package twopointers
import "sort"
func threeSumMulti(arr []int, target int) int {
mod := 1000000000
ans := 0
n := len(arr)
n1 := n - 1
n2 := n - 2
sort.Ints(arr)
for i := 0; i < n2; i++ {
t := target - arr[i]
j, k := i+1, n1
for j < k {
if arr[j]+arr[k] < t {
j++
} else if arr[j]+arr[k] > t {
k--
} else if arr[j] != arr[k] {
left, right := 1, 1
for j+left < k && arr[j] == arr[j+left] {
left++
}
j += left
for j < k-right && arr[k] == arr[k-right] {
right++
}
k -= right
ans += (left * right) % mod
} else {
m := k - j + 1
ans += (m * (m - 1) / 2) % mod
break
}
}
}
return ans % mod
}
/*
923. 3Sum With Multiplicity
Given an integer array arr, and an integer target, return the number of tuples i, j, k such that i < j < k and arr[i] + arr[j] + arr[k] == target.
As the answer can be very large, return it modulo 109 + 7.
Example 1:
Input: arr = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation:
Enumerating by the values (arr[i], arr[j], arr[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: arr = [1,1,2,2,2,2], target = 5
Output: 12
Explanation:
arr[i] = 1, arr[j] = arr[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.
Example 3:
Input: arr = [2,1,3], target = 6
Output: 1
Explanation: (1, 2, 3) occured one time in the array so we return 1.
Constraints:
3 <= arr.length <= 3000
0 <= arr[i] <= 100
0 <= target <= 300
923. 三数之和的多种可能 https://leetcode.cn/problems/3sum-with-multiplicity/
*/
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标签：arr,right,target,times,漏洞,occurs,left From： https://www.cnblogs.com/rsapaper/p/16964874.html