# AP Free Sequences solution codechef

## AP Free Sequences solution codechef

In a recent breakthrough, sequences which do not have Arithmetic Progressions (APs) of size 3, were studied. Let us try a simple problem related to that.

Consider a sequence �1≤�2≤…≤��. Three elements from it, say ��,��,��, where �<�<�, are said to form an AP, if (��−��)=(��−��). Call the sequence AP-Free, if no such three elements can be found.

Your goal is to find out whether the given sequence is AP-Free or not.

## AP Free Sequences solution codechef

• The first line of input will contain a single integer , denoting the number of test cases.
• Each test case consists of two lines of input.
• The first line of each test case contains a single integer  — the size of the sequence.
• The next line contains  integers – �1,�2,…,��, which is the sequence.

### Output Format

For each test case, output on a new line, “YES”, if the sequence is AP-Free. Output “NO” if it is not.

You may print each character in uppercase or lowercase. For example, YesyesYES, and yES are all considered identical.

• 1≤�≤10
• 3≤�≤100
• 1≤��≤1000000
• �1≤�2≤…≤��

### Sample 1:

Input

Output

4
3
1 2 3
3
1 2 4
3
10 25 40
4
1 10 11 21

No
Yes
No
No

### Explanation:

Testcase 1: The given sequence is (1,2,3). Here, you can select all the three elements, and they form an AP, since (2−1)=(3−2)=1. Hence, this sequence is not AP-Free.

Testcase 2: The given sequence is (1,2,4). No three elements form an AP. Hence, this sequence is AP-Free.

Testcase 3: The given sequence is (10,25,40). Here, you can select all the three elements, and they form an AP, since (25−10)=(40−25)=15. Hence, this sequence is not AP-Free.

Testcase 4: The given sequence is (1,10,11,12). Here, you can select (�1,�3,�4), and they form an AP, since (11−1)=(21−11)=10. Hence, this sequence is not AP-Free.