How do you solve Number of Excellent Pairs on LeetCode?

Number of Excellent Pairs (LeetCode #2354) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding bit manipulation is key to solving this problem efficiently.

What is the brute force solution for Number of Excellent Pairs?

The brute force approach for Number of Excellent Pairs is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach checks every possible pair of numbers in the array to see if they satisfy the excellent pair condition. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Excellent Pairs?

Number of Excellent Pairs uses the following concepts: Array, Hash Table, Binary Search, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Excellent Pairs?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as when k is very large or very small. Practice bit manipulation problems to become comfortable with operations like AND and OR.

What are common mistakes when solving Number of Excellent Pairs?

Not considering the frequency of numbers when counting pairs. Confusing the conditions for excellent pairs, especially with bitwise operations.

#2354

Number of Excellent Pairs

Hard

ArrayHash TableBinary SearchBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

By using a frequency map to count the occurrences of each number and leveraging the properties of bit manipulation, we can reduce the time complexity significantly. We can precompute the number of set bits for each unique number and then check combinations efficiently.

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Algorithm

4 steps

1Step 1: Create a frequency map to count occurrences of each number in nums.
2Step 2: Create a list of unique numbers and compute the number of set bits for each.
3Step 3: For each unique number, check how many other unique numbers can form an excellent pair based on their set bits.
4Step 4: Use binary search or two pointers to efficiently find pairs that meet the condition.

solution.py15 lines

1# Full working Python code
2from collections import Counter
3
4
5def countExcellentPairs(nums, k):
6    freq = Counter(nums)
7    unique_nums = list(freq.keys())
8    set_bits = [bin(num).count('1') for num in unique_nums]
9    count = 0
10    for i in range(len(unique_nums)):
11        for j in range(len(unique_nums)):
12            if set_bits[i] + set_bits[j] >= k:
13                count += freq[unique_nums[i]] * freq[unique_nums[j]]
14    return count
15

ℹ

Complexity note: The time complexity is O(n) for counting frequencies and O(n²) in the worst case for checking pairs, but since we leverage the frequency map, it can be optimized further. The space complexity is O(n) for storing the frequency map.

1Understanding bit manipulation is key to solving this problem efficiently.
2Using a frequency map allows us to count pairs without explicitly generating them.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.