923. 3Sum With Multiplicity #

Problem #

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:

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

Problem Summary #

This problem is an upgraded version of Problem 15. Given an array, find the number of combinations of 3 numbers whose sum equals target, with the requirement that i < j < k. The number of combinations of solutions does not need to be deduplicated; the same values with different indices count as different solutions (this is also the difference from Problem 15).

Solution Approach #

The general solution for this problem is the same as Problem 15, except that when counting all solution combinations, some knowledge of permutations and combinations is needed. If 3 identical numbers are chosen, calculate C n 3; when choosing 2 identical numbers, calculate C n 2; choosing one number is calculated normally. Finally, add up the number of all solutions.

Code #

package leetcode

import (
	"sort"
)

func threeSumMulti(A []int, target int) int {
	mod := 1000000007
	counter := map[int]int{}
	for _, value := range A {
		counter[value]++
	}

	uniqNums := []int{}
	for key := range counter {
		uniqNums = append(uniqNums, key)
	}
	sort.Ints(uniqNums)

	res := 0
	for i := 0; i < len(uniqNums); i++ {
		ni := counter[uniqNums[i]]
		if (uniqNums[i]*3 == target) && counter[uniqNums[i]] >= 3 {
			res += ni * (ni - 1) * (ni - 2) / 6
		}
		for j := i + 1; j < len(uniqNums); j++ {
			nj := counter[uniqNums[j]]
			if (uniqNums[i]*2+uniqNums[j] == target) && counter[uniqNums[i]] > 1 {
				res += ni * (ni - 1) / 2 * nj
			}
			if (uniqNums[j]*2+uniqNums[i] == target) && counter[uniqNums[j]] > 1 {
				res += nj * (nj - 1) / 2 * ni
			}
			c := target - uniqNums[i] - uniqNums[j]
			if c > uniqNums[j] && counter[c] > 0 {
				res += ni * nj * counter[c]
			}
		}
	}
	return res % mod
}