**Input:** arr1[] = [7, 1, 5, 2, 3, 6] , arr2[] = [3, 8, 6, 20, 7]

**Output:** [3, 6, 7]

**# Brute Force**

**Pseudo Code:**

- Initialize intersection array as empty.
- Do following for every element x of the first array.

a. If x is present in second array, then copy x to I.

b. Return Intersection.

**Code:**

```
#include <iostream>
#include <algorithm>
using namespace std;
void intersection(int arr1[], int arr2[], int n1, int n2)
{
int i, j;
for (i = 0; i < n1; i++)
{
for (j = 0; j < n2; j++)
{
if (arr1[i] == arr2[j])
{
cout << arr1[i] << " ";
}
}
}
}
int main()
{
int arr1[] = { 3, 11, 5, 22, 42, 476, 866, 34, 23, 78, 23, 8 };
int arr2[] = { 3, 11, 6, 866, 34 };
int m = sizeof(arr1) / sizeof(arr1[0]);
int n = sizeof(arr2) / sizeof(arr2[0]);
cout << "Intersection of two arrays is n";
intersection(arr1, arr2, m, n);
return 0;
}
```

**Space complexity** : O(1)

**Time complexity** : O(n1*n2), where n1 is the no. of elements of the first array and n2 is no. of elements of the second array.

**# Optimized Approach**

**Pseudo Code:**

- Initialize intersection I as empty.
- Find smaller of m and n and sort the smaller array.
- For every element x of larger array, do following

a. Binary Search x in smaller array. If x is present, then copy it to I.

b. Return I.

**Code:**

```
#include <iostream>
#include <algorithm>
using namespace std;
int binarySearch(int arr[], int l, int r, int x);
void printIntersection(int arr1[], int arr2[], int m, int n)
{
if (m > n)
{
int *tempp = arr1;
arr1 = arr2;
arr2 = tempp;
int temp = m;
m = n;
n = temp;
}
sort(arr1, arr1 + m);
for (int i = 0; i < n; i++)
if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1)
cout << arr2[i] << " ";
}
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l)
{
int mid = l + (r - l) / 2;
if (arr[mid] == x) return mid;
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
return binarySearch(arr, mid + 1, r, x);
}
return -1;
}
int main()
{
int arr1[] = { 3, 11, 5, 22, 42, 476, 866, 34, 23, 78, 23, 8 };
int arr2[] = { 3, 11, 6, 866, 34 };
int m = sizeof(arr1) / sizeof(arr1[0]);
int n = sizeof(arr2) / sizeof(arr2[0]);
cout << "Intersection of two arrays is n";
printIntersection(arr1, arr2, m, n);
return 0;
}
```

**Output:**

`Intersection of two arrays is n 3, 11, 866, 34`

This is how we solve the problem of Intersection of two arrays.

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