Problem Description

You are given an integer array nums. Your task is to find the maximum possible bitwise OR value that can be obtained from any subset of nums, and then count how many different non-empty subsets can produce this maximum OR value.

A subset is formed by selecting some (or all) elements from the original array, where the selected elements maintain their original values. Two subsets are considered different if they are formed by selecting elements at different indices, even if the resulting values are the same.

The bitwise OR of a subset is calculated by applying the OR operation to all elements in that subset. For example, if a subset contains elements [a, b, c], its bitwise OR would be a OR b OR c.

The solution uses a depth-first search (DFS) approach to explore all possible subsets. First, it calculates the maximum possible OR value mx by performing OR on all elements in the array (since OR operations can only maintain or increase the number of set bits, never decrease them). Then, it uses DFS to enumerate all possible subsets and counts those whose OR value equals mx.

The DFS function dfs(i, t) tracks:

• i: the current index in the array being considered
• t: the current OR value accumulated so far

At each step, the algorithm makes two choices:

1. Skip the current element: dfs(i + 1, t)
2. Include the current element: dfs(i + 1, t | nums[i])

When all elements have been considered (i == len(nums)), if the accumulated OR value t equals the maximum mx, the answer counter is incremented.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem involves an integer array and subsets, not nodes and edges as in graph problems.

Need to solve for kth smallest/largest?

• No: The problem is not about finding the kth smallest or largest element, but rather counting subsets with a specific property.

Involves Linked Lists?

• No: The problem deals with an integer array, not linked lists.

Does the problem have small constraints?

• Yes: Looking at typical constraints for this problem (array length usually ≤ 16), and the fact that we need to enumerate all possible subsets (2^n possibilities), the constraints are intentionally kept small to allow exhaustive enumeration.

Brute force / Backtracking?

• Yes: Since we need to explore all possible subsets to find those with maximum bitwise OR value, and the constraints are small enough to allow checking all 2^n subsets, a brute force/backtracking approach is appropriate.

Conclusion: The flowchart correctly leads us to use a Backtracking approach. This makes sense because:

1. We need to enumerate all possible subsets (a classic backtracking use case)
2. The small constraints make exhaustive search feasible
3. At each position in the array, we make a choice: include or exclude the current element
4. We track the OR value as we build each subset and count those matching the maximum

The DFS implementation in the solution is a form of backtracking where we explore both branches (include/exclude) at each decision point.

Intuition

The key insight is understanding what makes the maximum bitwise OR value. When we perform OR operations on numbers, we can only "turn on" bits (change 0 to 1), never "turn off" bits (change 1 to 0). This means the maximum possible OR value would have the most bits turned on.

Think about it this way: if we have nums = [3, 2, 1, 5], their binary representations are:

• 3 = 011
• 2 = 010
• 1 = 001
• 5 = 101

When we OR all of them together: 011 | 010 | 001 | 101 = 111 = 7, we get the maximum possible OR value because we've collected all the unique bit positions that appear anywhere in the array.

This leads to our first realization: the maximum OR value is simply the OR of all elements in the array. We can't get a higher value by selecting a subset because we can't introduce new 1-bits that don't exist in any element.

Now, the question becomes: how many subsets can achieve this maximum? Since we need to count all possible subsets, and with small constraints (typically n ≤ 16), we can afford to check all 2^n possible subsets.

The backtracking approach naturally fits here because subset generation is a series of binary decisions: for each element, we either include it or exclude it. We can visualize this as a binary tree where:

• At each level, we decide about one element
• Left branch = exclude the element
• Right branch = include the element

As we traverse this decision tree using DFS, we maintain the current OR value. When we reach a leaf (processed all elements), we check if our accumulated OR equals the maximum. If yes, we've found one valid subset.

The elegance of this solution is that it combines two simple ideas:

1. Calculate the target (maximum OR) upfront in O(n) time
2. Use DFS to systematically generate and check all subsets

This way, we avoid storing all subsets in memory - we just count them as we go.

Learn more about Backtracking patterns.

Solution Approach

The implementation follows a depth-first search (DFS) pattern to enumerate all possible subsets. Let's walk through the key components:

Step 1: Calculate the Maximum OR Value

mx = reduce(lambda x, y: x | y, nums)

We first compute the maximum possible OR value by performing bitwise OR on all elements. This gives us our target value since OR operations can only maintain or add 1-bits, never remove them.

Step 2: Define the DFS Function

def dfs(i, t):
    nonlocal ans, mx
    if i == len(nums):
        if t == mx:
            ans += 1
        return
    dfs(i + 1, t)
    dfs(i + 1, t | nums[i])

The DFS function takes two parameters:

• i: Current index in the array we're considering
• t: Current bitwise OR value accumulated so far

Step 3: Base Case When i == len(nums), we've made decisions for all elements. At this point:

• We check if our accumulated OR value t equals the maximum mx
• If yes, we increment our answer counter

Step 4: Recursive Cases For each element at index i, we explore two branches:

1. Exclude the element: dfs(i + 1, t) - Move to the next index without changing the OR value
2. Include the element: dfs(i + 1, t | nums[i]) - Move to the next index and update OR value by including nums[i]

Step 5: Initialize and Execute

ans = 0
mx = reduce(lambda x, y: x | y, nums)
dfs(0, 0)
return ans

We initialize:

• ans = 0: Counter for valid subsets
• Start DFS from index 0 with initial OR value of 0
• Return the final count

Time Complexity: O(2^n) where n is the length of nums, as we explore all possible subsets.

