by Priyanshu Tiwari

## Table Of Content

An array is a sequential collection of variables of same data type. It stores data elements in a continuous memory location. A string is a sequence of characters.

### Monk and Rotation #

Monk loves to preform different operations on arrays, and so being the principal of Hackerearth School, he assigned a task to his new student Mishki. Mishki will be provided with an integer array A of size N and an integer K , where she needs to rotate the array in the right direction by K steps and then print the resultant array. As she is new to the school, please help her to complete the task.

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;

void solve()
{
int n, k, temp;
cin >> n >> k;
int arr[n];
for (int i = 0; i < n; i++)
{
cin >> arr[i];
}
int x = n - (k % n);
for (int i = x; i < n; i++)
{
cout << arr[i] << " ";
}
for (int i = 0; i < x; i++)
{
cout << arr[i] << " ";
}
}

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
ll test;
cin >> test;
while (test--)
{
solve();
cout << endl;
}
}


### Monk and Inversions #

Monk’s best friend Micro, who happen to be an awesome programmer, got him an integer matrix M of size NxN for his birthday. Monk is taking coding classes from Micro. They have just completed array inversions and Monk was successful in writing a program to count the number of inversions in an array. Now, Micro has asked Monk to find out the number of inversion in the matrix M.

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;

int solve()
{
int n, i, j, ii, jj, ans = 0;
cin >> n;
int arr[n][n];
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
cin >> arr[i][j];
}
}

for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
for (ii = i; ii < n; ii++)
{
for (jj = j; jj < n; jj++)
{
if (arr[ii][jj] < arr[i][j])
ans++;
}
}
}
}

return ans;
}

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
ll test;
cin >> test;
while (test--)
{
cout << solve() << endl;
}
}


### Minimum AND xor OR #

Given an array A of N integers. Find out the minimum value of the following expression for all valid i,j. (Ai and Aj) xor (Ai or Aj), where i != j.

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;

int solve()
{
int n, i, ans = INT_MAX;
cin >> n;
int arr[n];
for (i = 0; i < n; i++)
{
cin >> arr[i];
}

sort(arr, arr + n);

for (i = 0; i < n - 1; i++)
{

ans = min(ans, arr[i] ^ arr[i + 1]);
}

return ans;
}

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
ll test;
cin >> test;
while (test--)
{
cout << solve() << endl;
}
}