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#include "header.hpp"
#include <algorithm>
#include <cmath>
#include <map>
#include <utility>
int fact(int n) {
int64_t result = 1;
for (int i = 1; i <= n; ++i) {
result *= i;
}
return result;
}
int selection(int n, int m) {
return (fact(m) / (fact(n) * fact(std::abs(m - n))));
}
class Solution {
public:
static int threeSumMulti(std::vector<int> &arr, int target) {
int MOD = 1000000007;
long ans = 0;
std::sort(arr.begin(), arr.end());
for (int i = 0; i < arr.size(); ++i) {
// We'll try to find the number of i < j < k
// with A[j] + A[k] == T, where T = target - A[i].
// The below is a "two sum with multiplicity".
int T = target - arr[i];
int j = i + 1, k = arr.size() - 1;
while (j < k) {
// These steps proceed as in a typical two-sum.
if (arr[j] + arr[k] < T)
j++;
else if (arr[j] + arr[k] > T)
k--;
else if (arr[j] != arr[k]) { // We have A[j] + A[k] == T.
// Let's count "left": the number of A[j] == A[j+1] == A[j+2] == ...
// And similarly for "right".
int left = 1, right = 1;
while (j + 1 < k && arr[j] == arr[j + 1]) {
left++;
j++;
}
while (k - 1 > j && arr[k] == arr[k - 1]) {
right++;
k--;
}
ans += left * right;
ans %= MOD;
j++;
k--;
} else {
// M = k - j + 1
// We contributed M * (M-1) / 2 pairs.
ans += (k - j + 1) * (k - j) / 2;
ans %= MOD;
break;
}
}
}
return (int)ans;
}
};
int main(int argc, char **argv) {
std::vector<int> arr = {1, 1, 2, 2, 3, 3, 4, 4, 5, 5};
auto target = 8;
std::cout << Solution::threeSumMulti(arr, target) << "\n";
return 0;
}
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