How do you solve Maximum Subarray With Equal Products on LeetCode?

Maximum Subarray With Equal Products (LeetCode #3411) 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: The relationship between product, GCD, and LCM is crucial for determining valid subarrays.

What is the time complexity of Maximum Subarray With Equal Products?

The optimal solution for Maximum Subarray With Equal Products runs in O(n) time and uses O(1) space. This complexity is achieved because we only traverse the array once using the sliding window technique.

What is the brute force solution for Maximum Subarray With Equal Products?

The brute force approach for Maximum Subarray With Equal Products is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible subarray to see if it meets the product equivalent condition. This involves calculating the product, GCD, and LCM for each subarray, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Subarray With Equal Products?

Maximum Subarray With Equal Products uses the following concepts: Array, Math, Sliding Window, Enumeration, Number Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Subarray With Equal Products?

Always explain your thought process clearly when discussing the brute force approach. Be prepared to discuss the trade-offs between brute force and optimal solutions. Practice implementing both approaches to solidify your understanding.

What are common mistakes when solving Maximum Subarray With Equal Products?

Failing to update GCD and LCM correctly when shrinking the window. Not considering edge cases where the product might overflow.

#3411

Maximum Subarray With Equal Products

Easy

ArrayMathSliding WindowEnumerationNumber TheoryHash MapArray

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)

Instead of checking all subarrays, we can use a sliding window approach to dynamically calculate the product, GCD, and LCM while maintaining a valid subarray. This reduces the number of calculations significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers for the sliding window and variables for product, GCD, and LCM.
2Step 2: Expand the window by moving the right pointer and update product, GCD, and LCM.
3Step 3: If the product is not equal to GCD * LCM, move the left pointer to shrink the window until the condition is satisfied or the window is empty.
4Step 4: Keep track of the maximum length of valid subarrays found.

solution.py31 lines

1# Full working Python code
2import math
3from functools import reduce
4
5def gcd(a, b):
6    while b:
7        a, b = b, a % b
8    return a
9
10def lcm(a, b):
11    return abs(a * b) // gcd(a, b)
12
13def maxProductEquivalentSubarray(nums):
14    n = len(nums)
15    max_length = 0
16    left = 0
17    product = 1
18    current_gcd = 0
19    current_lcm = 1
20    for right in range(n):
21        product *= nums[right]
22        current_gcd = gcd(current_gcd, nums[right]) if current_gcd else nums[right]
23        current_lcm = lcm(current_lcm, nums[right])
24        while product != current_gcd * current_lcm and left <= right:
25            product //= nums[left]
26            left += 1
27            current_gcd = reduce(gcd, nums[left:right + 1])
28            current_lcm = reduce(lcm, nums[left:right + 1])
29        max_length = max(max_length, right - left + 1)
30    return max_length
31

ℹ

Complexity note: This complexity is achieved because we only traverse the array once using the sliding window technique.

1The relationship between product, GCD, and LCM is crucial for determining valid subarrays.
2Using a sliding window can significantly reduce the number of calculations needed.

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