human
Solve in Python:
Given an integer array A, and an integer target, return the number of tuples i, j, k  such that i < j < k and A[i] + A[j] + A[k] == target.
As the answer can be very large, return it modulo 109 + 7.
 
Example 1:
Input: A = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation: 
Enumerating by the values (A[i], A[j], A[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: A = [1,1,2,2,2,2], target = 5
Output: 12
Explanation: 
A[i] = 1, A[j] = A[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 <= A.length <= 3000
0 <= A[i] <= 100
0 <= target <= 300

Write me a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def threeSumMulti(self, A: List[int], target: int) -> int:
        ADict = {}
        A = sorted(A)
        print(A)
        for a in A:
            if a in ADict:
                ADict[a]+=1
            else:
                ADict[a] = 1
        ijSame = False
        keys = sorted(ADict)
        totWays = 0
        # check all same:
        for i in range(len(keys)):
            if keys[i] > target:
                 break
            elif ADict[keys[i]]<3:
                continue
            elif keys[i]*3 == target:
                numOcc = ADict[keys[i]]
                totWays += numOcc*(numOcc-1)*(numOcc-2)/6
        
        if len(keys)==1:
            return int(totWays%(10**9 + 7))
        # 2 are same:
        for i in range(len(keys)):
            if keys[i] > target:
                 break
            elif ADict[keys[i]]<2:
                continue
            else:
                sum = keys[i]*2
                remain = target - sum
                if (not(remain==keys[i]) and remain in keys):
                    numDoubleOcc = ADict[keys[i]]
                    numSingleOcc = ADict[remain]
                    totWays += numDoubleOcc*(numDoubleOcc-1)/2*numSingleOcc
        if len(keys)==2:
            return int(totWays%(10**9 + 7))
        # all distinct
        for i in range(len(keys)):
            if keys[i] > target:
                 break
            for j in range(i+1, len(keys)):
                sum = keys[i] + keys[j]
                if sum > target:
                    break
                remaining = target - sum
                if remaining in keys[j+1:]:
                    totWays += ADict[keys[i]] *ADict[keys[j]] *ADict[remaining]
                
        return int(totWays%(10**9 + 7))
        
        
#         for i in range(len(keys)):
#             if keys[i] > target:
#                  return totWays
                
#             if ADict[keys[i]] ==1:
#                 startJ = i+1
#             else:
#                 startJ = i
#                ...