923. 3Sum With Multiplicity
阿新 • • 發佈:2018-11-18
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
思路:ijk順序什麼的,其實無所謂
因為就3個數,考慮以下3種情況:
1. 3個數都不同(注意除以6出掉重複)
2. 2個數相同
3. 3個數都相同
class Solution(object): def threeSumMulti(self, A, target): """ :type A: List[int] :type target: int :rtype: int """ mod=10**9+7 res=0 a=A d={} for i in a: d[i]=d.get(i,0)+1 for i in d: for j in d: if i==j or target-i-j==i or target-i-j==j or target-i-j not in d: continue res+=d[i]*d[j]*d[target-i-j] res//=6 # print(res) for i in d: if d[i]<2 or target-i-i==i or target-i-i not in d: continue res+=(d[i]*(d[i]-1)//2)*d[target-i-i] # print(res) for i in d: if d[i]<3 or target!=i*3: continue res+=d[i]*(d[i]-1)*(d[i]-2)//6 return res%mod s=Solution() print(s.threeSumMulti(A = [1,1,2,2,3,3,4,4,5,5], target = 8)) print(s.threeSumMulti(A = [1,1,2,2,2,2], target = 5)) print(s.threeSumMulti(A = [16,51,36,29,84,80,46,97,84,16], target = 171))