Space Complexity: O(n) for the recursion stack depth.

The beauty of this approach is that we don't need to store all subsets - we generate them implicitly through our recursive calls and only count those matching our criteria. The nonlocal keyword allows the nested function to modify the ans variable in the outer scope, accumulating our count across all recursive calls.

Example Walkthrough

Let's walk through a concrete example with nums = [3, 1] to illustrate how the solution works.

Step 1: Calculate Maximum OR Value

• Binary representations: 3 = 11, 1 = 01
• Maximum OR = 3 | 1 = 11 | 01 = 11 = 3
• So our target value mx = 3

Step 2: DFS Exploration Starting with dfs(0, 0), we'll trace through all recursive calls:

dfs(0, 0)  [Starting at index 0, current OR = 0]
├── dfs(1, 0)  [Skip nums[0]=3, OR stays 0]
│   ├── dfs(2, 0)  [Skip nums[1]=1, OR stays 0]
│   │   └── Base case: i=2, t=0 ≠ mx=3, don't count
│   └── dfs(2, 0|1=1)  [Include nums[1]=1, OR becomes 1]
│       └── Base case: i=2, t=1 ≠ mx=3, don't count
└── dfs(1, 0|3=3)  [Include nums[0]=3, OR becomes 3]
    ├── dfs(2, 3)  [Skip nums[1]=1, OR stays 3]
    │   └── Base case: i=2, t=3 = mx=3, count! ans=1
    └── dfs(2, 3|1=3)  [Include nums[1]=1, OR stays 3]
        └── Base case: i=2, t=3 = mx=3, count! ans=2

Step 3: Result Analysis We found 2 subsets that produce the maximum OR value of 3:

1. [3] - Just element at index 0
2. [3, 1] - Elements at indices 0 and 1

The answer is 2.

Notice how:

• The subset [1] gives OR = 1, which is less than maximum
• The empty subset gives OR = 0, which is less than maximum
• Only subsets containing the element 3 can achieve the maximum OR value of 3, since 3 already has all the bits we need

Solution Implementation

Time and Space Complexity

Time Complexity: O(2^n)

The algorithm uses a depth-first search (DFS) approach to explore all possible subsets of the array. At each index i, the function makes two recursive calls:

• One call excludes the current element: dfs(i + 1, t)
• One call includes the current element: dfs(i + 1, t | nums[i])

This creates a binary decision tree where each level represents whether to include or exclude an element. With n elements in the array, the tree has a height of n and contains 2^n leaf nodes, representing all possible subsets. Each subset is visited exactly once, resulting in O(2^n) time complexity.

Space Complexity: O(n)

The space complexity is determined by the maximum depth of the recursive call stack. Since the recursion goes as deep as the length of the array (from index 0 to n), the call stack can have at most n recursive calls active at any given time. Each recursive call uses O(1) space for its parameters and local variables. Therefore, the total space complexity is O(n).

Note that the variables ans and mx are shared across all recursive calls using nonlocal and don't contribute additional space for each call.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Handle Empty Subsets

A common mistake is accidentally counting the empty subset when the initial OR value is 0. While the problem specifies "non-empty subsets", the DFS approach naturally generates an empty subset when we skip all elements.

Problematic scenario:

# If nums = [0], the maximum OR is 0
# The DFS would count both the empty subset (OR = 0) and subset [0] (OR = 0)
# This gives count = 2, but the correct answer is 1 (only [0])

Solution: Add a check to ensure we're only counting non-empty subsets:

def dfs(index: int, current_or: int, selected_any: bool) -> None:
    nonlocal count, max_or
  
    if index == len(nums):
        # Only count if we selected at least one element
        if selected_any and current_or == max_or:
            count += 1
        return
  
    # Exclude current element
    dfs(index + 1, current_or, selected_any)
  
    # Include current element (mark that we've selected something)
    dfs(index + 1, current_or | nums[index], True)

# Start with selected_any = False
dfs(0, 0, False)

2. Integer Overflow in Other Languages

While Python handles arbitrary-precision integers automatically, implementing this in languages like C++ or Java could lead to overflow issues with bitwise operations on large numbers.

Solution for other languages: Use appropriate data types (e.g., long long in C++ or long in Java) and be mindful of the constraints on input values.

3. Inefficient Maximum OR Calculation

Some might try to calculate the maximum OR within the DFS itself, leading to unnecessary comparisons and potential errors:

# Inefficient approach - tracking maximum during DFS
def dfs(index, current_or):
    global max_or
    max_or = max(max_or, current_or)  # Extra work in every call
    # ... rest of the logic

Solution: Pre-calculate the maximum OR as shown in the original solution. Since OR operations can only add bits (never remove them), the OR of all elements gives the maximum possible value immediately.

4. Using Global Variables Instead of Nonlocal

Using global instead of nonlocal or passing mutable objects can lead to issues:

# Problematic if this is inside a class method
global count  # This references module-level variable, not instance variable

Solution: Use nonlocal for nested functions or pass a mutable container:

# Option 1: nonlocal (as shown in original)
nonlocal count

# Option 2: Mutable container
result = [0]  # Use list to store count
def dfs(index, current_or):
    if index == len(nums):
        if current_or == max_or:
            result[0] += 1
    # ...

5. Not Understanding Subset vs Subsequence

Some might confuse this with finding subsequences and try to maintain order, which is unnecessary for subsets.

Key distinction:

• Subset: Any selection of elements (order doesn't matter)
• Subsequence: Selection maintaining relative order

The DFS approach correctly handles subsets by considering include/exclude decisions for each index position.