Code Engineer

/* Stickler the thief wants to loot money from a society having n houses in a single line.  He is a weird person and follows a certain rule when looting the houses.  According to the rule, he will never loot two consecutive houses. At the same time,   he wants to maximize the amount he loots. The thief knows which house has what amount of money   but is unable to come up with an optimal looting strategy. He asks for your help to find the maximum money   he can get if he strictly follows the rule. Each house has a[i] amount of money present in it. */ #include<bits/stdc++.h> // this is clearly an consec include exclude problem using namespace std; int maxsumnotconsec(int a[], int n) {     int incl=a[0];     int exec=0;     for(int i=1;i<n;i++)     {         int inc_new=exec+a[i];         // we can incude the new element(latest ele) if and only if we have ex...

#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; } } }

// what if a product based  company asks that they want the minimum of stack in resonalble time.. // if you knew this program congrats... you got job in product based company #include<iostream> #include<stack> using namespace std; void push(int a); bool isFull(int n); bool isEmpty(); int pop(); int getMin(); stack<int> s; int main(){ int t; cin>>t; while(t--){ int n,a; cin>>n; while(!isEmpty()){     pop(); } while(!isFull(n)){ cin>>a; push(a); } cout<<getMin()<<endl; } } /*This is a function problem.You only need to complete the function given below*/ /*Complete the function(s) below*/ void push(int a) {      s.push(a); } bool isFull(int n) {     return (s.size()==n); } bool isEmpty() {return s.empty();     } int pop() {     int a=s.top();     s.pop();     return a ;     } in...

/ * 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();         } ...