For example, these are arithmetic sequences:

1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9

The following sequence is not arithmetic:

1, 1, 2, 5, 7

You are given an array of n integers, nums, and two arrays of m integers each, l and r, representing the m range queries, where the ith query is the range [l[i], r[i]]. All the arrays are 0-indexed.

Return a list of boolean elements answer, where answer[i] is true if the subarray nums[l[i]], nums[l[i]+1], ... , nums[r[i]] can be rearranged to form an arithmetic sequence, and false otherwise.

Example 1:

Input: nums = [4,6,5,9,3,7], l = [0,0,2], r = [2,3,5]
Output: [true,false,true]
Explanation:
In the 0th query, the subarray is [4,6,5]. This can be rearranged as [6,5,4], which is an arithmetic sequence.
In the 1st query, the subarray is [4,6,5,9]. This cannot be rearranged as an arithmetic sequence.
In the 2nd query, the subarray is [5,9,3,7]. This can be rearranged as [3,5,7,9], which is an arithmetic sequence.

Example 2:

Input: nums = [-12,-9,-3,-12,-6,15,20,-25,-20,-15,-10], l = [0,1,6,4,8,7], r = [4,4,9,7,9,10]
Output: [false,true,false,false,true,true]

Constraints:

• n == nums.length
• m == l.length
• m == r.length
• 2 <= n <= 500
• 1 <= m <= 500
• 0 <= l[i] < r[i] < n
• -105 <= nums[i] <= 105

Solution: Brute Force

Sort the range of each query and check.

Time complexity: O(nlogn * m)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) {
    vector<bool> ans(l.size(), true);
    for (int i = 0; i < l.size(); ++i) {
      vector<int> arr(begin(nums) + l[i], begin(nums) + r[i] + 1);
      sort(begin(arr), end(arr));
      const int d = arr[1] - arr[0];
      for (int j = 2; j < arr.size() && ans[i]; ++j)
        ans[i] = ans[i] && (arr[j] - arr[j - 1] == d);
    }
    return ans;
  }
};

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