What is the time complexity of Count Number of Maximum Bitwise-OR Subsets?

The optimal solution has a time complexity of O(2^n).

What data structures are used to solve Count Number of Maximum Bitwise-OR Subsets?

This problem uses Array, Backtracking, Bit Manipulation, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Number of Maximum Bitwise-OR Subsets?

Insight 1: The maximum bitwise OR can be found by taking the bitwise OR of all elements in the array. This is because the bitwise OR operation can only set bits to 1; it cannot clear them. Therefore, including all elements will maximize the OR value. Insight 2: A backtracking approach is suitable due to the constraint on the array length (n <= 16). This allows us to explore all possible subsets without exceeding reasonable time limits.

Which companies ask LeetCode 2044. Count Number of Maximum Bitwise-OR Subsets in interviews?

This problem is frequently asked at Citadel.

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Count Number of Maximum Bitwise-OR Subsets - LeetCode 2044 Solution

Q: How to solve Count Number of Maximum Bitwise-OR Subsets?

The solution employs a backtracking algorithm to explore all possible subsets of the given array. First, it calculates the maximum possible bitwise OR by OR-ing all elements in the array. Then, the backtracking function recursively explores subsets, calculating the bitwise OR of each subset. When a subset's OR equals the maximum OR, it means we have found a valid subset. Instead of just incrementing the count by 1, we add `2^(n - index)` to the count. This is because any combination of the remaining elements will also give the same max_or. If the current_or exceeds the max_or, the function stops exploring that branch because OR operation is non-decreasing. The base cases for the recursion are when `index` reaches the end of the array or when `current_or` already equals `max_or`.

Q: Which companies ask LeetCode 2044. Count Number of Maximum Bitwise-OR Subsets in interviews?

This problem is frequently asked at Citadel.

Count Number of Maximum Bitwise-OR Subsets - Complete Solution Guide

Count Number of Maximum Bitwise-OR Subsets is LeetCode problem 2044, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer array nums , find the maximum possible bitwise OR of a subset of nums and return the number of different non-empty subsets with the maximum bitwise OR . An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b . Two subsets are considered different if the indices of the elements chosen are different. The bitwise OR of an array a is equal to a[0] OR a[1] OR ... OR a[a.length - 1] ( 0-indexed ). Example 1: Input: nums = [3,1]

Detailed Explanation

The problem asks us to find the number of non-empty subsets of a given integer array `nums` such that the bitwise OR of the elements in each subset is equal to the maximum possible bitwise OR that can be achieved by any subset of `nums`. Essentially, we need to find the maximum OR value among all possible subsets and then count how many distinct subsets produce this maximum OR value. The constraints specify that the array length is at most 16, and the values in the array are between 1 and 10^5.

Solution Approach

The solution employs a backtracking algorithm to explore all possible subsets of the given array. First, it calculates the maximum possible bitwise OR by OR-ing all elements in the array. Then, the backtracking function recursively explores subsets, calculating the bitwise OR of each subset. When a subset's OR equals the maximum OR, it means we have found a valid subset. Instead of just incrementing the count by 1, we add `2^(n - index)` to the count. This is because any combination of the remaining elements will also give the same max_or. If the current_or exceeds the max_or, the function stops exploring that branch because OR operation is non-decreasing. The base cases for the recursion are when `index` reaches the end of the array or when `current_or` already equals `max_or`.

Step-by-Step Algorithm

Step 1: Calculate the `max_or` by taking the bitwise OR of all elements in the `nums` array.
Step 2: Initialize a `count` variable to 0. This variable will store the number of subsets with the maximum bitwise OR.
Step 3: Define a backtracking function `backtrack(index, current_or)` that takes the current index and the current bitwise OR as arguments.
Step 4: In the `backtrack` function, check if `current_or` is equal to `max_or`. If it is, it means that we have found a subset with the maximum bitwise OR. Increment the `count` by `2^(n - index)`. This represents the number of subsets including the remaining elements from index to the end of the array.
Step 5: Check if `index` is equal to `n` (the length of the array). If it is, it means we've reached the end of the array, and we should return.
Step 6: Recursively call the `backtrack` function twice: once without including the current element (`backtrack(index + 1, current_or)`) and once including the current element (`backtrack(index + 1, current_or | nums[index])`).
Step 7: Start the backtracking process by calling `backtrack(0, 0)`.
Step 8: Return the `count` after the backtracking is complete.

Key Insights

Insight 1: The maximum bitwise OR can be found by taking the bitwise OR of all elements in the array. This is because the bitwise OR operation can only set bits to 1; it cannot clear them. Therefore, including all elements will maximize the OR value.
Insight 2: A backtracking approach is suitable due to the constraint on the array length (n <= 16). This allows us to explore all possible subsets without exceeding reasonable time limits.
Insight 3: When a subset yields the maximum bitwise OR, we can optimize by recognizing that any further subset derived from that subset also contributes to the count. The number of such subsets is 2^(n - index) where index is current index in the backtracking call.

Complexity Analysis

Time Complexity: O(2^n)

Space Complexity: O(n)

Topics

This problem involves: Array, Backtracking, Bit Manipulation, Enumeration.

Companies

Asked at: Citadel.

