LeetCode #3757 — HARD

Number of Effective Subsequences

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Dynamic Programming · Bit Manipulation

Example 1

[1,2,3]

Example 2

[7,4,6]

Example 3

[8,8]

Step 02

Core Insight

What unlocks the optimal approach

The bitwise OR is reduced if a bit is completely removed from the array OR.
To completely remove a bit, all of its occurrences must be removed from the array.
Count such subsets and apply inclusion-exclusion on the bits.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3757: Number of Effective Subsequences
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3757: Number of Effective Subsequences
// package main
// 
// import "math/bits"
// 
// // https://space.bilibili.com/206214
// const mod = 1_000_000_007
// const maxN = 100_001
// 
// var pow2 = [maxN]int{1}
// 
// func init() {
// 	// 预处理 2 的幂
// 	for i := 1; i < maxN; i++ {
// 		pow2[i] = pow2[i-1] * 2 % mod
// 	}
// }
// 
// func countEffective(nums []int) int {
// 	or := 0
// 	for _, x := range nums {
// 		or |= x
// 	}
// 
// 	w := bits.Len(uint(or))
// 	f := make([]int, 1<<w)
// 	for _, x := range nums {
// 		f[x]++
// 	}
// 	for i := range w {
// 		if or>>i&1 == 0 { // 优化：or 中是 0 的比特位无需计算
// 			continue
// 		}
// 		for s := 0; s < 1<<w; s++ {
// 			s |= 1 << i
// 			f[s] += f[s^1<<i]
// 		}
// 	}
// 	// 计算完毕后，f[s] 表示 nums 中的是 s 的子集的元素个数
// 
// 	ans := pow2[len(nums)] // 所有子序列的个数
// 	// 枚举 or 的所有子集（包括空集）
// 	for sub, ok := or, true; ok; ok = sub != or {
// 		sign := 1 - bits.OnesCount(uint(or^sub))%2*2
// 		ans -= sign * pow2[f[sub]]
// 		sub = (sub - 1) & or
// 	}
// 	return (ans%mod + mod) % mod // 保证结果非负
// }

// Accepted solution for LeetCode #3757: Number of Effective Subsequences
package main

import "math/bits"

// https://space.bilibili.com/206214
const mod = 1_000_000_007
const maxN = 100_001

var pow2 = [maxN]int{1}

func init() {
	// 预处理 2 的幂
	for i := 1; i < maxN; i++ {
		pow2[i] = pow2[i-1] * 2 % mod
	}
}

func countEffective(nums []int) int {
	or := 0
	for _, x := range nums {
		or |= x
	}

	w := bits.Len(uint(or))
	f := make([]int, 1<<w)
	for _, x := range nums {
		f[x]++
	}
	for i := range w {
		if or>>i&1 == 0 { // 优化：or 中是 0 的比特位无需计算
			continue
		}
		for s := 0; s < 1<<w; s++ {
			s |= 1 << i
			f[s] += f[s^1<<i]
		}
	}
	// 计算完毕后，f[s] 表示 nums 中的是 s 的子集的元素个数

	ans := pow2[len(nums)] // 所有子序列的个数
	// 枚举 or 的所有子集（包括空集）
	for sub, ok := or, true; ok; ok = sub != or {
		sign := 1 - bits.OnesCount(uint(or^sub))%2*2
		ans -= sign * pow2[f[sub]]
		sub = (sub - 1) & or
	}
	return (ans%mod + mod) % mod // 保证结果非负
}

# Accepted solution for LeetCode #3757: Number of Effective Subsequences
#
# @lc app=leetcode id=3757 lang=python3
#
# [3757] Number of Effective Subsequences
#

# @lc code=start
class Solution:
    def countEffective(self, nums: List[int]) -> int:
        MOD = 10**9 + 7
        if not nums:
            return 0
        
        T = 0
        for x in nums:
            T |= x
            
        if T == 0:
            return 0
        
        # The maximum possible value is bounded by the smallest power of 2 > T
        limit_bits = T.bit_length()
        LIMIT = 1 << limit_bits
        
        dp = [0] * LIMIT
        for x in nums:
            dp[x] += 1
            
        # SOS DP (Sum Over Subsets)
        # dp[mask] will store the count of numbers in nums that are submasks of 'mask'
        for i in range(limit_bits):
            if (T >> i) & 1:
                bit = 1 << i
                # Iterate through pairs (k, k+bit) where k has the i-th bit unset
                for base in range(0, LIMIT, 2 * bit):
                    for k in range(base, base + bit):
                        dp[bit + k] += dp[k]
                        
        n = len(nums)
        pow2 = [1] * (n + 1)
        for i in range(1, n + 1):
            pow2[i] = (pow2[i-1] * 2) % MOD
            
        count_eq_T = 0
        t_pop = T.bit_count()
        
