LeetCode 3392: Count Subarrays of Length Three With a Condition

 

LeetCode problems are an excellent way to strengthen your problem-solving and algorithmic skills. In this blog post, we will dive deep into solving LeetCode 3392: Count Subarrays of Length Three With a Condition, a problem that requires us to analyze subarrays of length three with a specific mathematical condition.

By the end of this post, you will understand:

• The problem statement in detail

• Step-by-step approaches to solve the problem

• The C++ implementation of each approach

• Time and space complexity analysis

Problem Statement

Description

Given an integer array nums, you need to 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.

Key Details:

• A subarray is a contiguous, non-empty sequence of elements within an array.

• The length of the subarray should be exactly 3.

Examples

Example 1:

Input:

nums = [1,2,1,4,1]

Output:

1

Explanation:

• The only valid subarray is because:

Example 2:

Input:

nums = [1,1,1]

Output:

0

Explanation:

• The only subarray does not satisfy the condition .

---

Approaches to Solve the Problem

1. Brute Force Approach (Sliding Window Approach)

The simplest approach is to iterate through every subarray of length 3 and check the condition for each one.

Algorithm:

1. Loop through the array using an index i such that $i+2<\text{nums.length}$.
2. For each subarray [nums[i], nums[i+1], nums[i+2]], check if:
   $$\text{nums[i]} + \text{nums[i+2]} = \frac{\text{nums[i+1]}}{2}$$
3. If the condition is true, increment the counter.
4. Return the counter.

C++ Implementation

  int countSubarraysBruteForce(vector<int>& nums) 
  {
        int count = 0;
        for (int i = 0; i <= nums.size() - 3; ++i) 
        {
            if (nums[i] + nums[i+2] == nums[i+1] / 2.0)
            {
                ++count;
            }
        	return count;
    	}
  }
    

Time and Space Complexity

Time Complexity:

• The loop runs $O(n-2)$ times, where $n$ is the size of the array.
• Checking the condition is $O(1)$.

Overall: $O(n)$.

Space Complexity:

• We use a constant amount of extra space.

Overall: $O(1)$.

By thoroughly understanding the problem and implementing the solution step-by-step, you can strengthen your algorithmic skills.If you liked this explanation, don’t forget to share and bookmark for future reference!

Some Tags:

LeetCode Problem 3392 Solution, Count Subarrays of Length Three with a Condition, LeetCode Arrays Problem Solution, Subarray Length 3 Condition Explained, Algorithms for Array Problems, Sliding Window Technique Explained, Brute Force vs Optimized Algorithm, Subarray Manipulation in C++, LeetCode Easy Problems Solved, Array Algorithms in Competitive Programming, Time Complexity Analysis of Array Problems, Space Complexity Optimization Techniques, C++ Code for LeetCode Solutions, Learn Algorithms with Examples, Subarray Problems in Coding Interviews, Learn Coding with LeetCode, Algorithm Tutorials for Beginners, Problem-Solving with Arrays, C++ Programming for Beginners, Step-by-Step Coding Explanations.