Welcome to Subscribe On Youtube
3392. Count Subarrays of Length Three With a Condition
Description
Given an integer array nums, return the number of subarrays of length 3 such that the sum of the first and third numbers equals exactly half of the second number.
Example 1:
Input: nums = [1,2,1,4,1]
Output: 1
Explanation:
Only the subarray [1,4,1] contains exactly 3 elements where the sum of the first and third numbers equals half the middle number.
Example 2:
Input: nums = [1,1,1]
Output: 0
Explanation:
[1,1,1] is the only subarray of length 3. However, its first and third numbers do not add to half the middle number.
Constraints:
3 <= nums.length <= 100-100 <= nums[i] <= 100
Solutions
Solution 1: Single Pass
We traverse each subarray of length $3$ in the array $\textit{nums}$ and check if twice the sum of the first and third numbers equals the second number. If it does, we increment the answer by $1$.
After traversing, we return the answer.
The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
-
class Solution { public int countSubarrays(int[] nums) { int ans = 0; for (int i = 1; i + 1 < nums.length; ++i) { if ((nums[i - 1] + nums[i + 1]) * 2 == nums[i]) { ++ans; } } return ans; } } -
class Solution { public: int countSubarrays(vector<int>& nums) { int ans = 0; for (int i = 1; i + 1 < nums.size(); ++i) { if ((nums[i - 1] + nums[i + 1]) * 2 == nums[i]) { ++ans; } } return ans; } }; -
class Solution: def countSubarrays(self, nums: List[int]) -> int: return sum( (nums[i - 1] + nums[i + 1]) * 2 == nums[i] for i in range(1, len(nums) - 1) ) -
func countSubarrays(nums []int) (ans int) { for i := 1; i+1 < len(nums); i++ { if (nums[i-1]+nums[i+1])*2 == nums[i] { ans++ } } return } -
function countSubarrays(nums: number[]): number { let ans: number = 0; for (let i = 1; i + 1 < nums.length; ++i) { if ((nums[i - 1] + nums[i + 1]) * 2 === nums[i]) { ++ans; } } return ans; }