How do you solve Number of Effective Subsequences on LeetCode?

Number of Effective Subsequences (LeetCode #3757) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Effective subsequences remove bits contributing to strength.

What is the time complexity of Number of Effective Subsequences?

The optimal solution for Number of Effective Subsequences runs in O(n) time and uses O(n) space. Counting occurrences and calculating powers is linear with respect to the number of elements.

What is the brute force solution for Number of Effective Subsequences?

The brute force approach for Number of Effective Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible subsequences and check if removing each one decreases the strength. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Number of Effective Subsequences?

Understand bitwise operations well. Practice combinatorial problems. Discuss your thought process clearly.

What are common mistakes when solving Number of Effective Subsequences?

Not considering all occurrences of a number. Confusing bitwise operations.

#3757

Number of Effective Subsequences

Hard

ArrayMathDynamic ProgrammingBit ManipulationCombinatoricsHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Identify bits that contribute to the overall strength. Effective subsequences are those that remove all instances of a bit contributing to the strength.

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Algorithm

3 steps

1Step 1: Calculate the total strength using bitwise OR of all elements.
2Step 2: Count occurrences of each unique number in the array.
3Step 3: For each unique number, calculate effective subsequences as 2^(count) - 1 and sum them.

solution.py12 lines

1def countEffectiveSubsequences(nums):
2    from collections import Counter
3    MOD = 10**9 + 7
4    total_strength = 0
5    for num in nums:
6        total_strength |= num
7    count = Counter(nums)
8    total = 0
9    for num, cnt in count.items():
10        if (total_strength & num) == num:
11            total += (pow(2, cnt, MOD) - 1) % MOD
12    return total % MOD

ℹ

Complexity note: Counting occurrences and calculating powers is linear with respect to the number of elements.

1Effective subsequences remove bits contributing to strength.
2Count unique elements and their occurrences.

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