How do you solve 3Sum With Multiplicity on LeetCode?

3Sum With Multiplicity (LeetCode #923) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using a frequency map helps avoid redundant calculations.

What is the brute force solution for 3Sum With Multiplicity?

The brute force approach for 3Sum With Multiplicity is Brute Force, with O(n³) time complexity and O(1) space complexity. The brute-force approach involves checking every possible triplet in the array to see if their sum equals the target. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for 3Sum With Multiplicity?

Always clarify how to handle duplicates in the input. Discuss your thought process while writing code to show understanding. Consider edge cases, such as when all elements are the same.

What are common mistakes when solving 3Sum With Multiplicity?

Failing to account for duplicate elements correctly. Not using modulo operation in the final count.

#923

3Sum With Multiplicity

Medium

ArrayHash TableTwo PointersSortingCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n³)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal solution uses a combination of sorting and a hash map to efficiently count the occurrences of each number, allowing us to find valid triplets without redundant calculations.

⚙️

Algorithm

5 steps

1Step 1: Sort the array to enable the use of two pointers.
2Step 2: Create a frequency map to count occurrences of each number in the array.
3Step 3: Iterate through each unique number as the first element of the triplet.
4Step 4: For each pair of subsequent numbers, calculate the required sum and check if it exists in the frequency map.
5Step 5: Use combinatorial counting to determine how many valid triplets can be formed based on the frequency of the numbers.

solution.py19 lines

1def threeSumMulti(arr, target):
2    from collections import Counter
3    count = 0
4    mod = 10**9 + 7
5    freq = Counter(arr)
6    unique = sorted(freq.keys())
7    for i in range(len(unique)):
8        for j in range(i, len(unique)):
9            k = target - unique[i] - unique[j]
10            if k in freq:
11                if i == j == unique.index(k):
12                    count += freq[unique[i]] * (freq[unique[i]] - 1) * (freq[unique[i]] - 2) // 6
13                elif i == j:
14                    count += freq[unique[i]] * (freq[unique[i]] - 1) // 2 * freq[k]
15                elif j == unique.index(k):
16                    count += freq[unique[j]] * (freq[unique[j]] - 1) // 2 * freq[unique[i]]
17                else:
18                    count += freq[unique[i]] * freq[unique[j]] * freq[k]
19    return count % mod

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops for unique values, while the space complexity is O(n) for storing the frequency map.

1Using a frequency map helps avoid redundant calculations.
2Sorting the array allows for efficient two-pointer techniques.

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