Let's walk through a solution to the 3Sum Multiplicity problem.

Naive Solution

The most straightforward approach is to use three nested loops to iterate through all possible combinations of i, j, and k such that i < j < k. For each combination, we check if arr[i] + arr[j] + arr[k] equals the target. If it does, we increment our count. Finally, we return the count modulo 10^9 + 7.

def threeSumMultiplicity_naive(arr, target):
    n = len(arr)
    count = 0
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if arr[i] + arr[j] + arr[k] == target:
                    count += 1
    return count % (10**9 + 7)

Time Complexity: O(n^3), due to the three nested loops. Space Complexity: O(1), as we only use a constant amount of extra space.

Optimal Solution

We can optimize this solution by using a counting approach. We can use a hash map (or in this case, since the constraints limit arr[i] to be between 0 and 100, an array) to store the frequency of each number in arr. Then, we can iterate through all possible pairs of numbers and check if the third number needed to reach the target exists in our frequency map.

def threeSumMultiplicity(arr, target):
    MOD = 10**9 + 7
    count = 0
    freq = {}
    for num in arr:
        freq[num] = freq.get(num, 0) + 1
    
    nums = sorted(freq.keys())

    for i in range(len(nums)):
        a = nums[i]
        for j in range(i, len(nums)):
            b = nums[j]
            c = target - a - b

            if c in freq and c >= b:
                if a == b == c:
                    count = (count + freq[a] * (freq[a] - 1) * (freq[a] - 2) // 6) % MOD
                elif a == b != c:
                    count = (count + freq[a] * (freq[a] - 1) // 2 * freq[c]) % MOD
                elif a != b == c:
                    count = (count + freq[a] * freq[b] * (freq[b] - 1) // 2) % MOD
                elif a < b < c:
                    count = (count + freq[a] * freq[b] * freq[c]) % MOD
                    
    return count

Time Complexity: O(n + k^2), where n is the length of the array and k is the number of distinct elements in the array. The n comes from counting the frequencies. The k^2 comes from iterating through the possible pairs. Because k is bounded by 101 in this problem, we consider the runtime to be closer to O(n). Space Complexity: O(k), where k is the number of distinct elements in the array, which is the space used to store the frequency map.

Edge Cases

• Empty array: If the input array is empty or has fewer than 3 elements, the function should return 0, as no triplets can be formed.
• Target out of range: If the target value is less than 0 or greater than 300 (since each element is between 0 and 100), the function might return 0 quickly.
• Large counts: The problem statement specifies that the answer can be very large, so we need to return the result modulo 10^9 + 7 to avoid integer overflow.
• Duplicate numbers: The optimal solution handles cases with duplicate numbers correctly by considering the frequencies of the numbers when calculating the number of triplets.