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-rw-r--r--923/main.cpp76
1 files changed, 76 insertions, 0 deletions
diff --git a/923/main.cpp b/923/main.cpp
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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;
+}