0923.3 Sum With Multiplicity

923. 3Sum With Multiplicity#

题目 #

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

代码 #

``````
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
}

``````

Apr 8, 2023