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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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Intro to Hashing problems made easy

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/*--- Below are 3 problems i have used hasing technique, After seeing below problems you'll be 100% clear on the basic hasing needed for competative programming First Repeating Element  Given an array arr[] of size N. The task is to find the first repeating element in an array of integers, i.e., an element that occurs more than once and whose index of first occurrence is smallest. Input : The first line contains an integer T denoting the total number of test cases. In each test cases, First line is number of elements in array N and second its values. Output: In each separate line print the index of first repeating element, if there is not any repeating element then print “-1” (without quotes). Use 1 Based Indexing. Constraints: 1 <= T <= 500 1 <= N <= 106 0 <= Ai <= 106 Example: Input: 1 7 1 5 3 4 3 5 6 Output: 2 Explanation: Testcase 1: 5 is appearing twice and its first appearence is at index 2 which is less than 3 whose first occur...
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Stack using Queues

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#include<bits/stdc++.h> using namespace std; class Q { stack<int> s1,s2; public :void enque(int a) { s1.push(a); } public:int deque() { int j=-1; while(s1.size()<=1) { int a= s1.top(); s1.pop(); s2.push(a); } if(s1.size()==1) { j=s1.top(); s1.pop(); } while(!s2.empty()) { int a= s1.top(); s1.pop(); s2.push(a); } return j; } }; int main() { Q q1; int r=0; while(r!=3) { cout<<"1. Enque"<<endl; cout<<"2.Deque"<<endl; cout<<"3. exit"<<endl; cin>>r; if(r==1) {int e; cout<<"Enter an element to insert in queue"; cin>>e; q1.enque(e); } else if(r==2) { cout<<q1.deque()<<endl; } } }
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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();         } ...
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