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2229. Check if an Array Is Consecutive
Description
Given an integer array nums
, return true
if nums
is consecutive, otherwise return false
.
An array is consecutive if it contains every number in the range [x, x + n - 1]
(inclusive), where x
is the minimum number in the array and n
is the length of the array.
Example 1:
Input: nums = [1,3,4,2] Output: true Explanation: The minimum value is 1 and the length of nums is 4. All of the values in the range [x, x + n - 1] = [1, 1 + 4 - 1] = [1, 4] = (1, 2, 3, 4) occur in nums. Therefore, nums is consecutive.
Example 2:
Input: nums = [1,3] Output: false Explanation: The minimum value is 1 and the length of nums is 2. The value 2 in the range [x, x + n - 1] = [1, 1 + 2 - 1], = [1, 2] = (1, 2) does not occur in nums. Therefore, nums is not consecutive.
Example 3:
Input: nums = [3,5,4] Output: true Explanation: The minimum value is 3 and the length of nums is 3. All of the values in the range [x, x + n - 1] = [3, 3 + 3 - 1] = [3, 5] = (3, 4, 5) occur in nums. Therefore, nums is consecutive.
Constraints:
1 <= nums.length <= 105
0 <= nums[i] <= 105
Solutions
-
class Solution { public boolean isConsecutive(int[] nums) { int mi = nums[0]; int mx = nums[0]; Set<Integer> s = new HashSet<>(); for (int v : nums) { mi = Math.min(mi, v); mx = Math.max(mx, v); s.add(v); } int n = nums.length; return s.size() == n && mx == mi + n - 1; } }
-
class Solution { public: bool isConsecutive(vector<int>& nums) { unordered_set<int> s(nums.begin(), nums.end()); int mi = *min_element(nums.begin(), nums.end()); int mx = *max_element(nums.begin(), nums.end()); int n = nums.size(); return s.size() == n && mx == mi + n - 1; } };
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class Solution: def isConsecutive(self, nums: List[int]) -> bool: mi, mx = min(nums), max(nums) n = len(nums) return len(set(nums)) == n and mx == mi + n - 1
-
func isConsecutive(nums []int) bool { s := map[int]bool{} mi, mx := slices.Min(nums), slices.Max(nums) for _, x := range nums { s[x] = true } return len(s) == len(nums) && mx == mi+len(nums)-1 }