How do you solve Number of Different Subsequences GCDs on LeetCode?

Number of Different Subsequences GCDs (LeetCode #1819) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m log m) time complexity and O(m) space complexity. Key insight: Understanding the properties of GCD can help reduce the problem space significantly.

What is the brute force solution for Number of Different Subsequences GCDs?

The brute force approach for Number of Different Subsequences GCDs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible non-empty subsequences of the array and calculating their GCDs. This method is straightforward but inefficient for larger arrays due to the exponential number of subsequences. The optimal Optimal Solution approach improves this to O(n + m log m).

What algorithm or data structure is used to solve Number of Different Subsequences GCDs?

Number of Different Subsequences GCDs uses the following concepts: Array, Math, Counting, Number Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Different Subsequences GCDs?

Always think about the properties of the problem — in this case, the relationship between divisors and GCD. Start with a brute-force solution to understand the problem better before optimizing. Be prepared to explain your thought process and how you arrived at the optimal solution.

What are common mistakes when solving Number of Different Subsequences GCDs?

Failing to consider all possible GCDs, especially those that are not directly in the array. Not using a set to track unique GCDs, leading to incorrect counts.

#1819

Number of Different Subsequences GCDs

Hard

ArrayMathCountingNumber TheoryHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n + m log m)Space O(m)

Instead of generating all subsequences, we can determine the possible GCDs directly by checking divisors of the maximum number in the array. This is efficient because it reduces the problem to counting how many numbers in the array are divisible by each potential GCD.

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Algorithm

5 steps

1Step 1: Find the maximum number in the array.
2Step 2: Create a frequency array to count occurrences of each number up to the maximum number.
3Step 3: Iterate through each number from 1 to the maximum number and check if it can be a GCD by counting how many numbers in the array are divisible by it.
4Step 4: If a number can be a GCD, add it to a set to track unique GCDs.
5Step 5: Return the size of the set as the result.

solution.py13 lines

1def countDifferentGCDs(nums):
2    max_num = max(nums)
3    freq = [0] * (max_num + 1)
4    for num in nums:
5        freq[num] += 1
6    gcd_set = set()
7    for g in range(1, max_num + 1):
8        count = 0
9        for multiple in range(g, max_num + 1, g):
10            count += freq[multiple]
11        if count > 0:
12            gcd_set.add(g)
13    return len(gcd_set)

ℹ

Complexity note: The time complexity is O(n + m log m) where n is the length of the input array and m is the maximum number in the array. The space complexity is O(m) for the frequency array.

1Understanding the properties of GCD can help reduce the problem space significantly.
2Using a frequency array allows us to efficiently count multiples and potential GCDs.

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