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.
Constraints:
3 <= arr.length <= 30000 <= arr[i] <= 1000 <= target <= 300struct Solution;
const MOD: usize = 1_000_000_007;
impl Solution {
fn three_sum_multi(a: Vec<i32>, target: i32) -> i32 {
let mut count = vec![0; 101];
for x in a {
count[x as usize] += 1;
}
let mut res = 0;
for x in 0..101 {
for y in x + 1..101 {
if x + y >= target as usize {
break;
}
for z in y + 1..101 {
if x + y + z > target as usize {
break;
}
if x + y + z == target as usize {
res += count[x] * count[y] * count[z];
res %= MOD;
}
}
}
}
for x in 0..101 {
for y in x + 1..101 {
if x + x + y != target as usize {
continue;
}
if count[x] > 1 {
res += count[x] * (count[x] - 1) / 2 * count[y];
res %= MOD;
}
}
}
for x in 0..101 {
for y in x + 1..101 {
if x + y + y != target as usize {
continue;
}
if count[y] > 1 {
res += count[x] * count[y] * (count[y] - 1) / 2;
res %= MOD;
}
}
}
for x in 0..101 {
if x + x + x != target as usize {
continue;
}
if count[x] > 2 {
res += count[x] * (count[x] - 1) * (count[x] - 2) / 6;
res %= MOD;
}
}
res as i32
}
}
#[test]
fn test() {
let a = vec![1, 1, 2, 2, 3, 3, 4, 4, 5, 5];
let target = 8;
let res = 20;
assert_eq!(Solution::three_sum_multi(a, target), res);
let a = vec![1, 1, 2, 2, 2, 2];
let target = 5;
let res = 12;
assert_eq!(Solution::three_sum_multi(a, target), res);
}