# 923. 3Sum With Multiplicity

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
``````

• code
``````class Solution(object):
def threeSumMulti(self, arr: List[int], target: int) -> int:
MOD = 10**9 + 7
count = Counter(arr)
ans = 0

# All different
for x in range(101):
for y in range(x+1, 101):
z = target - x - y
if y < z <= 100:
ans += count[x] * count[y] * count[z]

# x == y
for x in range(101):
z = target - 2*x
if x < z <= 100:
ans += count[x] * (count[x] - 1) // 2 * count[z]

# y == z
for x in range(101):
if (target - x) % 2 == 0:
y = (target - x) // 2
if x < y <= 100:
ans += count[x] * count[y] * (count[y] - 1) // 2

# x == y == z
if target % 3 == 0:
x = target // 3
if 0 <= x <= 100:
ans += count[x] * (count[x] - 1) * (count[x] - 2) // 6

return ans % MOD
``````
• code
``````class Solution:
def threeSumMulti(self, arr: List[int], target: int) -> int:
c = collections.Counter(arr)
res = 0
for i, j in itertools.combinations_with_replacement(c, 2):
# print(i, j)
k = target - i - j
if i == j == k: res += c[i] * (c[i] - 1) * (c[i] - 2) // 6
elif i == j != k: res += c[i] * (c[i] - 1) // 2 * c[k]
elif k > i and k > j: res += c[i] * c[j] * c[k]
return res % (10**9 + 7)
``````