1630. Arithmetic Subarrays

Problem

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

Solution

/**
 * @param {number[]} nums
 * @param {number[]} l
 * @param {number[]} r
 * @return {boolean[]}
 */
var checkArithmeticSubarrays = function(nums, l, r) {
    return l.map((_, i) => check(nums, l[i], r[i]));
};

var check = function(nums, l, r) {
    var map = {};
    var min = Number.MAX_SAFE_INTEGER;
    var max = Number.MIN_SAFE_INTEGER;
    for (var i = l; i <= r; i++) {
        min = Math.min(min, nums[i]);
        max = Math.max(max, nums[i]);
        map[nums[i]] = true;
    }
    var diff = (max - min) / (r - l);
    for (var num = min; num < max; num += diff) {
        if (map[num] === undefined) {
            return false;
        }
    }
    return true;
};

Explain:

nope.

Complexity:

• Time complexity : O(n * m).
• Space complexity : O(n).