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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

Solution Explanation

The problem asks to determine if subarrays within a given array nums can be rearranged to form an arithmetic sequence. An arithmetic sequence is defined as a sequence with at least two elements where the difference between consecutive elements is constant.

The solution employs a function check (or a similar function depending on the programming language) to efficiently verify if a subarray satisfies this condition. This function is called repeatedly for each query range specified by l and r.

Algorithm:

  1. check Function:

    • Extracts the subarray defined by l and r from nums.
    • Creates a Set (or equivalent data structure in other languages) to store the unique elements of the subarray, efficiently allowing checking for presence of elements later.
    • Finds the minimum (a1) and maximum (an) values in the subarray.
    • Calculates the common difference d (if it exists) as (an - a1) / (n - 1), where n is the subarray length. If the division results in a remainder, it means a constant difference doesn't exist, and the function returns false.
    • Iterates from the second smallest element (a1 + d) up to the largest element (an), checking if each element a1 + (i - 1) * d is present in the set. If any element is missing, it's not an arithmetic sequence, and false is returned.
    • If all elements are found in the set, it is an arithmetic sequence, so the function returns true.

  2. Main Loop:

    • Iterates through the query ranges specified by l and r (using zip in Python).
    • For each range, calls the check function to determine if the corresponding subarray can be rearranged into an arithmetic sequence.
    • Appends the boolean result to the ans list.
    • Returns the ans list.

Time Complexity:

  • The check function iterates through the subarray once to find the min/max and build the set, which takes O(n) time where n is the length of the subarray. The check for consecutive elements also takes O(n). Thus, the total time complexity of check is O(n).
  • The main loop iterates through m queries, and for each query, the check function is called. Therefore, the overall time complexity is O(m * n), where m is the number of queries and n is the maximum length of any subarray.

Space Complexity:

  • The check function uses a set to store subarray elements, taking O(n) space.
  • The ans list has a size equal to the number of queries m, requiring O(m) space.
  • Therefore, the overall space complexity is O(m + n).

Example Code (Python):

class Solution:
    def checkArithmeticSubarrays(self, nums: List[int], l: List[int], r: List[int]) -> List[bool]:
        def check(nums, l, r):
            n = r - l + 1
            s = set(nums[l:l+n])
            a1, an = min(nums[l:l+n]), max(nums[l:l+n])
            d, mod = divmod(an - a1, n - 1)
            return mod == 0 and all((a1 + (i - 1) * d) in s for i in range(1, n))
 
        return [check(nums, left, right) for left, right in zip(l, r)]

The code in other languages follows a similar structure, with minor syntactic variations to accommodate language-specific features. The core algorithm remains consistent.