Largest Combination With Bitwise AND Greater Than Zero — LeetCode #2275 (Medium)
Tags: Array, Hash Table, Bit Manipulation, Counting
Related patterns: Hash Map, Array
Brute Force approach
Time complexity: O(n²). Space complexity: O(1).
The brute force approach involves generating all possible combinations of the array elements and calculating their bitwise AND. This is straightforward but inefficient, especially for larger arrays.
The time complexity is O(n²) because we are generating combinations and calculating the AND for each one, which can take up to n operations for each combination.
- Step 1: Generate all possible combinations of the elements in the candidates array.
- Step 2: For each combination, calculate the bitwise AND of its elements.
- Step 3: Keep track of the size of the largest combination that has a bitwise AND greater than 0.
For candidates = [16, 17, 71, 62, 12, 24, 14]:
1. r = 1: combinations = [16], AND = 16 (size = 1)
2. r = 2: combinations = [16, 17], AND = 0 (size = 2)
3. r = 3: combinations = [16, 17, 71], AND = 0 (size = 3)
4. r = 4: combinations = [16, 17, 62, 24], AND = 16 (size = 4)
5. r = 5: combinations = [16, 17, 71, 62, 12], AND = 0 (size = 5)
6. r = 6: combinations = [16, 17, 71, 62, 12, 24], AND = 0 (size = 6)
Final max size = 4.
Optimal Solution approach
Time complexity: O(n). Space complexity: O(1).
The optimal solution leverages the fact that for the bitwise AND to be greater than zero, we can focus on each bit position and count how many numbers have that bit set. The maximum count for any bit position gives us the size of the largest combination.
The time complexity is O(n) because we iterate through the candidates once and check each of the 24 bits, which is a constant factor.
- Step 1: Initialize an array to count occurrences of each bit position (up to 24 bits).
- Step 2: For each number in the candidates, update the count for each bit position that is set (1).
- Step 3: The maximum value in the count array is the size of the largest combination with a bitwise AND greater than 0.
For candidates = [16, 17, 71, 62, 12, 24, 14]:
1. count = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
2. num = 16: count[4] = 1
3. num = 17: count[0] = 1, count[4] = 2
4. num = 71: count[0] = 2, count[1] = 1, count[2] = 1, count[4] = 3
5. num = 62: count[1] = 2, count[2] = 2, count[4] = 4
6. num = 12: count[2] = 3, count[3] = 1, count[4] = 5
Final max count = 5.
Key Insights
- The bitwise AND operation requires at least one bit to be 1 in all numbers for the result to be greater than 0.
- Focusing on individual bit positions allows us to efficiently count potential combinations.
Common Mistakes
- Students often try to brute-force combinations without realizing the inefficiency.
- Failing to consider the properties of bitwise operations can lead to incorrect assumptions about the problem.
Interview Tips
- Always start with a brute force solution to understand the problem better.
- Look for patterns in the data, especially with bit manipulation problems.
- Discuss your thought process out loud during the interview to demonstrate your understanding.