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Added solutions of Different DS and Algos asked in Coding Round of Comapanies #12

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23 changes: 23 additions & 0 deletions divide and conquer/Modular Exponentation.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
#include<bits/stdc++.h>
using namespace std;
int main() {
long long int t;cin>>t;
while(t--)
{
long long int x,y,p,res=1;
cin>>x>>y>>p;
if(x==0)
cout<<"0\n";
else
{
while(y)
{
if(y%2!=0)
res=(res*x)%p;
y=y/2;
x=(x*x)%p;
}
cout<<res<<"\n";
}
}
}
49 changes: 49 additions & 0 deletions dynamic programming/LastDigit of Nth Fibonacci.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,49 @@

/* Program to print last digit of nth fibonacci
Input:
N = 14
Output:
7
Explanation:
14th Fibonacci number is 377
It's last digit is 7

*/

#include<bits/stdc++.h>
using namespace std;

class Solution{
public:
int fib(int n){
if(n==0)
return 0;
else if(n==1 || n==2)
return 1;
else
{
int a[n];
a[0]=1;
a[1]=1;
for(int i=2;i<n;i++)
a[i]=(a[i-1]+a[i-2])%10;
return a[n-1];
}

}
};


int main()
{
int t;
cin>>t;
while(t--)
{
int N;
cin>>N;
Solution ob;
cout << ob.fib(N) << endl;
}
return 0;
}
74 changes: 74 additions & 0 deletions stack/paranthesisChecker.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,74 @@
/* The famous problem of checking Paranthesis is 'Balanced' or 'Not Balanced' using stacks.
For example, the program should print 'balanced' for exp = �[()]{}{[()()]()}�
and 'not balanced' for exp = �[(])�
*/

#include<bits/stdc++.h>
using namespace std;
int main() {
int t;cin>>t;
while(t--)
{
string n;
cin>>n;
stack<int>s;
int flag=0;
for(int i=0;i<n.length();i++)
{
/* While traversing the string if found symbols like '(', '{' and '[' push them in stack*/
if(n[i]=='(' || n[i]=='{' || n[i]=='[')
s.push(n[i]);
else
{
/* If string starts with ')', '}, or ']', then the string can never be balances, so raise the flag*/
if(s.empty())
{
flag=1;
break;
}
else if(n[i]==')')
{
/* If string starts with ')' then the top of the stack must contain '(' for the string to be valid.
Pop the top of stack.
*/
int t=s.top();
s.pop();
if(t!='(')
{
flag=1;break;
}
}
else if(n[i]=='}')
{
/* If string starts with '}' then the top of the stack must contain '(' for the string to be valid.
Pop the top of stack.
*/
int t=s.top();
s.pop();
if(t!='{')
{
flag=1;break;
}
}
else if(n[i]==']')
{
/* If string starts with ']' then the top of the stack must contain '(' for the string to be valid.
Pop the top of stack.
*/
int t=s.top();
s.pop();
if(t!='[')
{
flag=1;break;
}
}
}
}
/* If at last the stack is empty and the flag value is also 1, then it is balanced else unbalanced.
*/
if(flag==0 && s.empty())
cout<<"balanced\n";
else
cout<<"not balanced\n";
}
}