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# C++ Program To Find The Union And Intersection Of Two Sorted Array In Increasing Order.

## Find The Union And Intersection Of Two Sorted Array In Increasing Order.

Union of arrays arr1[] and arr2[]

To find union of two sorted arrays, follow the following merge procedure :

1) Use two index variables i and j, initial values i = 0, j = 0
2) If arr1[i] is smaller than arr2[j] then print arr1[i] and increment i.
3) If arr1[i] is greater than arr2[j] then print arr2[j] and increment j.
4) If both are same then print any of them and increment both i and j.
5) Print remaining elements of the larger array.

Below is the implementation of the above approach :

``````// C++ program to find union of
// two sorted arrays
#include <bits/stdc++.h>
using namespace std;

/* Function prints union of arr1[] and arr2[]
m is the number of elements in arr1[]
n is the number of elements in arr2[] */
void printUnion(int arr1[], int arr2[], int m, int n)
{
int i = 0, j = 0;
while (i < m && j < n) {
if (arr1[i] < arr2[j])
cout << arr1[i++] << " ";

else if (arr2[j] < arr1[i])
cout << arr2[j++] << " ";

else {
cout << arr2[j++] << " ";
i++;
}
}

/* Print remaining elements of the larger array */
while (i < m)
cout << arr1[i++] << " ";

while (j < n)
cout << arr2[j++] << " ";
}

/* Driver program to test above function */
int main()
{
int arr1[] = { 1, 2, 4, 5, 6 };
int arr2[] = { 2, 3, 5, 7 };

int m = sizeof(arr1) / sizeof(arr1);
int n = sizeof(arr2) / sizeof(arr2);

// Function calling
printUnion(arr1, arr2, m, n);

return 0;
}
``````

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Time Complexity : O(m + n)

Handling duplicates in any of the array : Above code does not handle duplicates in any of the array. To handle the duplicates, just check for every element whether adjacent elements are equal.

Below is the implementation of this approach.

``````// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

static void UnionArray(int arr1[],
int arr2[], int l1, int l2)
{
// Taking max element present in either array
int m = arr1[l1 - 1];
int n = arr2[l2 - 1];

int ans = 0;

if (m > n) {
ans = m;
}
else
ans = n;

// Finding elements from 1st array
// (non duplicates only). Using
// another array for storing union
// elements of both arrays
// Assuming max element present
// in array is not more than 10^7
int newtable[ans + 1];
memset(newtable,0,sizeof(newtable));
// First element is always
cout << arr1 << " ";

// Incrementing the First element's count
// in it's corresponding index in newtable
++newtable[arr1];

// Starting traversing the first
// array from 1st index till last
for (int i = 1; i < l1; i++) {
// Checking whether current element
// is not equal to it's previous element
if (arr1[i] != arr1[i - 1]) {
cout << arr1[i] << " ";
++newtable[arr1[i]];
}
}

// Finding only non common
// elements from 2nd array
for (int j = 0; j < l2; j++) {
// By checking whether it's already
// present in newtable or not
if (newtable[arr2[j]] == 0) {
cout << arr2[j] << " ";
++newtable[arr2[j]];
}
}
}

// Driver Code
int main()
{
int arr1[] = { 1, 2, 2, 2, 3 };
int arr2[] = { 2, 3, 4, 5 };
int n = sizeof(arr1) / sizeof(arr1);
int m = sizeof(arr2) / sizeof(arr2);

UnionArray(arr1, arr2, n, m);

return 0;
}

// This code is contributed by splevel62.
``````

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## Conclusion

Here, we have seen how to implement a C++ program to Find The Union And Intersection Of Two Sorted Array In Increasing Order.