How do you solve Number of Subarrays With LCM Equal to K on LeetCode?

Number of Subarrays With LCM Equal to K (LeetCode #2470) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the properties of LCM and how it relates to subarrays is crucial.

What is the brute force solution for Number of Subarrays With LCM Equal to K?

The brute force approach for Number of Subarrays With LCM Equal to K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray of the given array and calculating its LCM. If the LCM equals k, we count that subarray. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Subarrays With LCM Equal to K?

Number of Subarrays With LCM Equal to K uses the following concepts: Array, Math, Number Theory. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Number of Subarrays With LCM Equal to K?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process and approach as you code, especially when optimizing. Practice calculating LCM and GCD manually to strengthen your understanding.

What are common mistakes when solving Number of Subarrays With LCM Equal to K?

Failing to handle cases where the LCM exceeds k early, leading to unnecessary calculations. Not considering the impact of the order of elements in the subarray when calculating LCM.

#2470

Number of Subarrays With LCM Equal to K

Medium

ArrayMathNumber TheorySliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to efficiently find subarrays with the desired LCM. By maintaining a window of elements, we can calculate the LCM incrementally, reducing the need for redundant calculations.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter to zero and two pointers for the sliding window.
2Step 2: Expand the right pointer to include elements in the window and calculate the LCM.
3Step 3: If the LCM exceeds k, move the left pointer to shrink the window until the LCM is less than or equal to k.
4Step 4: If the LCM equals k, count all valid subarrays ending at the current right pointer.
5Step 5: Continue until all elements are processed.

solution.py24 lines

1# Full working Python code
2from math import gcd
3
4# Function to calculate LCM
5def lcm(a, b):
6    return a * b // gcd(a, b)
7
8# Main function to count subarrays with LCM equal to k
9def countSubarraysWithLCM(nums, k):
10    count = 0
11    n = len(nums)
12    left = 0
13    current_lcm = 1
14    for right in range(n):
15        current_lcm = lcm(current_lcm, nums[right])
16        while left <= right and current_lcm > k:
17            current_lcm = lcm(current_lcm, nums[left])
18            left += 1
19        if current_lcm == k:
20            count += (right - left + 1)
21    return count
22
23# Example usage
24print(countSubarraysWithLCM([3, 6, 2, 7, 1], 6))  # Output: 4

ℹ

Complexity note: The time complexity is O(n) because we are using a single loop to traverse the array, and the LCM calculation is constant time due to the limited range of numbers. The space complexity is O(1) as we are using a constant amount of space.

1Understanding the properties of LCM and how it relates to subarrays is crucial.
2Using a sliding window technique can significantly reduce the time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.