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LeetCode - 1630. Arithmetic Subarrays

1630. Arithmetic Subarrays - Medium

문제

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:

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

The following sequence is not arithmetic:

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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.

제한사항`

  • 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

입출력 예

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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.
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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]

풀이

  • Sort
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class Solution {
public:
    vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) {
        std::vector<bool> res;
        for (auto i = 0 ; i < l.size() ; ++i) {
            std::vector<int> sub(nums.begin()+l[i], nums.begin()+r[i]+1);
            std::sort(sub.begin(), sub.end());
            
            bool flag = true;
            int val = sub[1] - sub[0];
            for (auto j = 1 ; j < sub.size() - 1 ; ++j) {
                int temp = sub[j+1] - sub[j];                
                if (val != temp) {
                    flag = false;
                    break;
                }
            }
            res.push_back(flag);
        }
        return res;
    }
};
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