Count Number of Maximum Bitwise-OR Subsets | LeetCode 2044

Medium

Array Backtracking Bit Manipulation Enumeration

Time: O(2^n)Space: O(n)

Solution Code

Available in:

class Solution {
    private int count = 0;
    private int max_or = 0;
    private int n;
    private int[] nums;

    public int countMaxOrSubsets(int[] nums) {
        this.nums = nums;
        n = nums.length;
        max_or = 0;
        for (int num : nums) {
            max_or |= num;
        }

        count = 0;

        backtrack(0, 0);
        return count;
    }

    private void backtrack(int index, int current_or) {
        if (current_or == max_or) {
            count += 1 << (n - index);
            return;
        }

        if (index == n) {
            return;
        }

        backtrack(index + 1, current_or);

        backtrack(index + 1, current_or | nums[index]);
    }
}

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Problem Info

Description

Problem 2044: Count Number of Maximum Bitwise-OR Subsets - Detailed explanation and solution approaches.

Topics

Array Backtracking Bit Manipulation Enumeration

Companies

Citadel

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2044

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Calculate the `max_or` by taking the bitwise OR of all elements in the `nums` array.
2
Step 2: Initialize a `count` variable to 0. This variable will store the number of subsets with the maximum bitwise OR.
3
Step 3: Define a backtracking function `backtrack(index, current_or)` that takes the current index and the current bitwise OR as arguments.
4
Step 4: In the `backtrack` function, check if `current_or` is equal to `max_or`. If it is, it means that we have found a subset with the maximum bitwise OR. Increment the `count` by `2^(n - index)`. This represents the number of subsets including the remaining elements from index to the end of the array.
5
Step 5: Check if `index` is equal to `n` (the length of the array). If it is, it means we've reached the end of the array, and we should return.
6
Step 6: Recursively call the `backtrack` function twice: once without including the current element (`backtrack(index + 1, current_or)`) and once including the current element (`backtrack(index + 1, current_or | nums[index])`).
7
Step 7: Start the backtracking process by calling `backtrack(0, 0)`.
8
Step 8: Return the `count` after the backtracking is complete.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(2^n)

🐌 Exponential - Very slow! Often needs optimization.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the fact that in the worst case, the backtracking function explores every possible subset. For each element in `nums`, we have two choices: include it or exclude it. This yields a binary tree structure with a depth of n, resulting in 2^n possible paths. The space complexity stems from the depth of the recursion, which is at most n, resulting in O(n) space for the call stack.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative approach using bit masking

Instead of recursion, we can iterate through all possible subsets using bit masking. Each bit in a number from 1 to 2^n - 1 represents whether an element is included in the subset. This method also exhaustively enumerates all subsets.

Time: O(n * 2^n), where n is the length of the array.Space: O(1), excluding the input array.

Trade-offs: The iterative approach might be slightly faster in practice due to the elimination of recursion overhead, but it has the same overall time complexity. The space complexity might be slightly better due to no recursive call stack, but this advantage is negligible for the constraint of n<=16. The backtracking version is arguably easier to read.

✨

Key Insights

1Insight 1: The maximum bitwise OR can be found by taking the bitwise OR of all elements in the array. This is because the bitwise OR operation can only set bits to 1; it cannot clear them. Therefore, including all elements will maximize the OR value.
2Insight 2: A backtracking approach is suitable due to the constraint on the array length (n <= 16). This allows us to explore all possible subsets without exceeding reasonable time limits.
3Insight 3: When a subset yields the maximum bitwise OR, we can optimize by recognizing that any further subset derived from that subset also contributes to the count. The number of such subsets is 2^(n - index) where index is current index in the backtracking call.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: An empty input array. The problem statement specifies that the array is non-empty, so this case isn't explicitly handled, but it would return 0 if the input was an empty list.
Edge case 2: An array with all identical elements. The algorithm will correctly identify that all non-empty subsets have the same bitwise OR and count them appropriately. Consider the array [2,2,2] as in the examples.
Edge case 3: An array with elements that when ORed together don't increase the overall OR. This means some elements effectively contribute nothing when ORed with the existing elements in the subset. The algorithm still correctly explores all subsets and counts those that achieve the `max_or`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not initializing `max_or` correctly. The initial value should be 0.

✗Common mistake 2: Incorrectly calculating the number of subsets with the maximum bitwise OR. When `current_or == max_or`, it's crucial to add `2^(n - index)` because including any combination of the remaining elements will still result in the same maximum OR.

✗Common mistake 3: Overlooking the bitwise OR operations precedence which is lower than comparison operators. Therefore, it's essential to wrap the bitwise OR operation in parenthesis when comparing it with `max_or` like so: `(current_or | nums[index])`

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Feature selection in machine learning. You might want to identify subsets of features that maximize a certain metric (e.g., information gain). This problem is conceptually similar to finding subsets with a maximum OR.

🏦Real-world scenario 2: Network configuration. Selecting optimal combinations of network devices or settings to maximize network performance or security, which could be modeled as a bitwise OR operation based on features provided by configurations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What is the difference between backtracking and brute force?

Backtracking prunes invalid paths early using constraints, while brute force explores all possibilities. In backtracking, you "choose" an option, "explore" it recursively, then "unchoose" (backtrack) if it doesn't lead to a solution. This eliminates entire branches of the search tree.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Number of Maximum Bitwise-OR Subsets"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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