923. 3Sum With Multiplicity

Medium (HARD)

---

Given an integer array arr, and an integer target, return the number of tuples i, j, k such that i < j < k and arr[i] + arr[j] + arr[k] == target.

As the answer can be very large, return it modulo 109 + 7.

Example 1:

Input: arr = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation: 
Enumerating by the values (arr[i], arr[j], arr[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.

Example 2:

Input: arr = [1,1,2,2,2,2], target = 5
Output: 12
Explanation: 
arr[i] = 1, arr[j] = arr[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.

Constraints:

• 3 <= arr.length <= 3000

• 0 <= arr[i] <= 100

• 0 <= target <= 300

class Solution:
    def threeSumMulti(self, arr: List[int], target: int) -> int:
        result = 0
        MOD = 10**9+7
        d = defaultdict(int)
        for i in range(len(arr)):
            for j in range(i+1, len(arr)):
                remainder = target - arr[i] - arr[j]
                # If `remainder` is in dict then simply get its count,
                # Because that will be the number of way this total sum can be acheived
                if remainder in d:
                    result += d[remainder]
            # Simply Counting Seen Elements so far
            d[arr[i]] += 1
        return result % MOD