Tags: "leetcode", "two-pointers", access_time 2-min read

3sum With Multiplicity

Created: August 24, 2019 by [lek-tin]

Last updated: August 24, 2019

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

Solution

class Solution:
    def threeSumMulti(self, nums: List[int], target: int) -> int:
        if not nums or len(nums)  < 3:
            return -1

        counter = collections.Counter(nums)
        res = 0

        for a in range(101):
            for b in range(a, 101):
                c = target - a - b
                if c > 100 or c < 0:
                    continue
                if a == b and b == c:
                    res += counter[a] * (counter[a] - 1) * (counter[a] - 2) / (1 * 2 * 3)
                elif a == b and b != c:
                    res += counter[a] * (counter[a] - 1) / (1 * 2) * counter[c]
                elif b < c:
                    res += counter[a] * counter[b] * counter[c]

        return int(res % (1e9 + 7))