> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/hashmap-and-hashset-and-sliding-window/check-arithmetic-progression.md).

# Check Arithmetic Progression

Given an array of **N** integers. Write a program to check whether an arithmetic progression can be formed using all the given elements. If possible print “YES”, else print “NO”.\
\
**Input:**\
First line of input contains a single integer T which denotes the number of test cases. Then T test cases follows. First line of each test case contains a single integer N which denotes number of elements in the array. Second line of each test case contains N space separated integers which denotes elements of the array.\
\
**Output:**\
For each test case, print "YES" without quotes if an arithmetic progression can be formed using all the given elements, else print "NO" without quotes.\
\
**Constraints:**\
1<=T<=100\
1<=N<=105\
1<=Arr\[i]<=105\
\
**Example:**\
**Input:**\
2\
4\
0 12 4 8\
4\
12 40 11 20\
**Output:**\
YES\
NO\
&#x20;

```java
class Solution {
    // O(NlogN) Time & O(1) Space complexity
    public static boolean isAP(int[] arr, int n) {
        if (n <= 2)
            return true;
        Arrays.sort(arr);
        int d = arr[1] - arr[0];
        for (int i = 2; i < n; i++) {
            if (arr[i] - arr[i - 1] != d)
                return false;
        }
        return true;
    }

    // O(N) Time & O(N) Space Complexity
    public static boolean isAP(int[] arr, int n) {
        if (n <= 2)
            return true;
        Set<Integer> set = new HashSet<>();
        int smallest = Integer.MAX_VALUE, smallest2 = Integer.MAX_VALUE;
        for (int x : arr) {
            if (x < smallest) {
                smallest2 = smallest;
                smallest = x;
            } else if (x < smallest2)
                smallest2 = x;
            if (set.contains(x))
                return false;
            set.add(x);
        }
        int diff = smallest2 - smallest;
        int current = smallest, count = 0;
        while (set.contains(current)) {
            current = current + diff;
            count++;
        }
        return count == n;
    }
}
```