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1630. Arithmetic Subarrays

A sequence of numbers is called arithmetic if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence s is arithmetic if and only if s[i+1] - s[i] == s[1] - s[0] for all valid i.

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

Solutions (Ruby)

1. Sort

# @param {Integer[]} nums
# @param {Integer[]} l
# @param {Integer[]} r
# @return {Boolean[]}
def check_arithmetic_subarrays(nums, l, r)
  ret = [false] * l.size

  (0...l.size).each do |i|
    sub = nums[l[i]..r[i]].sort
    ret[i] = sub.size > 1 && (2...sub.size).all? { |j| sub[j] - sub[j - 1] == sub[1] - sub[0] }
  end

  ret
end

Solutions (Rust)

1. Sort

impl Solution {
    pub fn check_arithmetic_subarrays(nums: Vec<i32>, l: Vec<i32>, r: Vec<i32>) -> Vec<bool> {
        let mut ret = vec![false; l.len()];

        for i in 0..l.len() {
            let mut sub = nums[l[i] as usize..=r[i] as usize].to_vec();
            sub.sort_unstable();
            ret[i] =
                sub.len() > 1 && (2..sub.len()).all(|j| sub[j] - sub[j - 1] == sub[1] - sub[0]);
        }

        ret
    }
}