# Leetcode: 3Sum With Multiplicity

3Sum With Multiplicity

Similar Problems:

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 10^9 + 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.
```

Note:

1. 3 <= A.length <= 3000
2. 0 <= A[i] <= 100
3. 0 <= target <= 300

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:
```// Blog link: https://code.dennyzhang.com/3sum-with-multiplicity
// Basic Ideas: fixed value range
// If a + b + c == target, and a, b, c are all different
//      count(a) * count(b) * count(c)
//
// Complexity: Time O(n), Space O(1)
import "math"
func threeSumMulti(A []int, target int) int {
mod := int(math.Pow(10, 9)) + 7
l := make([]int, 101)
for _, num := range A { l[num]++ }
res := 0
for i:=0; i<101; i++ {
if l[i]==0 { continue }
for j:=i; j<101; j++ {
k := target-i-j
if k>=j && k<101 {
// find a match: i+j+k
if i==j {
if j == k {
res += (l[i]*(l[i]-1)*(l[i]-2))/6
} else {
res += (l[i]*(l[i]-1)*l[k])/2
}
} else {
if j == k {
res += (l[i]*l[j]*(l[j]-1))/2
} else {
res += l[i]*l[j]*l[k]
}
}
}
res = res % mod
}
}
return res
}
```

Share It, If You Like It.