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Formatted question description: https://leetcode.ca/all/923.html

# 923. 3Sum With Multiplicity (Medium)

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

Companies:
Quora

Related Topics:
Two Pointers

## Solution 1.

• class Solution {
public int threeSumMulti(int[] A, int target) {
final int MODULO = 1000000007;
long count = 0;
Arrays.sort(A);
int length = A.length;
int leftStart = 0, leftEnd = length - 3;
for (int left = leftStart; left <= leftEnd; left++) {
if (A[left] > target)
break;
int remainTarget = target - A[left];
int mid = left + 1, right = length - 1;
while (mid < right) {
int sum = A[mid] + A[right];
if (sum == remainTarget) {
if (A[mid] != A[right]) {
int midCount = 1, rightCount = 1;
while (mid + 1 < right && A[mid] == A[mid + 1]) {
mid++;
midCount++;
}
while (right - 1 > mid && A[right] == A[right - 1]) {
right--;
rightCount++;
}
count += midCount * rightCount;
count %= MODULO;
mid++;
right--;
} else {
int curCount = right - mid + 1;
count += curCount * (curCount - 1) / 2;
count %= MODULO;
break;
}
} else if (sum < remainTarget)
mid++;
else
right--;
}
}
return (int) count;
}
}

############

class Solution {
private static final int MOD = (int) 1e9 + 7;

public int threeSumMulti(int[] arr, int target) {
int[] cnt = new int[101];
for (int v : arr) {
++cnt[v];
}
long ans = 0;
for (int j = 0; j < arr.length; ++j) {
int b = arr[j];
--cnt[b];
for (int i = 0; i < j; ++i) {
int a = arr[i];
int c = target - a - b;
if (c >= 0 && c <= 100) {
ans = (ans + cnt[c]) % MOD;
}
}
}
return (int) ans;
}
}

• // OJ: https://leetcode.com/problems/3sum-with-multiplicity/solution/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
int threeSumMulti(vector<int>& A, int target) {
unordered_map<int, long> m;
long ans = 0, mod = 1e9 + 7;
for (int n : A) m[n]++;
for (auto &[a, ca] : m) {
for (auto &[b, cb] : m) {
int c = target - a - b;
if (m.count(c) == 0) continue;
int cc = m[c];
if (a == b && b == c) ans += ca * (ca - 1) * (ca - 2) / 6; // all three are equal
else if (a == b && b != c) ans += ca * (ca - 1) / 2 * cc; // first two are equal and the 3rd is not
else if (a < b && b < c) ans += ca * cb * cc; // all three are not equal
}
}
return ans % mod;
}
};

• class Solution:
def threeSumMulti(self, arr: List[int], target: int) -> int:
cnt = Counter(arr)
ans = 0
mod = 10**9 + 7
for j, b in enumerate(arr):
cnt[b] -= 1
for i in range(j):
a = arr[i]
c = target - a - b
ans = (ans + cnt[c]) % mod
return ans

############

class Solution(object):
def threeSumMulti(self, A, target):
"""
:type A: List[int]
:type target: int
:rtype: int
"""
count = collections.Counter(A)
Aset = set(A)
Alist = list(Aset)
Alist.sort()
_lenA = len(Alist)
res = 0
for i in range(_lenA):
for j in range(i, _lenA):
c = target - Alist[i] - Alist[j]
if c >= Alist[j] and c in Aset:
if Alist[i] != Alist[j] != c:
res += count[Alist[i]] * count[Alist[j]] * count[c]
elif Alist[i] == Alist[j] and Alist[j] != c:
res += count[c] * self.caculate(count[Alist[i]], 2)
elif Alist[j] == c and Alist[i] != Alist[j]:
res += count[Alist[i]] * self.caculate(count[Alist[j]], 2)
elif Alist[i] == c and Alist[j] != c:
res += count[Alist[j]] * self.caculate(count[c], 2)
else:
res += self.caculate(count[Alist[i]], 3)
return res % (10 ** 9 + 7)

def caculate(self, x, i):
if i == 2:
return x * (x - 1) / 2
elif i == 3:
return x * (x - 1) * (x - 2) / 6

• func threeSumMulti(arr []int, target int) int {
const mod int = 1e9 + 7
cnt := [101]int{}
for _, v := range arr {
cnt[v]++
}
ans := 0
for j, b := range arr {
cnt[b]--
for i := 0; i < j; i++ {
a := arr[i]
c := target - a - b
if c >= 0 && c <= 100 {
ans += cnt[c]
ans %= mod
}
}
}
return ans
}