        # Principle of Inclusion-Exclusion to find number of subsequences with OR sum exactly T
        # We iterate over all submasks of T
        sub = T
        while True:
            # dp[sub] is the count of numbers in nums that are submasks of 'sub'
            # The number of subsequences using only these numbers is 2^dp[sub]
            term = pow2[dp[sub]]
            
            # PIE sign depends on the parity of the difference in set bits
            if (t_pop - sub.bit_count()) & 1:
                count_eq_T = (count_eq_T - term + MOD) % MOD
            else:
                count_eq_T = (count_eq_T + term) % MOD
            
            if sub == 0:
                break
            # Efficiently iterate to the next submask
            sub = (sub - 1) & T
            
        # The answer is Total Subsequences - Subsequences with OR sum equal to T
        # Total subsequences is 2^n
        return (pow2[n] - count_eq_T + MOD) % MOD
# @lc code=end

// Accepted solution for LeetCode #3757: Number of Effective Subsequences
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3757: Number of Effective Subsequences
// package main
// 
// import "math/bits"
// 
// // https://space.bilibili.com/206214
// const mod = 1_000_000_007
// const maxN = 100_001
// 
// var pow2 = [maxN]int{1}
// 
// func init() {
// 	// 预处理 2 的幂
// 	for i := 1; i < maxN; i++ {
// 		pow2[i] = pow2[i-1] * 2 % mod
// 	}
// }
// 
// func countEffective(nums []int) int {
// 	or := 0
// 	for _, x := range nums {
// 		or |= x
// 	}
// 
// 	w := bits.Len(uint(or))
// 	f := make([]int, 1<<w)
// 	for _, x := range nums {
// 		f[x]++
// 	}
// 	for i := range w {
// 		if or>>i&1 == 0 { // 优化：or 中是 0 的比特位无需计算
// 			continue
// 		}
// 		for s := 0; s < 1<<w; s++ {
// 			s |= 1 << i
// 			f[s] += f[s^1<<i]
// 		}
// 	}
// 	// 计算完毕后，f[s] 表示 nums 中的是 s 的子集的元素个数
// 
// 	ans := pow2[len(nums)] // 所有子序列的个数
// 	// 枚举 or 的所有子集（包括空集）
// 	for sub, ok := or, true; ok; ok = sub != or {
// 		sign := 1 - bits.OnesCount(uint(or^sub))%2*2
// 		ans -= sign * pow2[f[sub]]
// 		sub = (sub - 1) & or
// 	}
// 	return (ans%mod + mod) % mod // 保证结果非负
// }

// Accepted solution for LeetCode #3757: Number of Effective Subsequences
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3757: Number of Effective Subsequences
// package main
// 
// import "math/bits"
// 
// // https://space.bilibili.com/206214
// const mod = 1_000_000_007
// const maxN = 100_001
// 
// var pow2 = [maxN]int{1}
// 
// func init() {
// 	// 预处理 2 的幂
// 	for i := 1; i < maxN; i++ {
// 		pow2[i] = pow2[i-1] * 2 % mod
// 	}
// }
// 
// func countEffective(nums []int) int {
// 	or := 0
// 	for _, x := range nums {
// 		or |= x
// 	}
// 
// 	w := bits.Len(uint(or))
// 	f := make([]int, 1<<w)
// 	for _, x := range nums {
// 		f[x]++
// 	}
// 	for i := range w {
// 		if or>>i&1 == 0 { // 优化：or 中是 0 的比特位无需计算
// 			continue
// 		}
// 		for s := 0; s < 1<<w; s++ {
// 			s |= 1 << i
// 			f[s] += f[s^1<<i]
// 		}
// 	}
// 	// 计算完毕后，f[s] 表示 nums 中的是 s 的子集的元素个数
// 
// 	ans := pow2[len(nums)] // 所有子序列的个数
// 	// 枚举 or 的所有子集（包括空集）
// 	for sub, ok := or, true; ok; ok = sub != or {
// 		sign := 1 - bits.OnesCount(uint(or^sub))%2*2
// 		ans -= sign * pow2[f[sub]]
// 		sub = (sub - 1) & or
// 	}
// 	return (ans%mod + mod) % mod // 保证结果非负
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.