A sequence of numbers is called an arithmetic progression if the difference between any two consecutive elements is the same.
Given an array of numbers arr, return trueif the array can be rearranged to form an arithmetic progression. Otherwise, returnfalse.
Example 1:
Input: arr = [3,5,1]
Output: true
Explanation: We can reorder the elements as [1,3,5] or [5,3,1] with differences 2 and -2 respectively, between each consecutive elements.
Example 2:
Input: arr = [1,2,4]
Output: false
Explanation: There is no way to reorder the elements to obtain an arithmetic progression.
Constraints:
2 <= arr.length <= 1000
-106 <= arr[i] <= 106
Solutions
Solution 1: Sorting + Traversal
We can first sort the array arr, then traverse the array, and check whether the difference between adjacent items is equal.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array arr.
We first find the minimum value $a$ and the maximum value $b$ in the array $arr$. If the array $arr$ can be rearranged into an arithmetic sequence, then the common difference $d = \frac{b - a}{n - 1}$ must be an integer.
We can use a hash table to record all elements in the array $arr$, then traverse $i \in [0, n)$, and check whether $a + d \times i$ is in the hash table. If not, it means that the array $arr$ cannot be rearranged into an arithmetic sequence, and we return false. Otherwise, after traversing the array, we return true.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array arr.