# Find pair with given difference

We have an unsorted array and a number n, Our task is to find if there exists a pair of elements in the array whose difference is n.

##### Example 1

Input: arr[] = {5, 20, 3, 4, 50, 40}, n = 36

Output: Pair Found: (4, 40)

##### Example 2

Input: arr[] = {90, 10, 20, 12, 50}, n = 45

Output: No Such Pair

#### Approach

- Create a HashSet called elementsSet to store elements from the array.
- Iterate through each element in the input array.
- For each element in the array, calculate the potential partner element by adding n to the current element.
- Check if the potential partner element exists in the elementsSet by using the contains method. If it does, return true because you've found a pair with the desired difference.
- If the potential partner element is not found in the elementsSet, add the current element to the elementsSet. This is done to keep track of elements you've seen so far.
- Continue this process for all elements in the array.
- If you reach the end of the array without finding a pair with the desired difference, return false.

#### Complexity

- Time Complexity: O(N), where N is the number of elements in the array.
- Space Complexity: O(N), where N is the number of elements in the array.

#### Code

```
#Java Code
import java.util.HashSet;
public class PairWithDifference {
public static boolean hasPairWithDifference(int[] arr, int n) {
if (arr == null || arr.length < 2) {
// There must be at least two elements in the array to form a pair.
return false;
}
HashSet<Integer> elementsSet = new HashSet<>();
for (int element : arr) {
int partner1 = element + n;
int partner2 = element - n;
if (elementsSet.contains(partner1)) {
System.out.println("Pair found: (" + element + ", " + partner1 + ")");
return true;
} else if (elementsSet.contains(partner2)) {
System.out.println("Pair found: (" + element + ", " + partner2 + ")");
return true;
}
elementsSet.add(element);
}
return false;
}
public static void main(String[] args) {
int[] arr = {5, 20, 3, 2, 50, 80};
int n = 78;
if (hasPairWithDifference(arr, n)) {
System.out.println("A pair with a difference of " + n + " exists.");
} else {
System.out.println("No pair with a difference of " + n + " found.");
}
}
}
```

##### Output

Pair found: (2, 80)

A pair with a difference of 78 exists.

Thanks for feedback.