[ Leetcode ] 1630. Arithmetic Subarrays - Java

 

 

 

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

 

 
  class Solution {
    public List<Boolean> checkArithmeticSubarrays(int[] nums, int[] l, int[] r) {
        List<Boolean> list = new ArrayList<>();
        
        for(int i=0; i< l.length; i++){
            list.add(isArithmetic(Arrays.copyOfRange(nums, l[i], r[i]+1)));
        }
        
        return list;
    }
    
    public boolean isArithmetic(int[] nums){
        if(nums.length < 3){
            return true;
        }
        Arrays.sort(nums);
        int diff = nums[1] - nums[0];
        for(int i=2; i< nums.length; i++){
            if(diff != (nums[i] - nums[i-1])){
                return false;
            }
        }
        return true;
    }
 }
 

1. nums의 sublist를 만든다 (l[i] 부터 r[i]까지 닫히도록 r[i]에 1을 더한다)

2. sublist의 길이가 2 이하인경우는 true로 리턴

3. sort 한 후 첫 번째의 차이값과 다르면 return false;