ^= true;               n = (3 * x * x) + (y * y);             if (n <= limit &
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Linked List 10 operations from which problems are made ( complete guide for you)

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/* These are the functions we will implement make your seatbelts tight and start progrmming cout<<" 1. Insert node at begining"<<endl; cout<<" 2. Append a node"<<endl; cout<<" 3. display the l.l"<<endl; cout<<" 4. Insert before a selected node"<<endl; cout<<" 5. Insert after a selected node "<<endl; cout<<" 6.Delete a particular node"<<endl; cout<<" 7.Reverse A L.L "<<endl; cout<<" 8. Do bot reverse the l.l but display the reversed order (using recursion)"<<endl; cout<<"9. find middle element in a list using slow and fast pointer"<<endl; cout<<"10. check if linked list is palindrome or not (we know if even f=NULL odd f->next=NULL)"<<endl; cout<<" 11. exit"<<endl; */ #include<iostream> using namespace std; struct node{
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Stack - print bracket number

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/ * example 2 (0+1)*0(0+1)*0(0+1)* (1*0)*1* o/p 1 1 2 2 3 3  1 1  Example: Input: 3 (a+(b*c))+(d/e) ((())(())) ((((() Output: 1 2 2 1 3 3 1 2 3 3 2 4 5 5 4 1 1 2 3 4 5 5 */ #include<bits/stdc++.h> using namespace std; int main()  { int t; cin>>t; while(t--) {     string s;     stack <int> st;     int c=0; // we maintain the count of open brackets     cin>>s;     for(int i=0;i<s.length();i++)     {         if(s[i]=='(')         {c++;         cout<<c<<" ";         st.push(c);         }         if(s[i]==')')         {             cout<<st.top()<<" ";             st.pop();         }     }     cout<<endl; } return 0; }
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