Description
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.
给定一个整数数组 A，以及一个整数 target 作为目标值，返回满足 i < j < k 且 A[i] + A[j] + A[k] == target 的元组 i, j, k 的数量。
由于结果会非常大，请返回 结果除以 10^9 + 7 的余数。
题目链接：https://leetcode.com/problems/3sum-with-multiplicity/description/
Difficulty: medium
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
分析
- 如果直接在A中选取三个数字，然后计算和是否为target的话，时间复杂度为 $O(n^3)$
- 对A进行压缩计数，得到{1:2,2:2,3:3,4:2,5:2}(如例一)
- 从[1,2,3,4,5]中选取三个数，但是每个数被选中的次数不能超过它在A中出现的次数（在judge函数中实现，judge(a,b)相当于从b个中选取a个的可能性）
- 三层循环得到三个数字，如和为target，则返回所有数字出现的可能性的乘积，即当前数字的可能的组合个数；
- 反之，继续；
- 返回所有满足条件的组合总数。
参考代码
class Solution:
def threeSumMulti(self, A, target):
import collections
count=collections.Counter(A)
s=0
def judge(a,b):
if(a>b):
return 0
s=1
for i in range(b,a,-1):
s*=i
for i in range(2,b-a+1):
s//=i
return s
li=list(count.keys())
for i in range(len(li)):
for j in range(i,len(li)):
for k in range(j,len(li)):
index1=li[i]
index2=li[j]
index3=li[k]
if(index1+index2+index3!=target):
continue
countIndex=collections.Counter([index1,index2,index3])
curr = 1
for index in countIndex:
curr*=judge(countIndex[index],count[index])
s+=curr
s%=(10**9+7)
return s