Is Number of Effective Subsequences easy or hard?

Number of Effective Subsequences is categorized as a Hard problem because it mixes multiple concepts: subsequence counting, dynamic programming, bit manipulation, and combinatorics. The main difficulty comes from avoiding exponential enumeration while still counting all valid subsequences correctly.

Number of Effective Subsequences Python/Java solution

Implementations in Python, Java, C++, or Go usually maintain a DP structure indexed by bitmasks or bit counts. Each element updates existing states using bitwise operations, and valid states are accumulated into the final answer. The optimized implementation runs in near-linear time with respect to the array size.

How to solve Number of Effective Subsequences in O(n)?

Linear-style solutions rely on counting contributions per bit rather than generating subsequences. Track how many numbers contain each relevant bit and use precomputed powers of two to determine how many subsequences can be formed. This converts exponential enumeration into aggregated counting across the array.

What is the best approach for Number of Effective Subsequences?

The most efficient approach combines bit manipulation with combinatorics or compressed dynamic programming. Instead of enumerating every subsequence, it aggregates counts based on bit patterns contributed by array elements. This reduces the complexity to roughly O(n * B), where B is the number of relevant bits.

Is Number of Effective Subsequences asked at Google/Amazon/Meta?

Problems involving subsequence counting, bit manipulation, and combinatorics frequently appear in interviews at companies like Google, Amazon, and Meta. Hard-level questions often require transforming exponential subsequence problems into dynamic programming or mathematical counting solutions.

What data structure is used in Number of Effective Subsequences?

Typical solutions use arrays or hash maps to store DP states indexed by bitmasks. Bitwise operations such as AND, OR, and bit checks are used to combine subsequence states efficiently. Precomputed power arrays are also common for combinatorial counting.

What is the time complexity of Number of Effective Subsequences?

A naive subsequence enumeration requires O(2^n * B) time because every subset of the array must be checked. Optimized solutions use bitmask dynamic programming or combinatorics to reduce the complexity to around O(n * B) or O(n * M) depending on the number of tracked bit states.

Description Hints 3 Video 21:00

3757. Number of Effective Subsequences

Hard30.4% acceptance

You are given an integer array nums.

The strength of the array is defined as the bitwise OR of all its elements.

A subsequence is considered effective if removing that subsequence strictly decreases the strength of the remaining elements.

Return the number of effective subsequences in nums. Since the answer may be large, return it modulo 10⁹ + 7.

The bitwise OR of an empty array is 0.

Example 1:

Input: nums = [1,2,3]

Output: 3

Explanation:

The Bitwise OR of the array is 1 OR 2 OR 3 = 3.
Subsequences that are effective are:
- [1, 3]: The remaining element [2] has a Bitwise OR of 2.
- [2, 3]: The remaining element [1] has a Bitwise OR of 1.
- [1, 2, 3]: The remaining elements [] have a Bitwise OR of 0.
Thus, the total number of effective subsequences is 3.

Example 2:

Input: nums = [7,4,6]

Output: 4

Explanation:

The Bitwise OR of the array is 7 OR 4 OR 6 = 7.
Subsequences that are effective are:
- [7]: The remaining elements [4, 6] have a Bitwise OR of 6.
- [7, 4]: The remaining element [6] has a Bitwise OR of 6.
- [7, 6]: The remaining element [4] has a Bitwise OR of 4.
- [7, 4, 6]: The remaining elements [] have a Bitwise OR of 0.
Thus, the total number of effective subsequences is 4.

Example 3:

Input: nums = [8,8]

Output: 1

Explanation:

The Bitwise OR of the array is 8 OR 8 = 8.
Only the subsequence [8, 8] is effective since removing it leaves [] which has a Bitwise OR of 0.
Thus, the total number of effective subsequences is 1.

Example 4:

Input: nums = [2,2,1]

Output: 5

Explanation:

The Bitwise OR of the array is 2 OR 2 OR 1 = 3.
Subsequences that are effective are:
- [1]: The remaining elements [2, 2] have a Bitwise OR of 2.
- [2, 1] (using nums[0], nums[2]): The remaining element [2] has a Bitwise OR of 2.
- [2, 1] (using nums[1], nums[2]): The remaining element [2] has a Bitwise OR of 2.
- [2, 2]: The remaining element [1] has a Bitwise OR of 1.
- [2, 2, 1]: The remaining elements [] have a Bitwise OR of 0.
Thus, the total number of effective subsequences is 5.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Topics

Hints open one at a time

Hint 1

The bitwise OR is reduced if a bit is completely removed from the array OR.

Hint 2

To completely remove a bit, all of its occurrences must be removed from the array.

Hint 3

Count such subsets and apply inclusion-exclusion on the bits.

LanguageLoading editor…

1class Solution:
2    def countEffective(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3Case 4

Input

[1,2,3]

Input

[7,4,6]

Input

[8,8]

Input

[2,2,1]

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