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1630. Arithmetic Subarrays

Description

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

  • class Solution {
    public List<Boolean> checkArithmeticSubarrays(int[] nums, int[] l, int[] r) {
        List<Boolean> ans = new ArrayList<>();
        for (int i = 0; i < l.length; ++i) {
            ans.add(check(nums, l[i], r[i]));
        }
        return ans;
    }

    private boolean check(int[] nums, int l, int r) {
        Set<Integer> s = new HashSet<>();
        int n = r - l + 1;
        int a1 = 1 << 30, an = -a1;
        for (int i = l; i <= r; ++i) {
            s.add(nums[i]);
            a1 = Math.min(a1, nums[i]);
            an = Math.max(an, nums[i]);
        }
        if ((an - a1) % (n - 1) != 0) {
            return false;
        }
        int d = (an - a1) / (n - 1);
        for (int i = 1; i < n; ++i) {
            if (!s.contains(a1 + (i - 1) * d)) {
                return false;
            }
        }
        return true;
    }
}

  • class Solution {
public:
    vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) {
        vector<bool> ans;
        auto check = [](vector<int>& nums, int l, int r) {
            unordered_set<int> s;
            int n = r - l + 1;
            int a1 = 1 << 30, an = -a1;
            for (int i = l; i <= r; ++i) {
                s.insert(nums[i]);
                a1 = min(a1, nums[i]);
                an = max(an, nums[i]);
            }
            if ((an - a1) % (n - 1)) {
                return false;
            }
            int d = (an - a1) / (n - 1);
            for (int i = 1; i < n; ++i) {
                if (!s.count(a1 + (i - 1) * d)) {
                    return false;
                }
            }
            return true;
        };
        for (int i = 0; i < l.size(); ++i) {
            ans.push_back(check(nums, l[i], r[i]));
        }
        return ans;
    }
};

  • 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)]


  • func checkArithmeticSubarrays(nums []int, l []int, r []int) (ans []bool) {
	check := func(nums []int, l, r int) bool {
		s := map[int]struct{}{}
		n := r - l + 1
		a1, an := 1<<30, -(1 << 30)
		for _, x := range nums[l : r+1] {
			s[x] = struct{}{}
			if a1 > x {
				a1 = x
			}
			if an < x {
				an = x
			}
		}
		if (an-a1)%(n-1) != 0 {
			return false
		}
		d := (an - a1) / (n - 1)
		for i := 1; i < n; i++ {
			if _, ok := s[a1+(i-1)*d]; !ok {
				return false
			}
		}
		return true
	}
	for i := range l {
		ans = append(ans, check(nums, l[i], r[i]))
	}
	return
}

  • function checkArithmeticSubarrays(nums: number[], l: number[], r: number[]): boolean[] {
    const check = (nums: number[], l: number, r: number): boolean => {
        const s = new Set<number>();
        const n = r - l + 1;
        let a1 = 1 << 30;
        let an = -a1;
        for (let i = l; i <= r; ++i) {
            s.add(nums[i]);
            a1 = Math.min(a1, nums[i]);
            an = Math.max(an, nums[i]);
        }
        if ((an - a1) % (n - 1) !== 0) {
            return false;
        }
        const d = Math.floor((an - a1) / (n - 1));
        for (let i = 1; i < n; ++i) {
            if (!s.has(a1 + (i - 1) * d)) {
                return false;
            }
        }
        return true;
    };
    const ans: boolean[] = [];
    for (let i = 0; i < l.length; ++i) {
        ans.push(check(nums, l[i], r[i]));
    }
    return ans;
}


  • class Solution {
    public bool Check(int[] arr) {
        Array.Sort(arr);
        int diff = arr[1] - arr[0];
        for (int i = 2; i < arr.Length; i++) {
            if (arr[i] - arr[i - 1] != diff) {
                return false;
            }
        }
        return true;
    }

    public IList<bool> CheckArithmeticSubarrays(int[] nums, int[] l, int[] r) {
        List<bool> ans = new List<bool>();
        for (int i = 0; i < l.Length; i++) {
            int[] arr = new int[r[i] - l[i] + 1];
            for (int j = 0; j < arr.Length; j++) {
                arr[j] = nums[l[i] + j];
            }
            ans.Add(Check(arr));
        }
        return ans;
    }
}



  • impl Solution {
    pub fn check_arithmetic_subarrays(nums: Vec<i32>, l: Vec<i32>, r: Vec<i32>) -> Vec<bool> {
        let m = l.len();
        let mut res = vec![true; m];
        for i in 0..m {
            let mut arr = nums[l[i] as usize..=r[i] as usize].to_vec();
            arr.sort();
            for j in 2..arr.len() {
                if arr[j - 2] - arr[j - 1] != arr[j - 1] - arr[j] {
                    res[i] = false;
                    break;
                }
            }
        }
        res
    }
}

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