LeetCode 3392: Count Subarrays of Length Three With a Condition Solution
Master LeetCode problem 3392 (Count Subarrays of Length Three With a Condition), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
3392. Count Subarrays of Length Three With a Condition
Problem Explanation
Explanation:
To solve this problem, we can iterate through the array and for each subarray of length 3, check if the sum of the first and third elements is equal to half of the second element. We can do this by iterating through the array and checking the conditions for each subarray of length 3.
Algorithm:
- Initialize a variable
countto keep track of the number of valid subarrays. - Iterate through the array from index 1 to
nums.length - 2. - For each index
i, check if the sum ofnums[i-1]andnums[i+1]is equal to half ofnums[i]. - If the condition is met, increment
count. - Return the final value of
count.
Time Complexity:
The time complexity of this solution is O(n), where n is the size of the input array nums.
Space Complexity:
The space complexity of this solution is O(1) as we are using only a constant amount of extra space.
Solution Code
class Solution {
public int countSubarrays(int[] nums) {
int count = 0;
for (int i = 1; i < nums.length - 1; i++) {
if (nums[i-1] + nums[i+1] == nums[i] / 2) {
count++;
}
}
return count;
}
}Try It Yourself
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What is the time complexity of LeetCode 3392 (Count Subarrays of Length Three With a Condition)?
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