From 50dcc51aa192049484e4aaf8b260e8d900ec520d Mon Sep 17 00:00:00 2001
From: terminaldweller <thabogre@gmail.com>
Date: Wed, 6 Apr 2022 17:26:41 +0430
Subject: update

---
 923/main.cpp | 76 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 76 insertions(+)
 create mode 100644 923/main.cpp

(limited to '923/main.cpp')

diff --git a/923/main.cpp b/923/main.cpp
new file mode 100644
index 0000000..3e0349e
--- /dev/null
+++ b/923/main.cpp
@@ -0,0 +1,76 @@
+#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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