What is the time complexity of Maximum Product of Two Integers With No Common Bits?

The optimal solution for Maximum Product of Two Integers With No Common Bits runs in O(n + B²) time and uses O(B) space. We process each number once and check pairs of masks, where B is the bit-width (up to 20).

What is the brute force solution for Maximum Product of Two Integers With No Common Bits?

The brute force approach for Maximum Product of Two Integers With No Common Bits is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every pair of numbers to see if they share common bits. If they don't, calculate their product and keep track of the maximum. The optimal Optimal Solution approach improves this to O(n + B²).

What algorithm or data structure is used to solve Maximum Product of Two Integers With No Common Bits?

Maximum Product of Two Integers With No Common Bits uses the following concepts: Array, Dynamic Programming, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Dynamic Programming.

What are the interview tips for Maximum Product of Two Integers With No Common Bits?

Explain your thought process clearly. Discuss edge cases, like all numbers sharing bits. Practice bit manipulation problems.

What are common mistakes when solving Maximum Product of Two Integers With No Common Bits?

Not checking for distinct indices. Overlooking the bitmask representation.

#3670

Maximum Product of Two Integers With No Common Bits

Medium

ArrayDynamic ProgrammingBit ManipulationBit ManipulationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + B²)
Space	O(1)	O(B)

💡

Intuition

Time O(n + B²)Space O(B)

Use bit masks to categorize numbers and find the maximum product without common bits efficiently.

⚙️

Algorithm

3 steps

1Step 1: Create a list to store the maximum number for each bitmask (up to 2^20 for 1 ≤ nums[i] ≤ 10^6).
2Step 2: For each number, update the corresponding bitmask with the maximum value found.
3Step 3: Iterate through all pairs of masks, checking for no common bits and calculating the product.

solution.py10 lines

1def maxProduct(nums):
2    max_mask = [0] * (1 << 20)
3    for num in nums:
4        max_mask[num] = max(max_mask[num], num)
5    max_product = 0
6    for i in range(len(max_mask)):
7        for j in range(i + 1, len(max_mask)):
8            if (i & j) == 0:
9                max_product = max(max_product, max_mask[i] * max_mask[j])
10    return max_product

ℹ

Complexity note: We process each number once and check pairs of masks, where B is the bit-width (up to 20).

1Bit manipulation helps efficiently check for common bits.
2Using masks allows us to categorize numbers and maximize products.

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