Add solution #825

Author: JoshCrozier

README.md

@@ -633,6 +633,7 @@
 822|[Card Flipping Game](./0822-card-flipping-game.js)|Medium|
 823|[Binary Trees With Factors](./0823-binary-trees-with-factors.js)|Medium|
 824|[Goat Latin](./0824-goat-latin.js)|Easy|
+825|[Friends Of Appropriate Ages](./0825-friends-of-appropriate-ages.js)|Medium|
 827|[Making A Large Island](./0827-making-a-large-island.js)|Hard|
 830|[Positions of Large Groups](./0830-positions-of-large-groups.js)|Easy|
 831|[Masking Personal Information](./0831-masking-personal-information.js)|Medium|

solutions/0825-friends-of-appropriate-ages.js

@@ -0,0 +1,51 @@
+/**
+ * 825. Friends Of Appropriate Ages
+ * https://leetcode.com/problems/friends-of-appropriate-ages/
+ * Difficulty: Medium
+ *
+ * There are n persons on a social media website. You are given an integer array ages where ages[i]
+ * is the age of the ith person.
+ *
+ * A Person x will not send a friend request to a person y (x != y) if any of the following
+ * conditions is true:
+ * - age[y] <= 0.5 * age[x] + 7
+ * - age[y] > age[x]
+ * - age[y] > 100 && age[x] < 100
+ *
+ * Otherwise, x will send a friend request to y.
+ *
+ * Note that if x sends a request to y, y will not necessarily send a request to x. Also, a person
+ * will not send a friend request to themself.
+ *
+ * Return the total number of friend requests made.
+ */
+
+/**
+ * @param {number[]} ages
+ * @return {number}
+ */
+var numFriendRequests = function(ages) {
+  const ageCount = new Array(121).fill(0);
+
+  for (const age of ages) {
+    ageCount[age]++;
+  }
+
+  let result = 0;
+  for (let ageX = 1; ageX <= 120; ageX++) {
+    if (ageCount[ageX] === 0) continue;
+    for (let ageY = 1; ageY <= 120; ageY++) {
+      if (ageCount[ageY] === 0) continue;
+      if (ageY <= 0.5 * ageX + 7) continue;
+      if (ageY > ageX) continue;
+      if (ageY > 100 && ageX < 100) continue;
+      if (ageX === ageY) {
+        result += ageCount[ageX] * (ageCount[ageY] - 1);
+      } else {
+        result += ageCount[ageX] * ageCount[ageY];
+      }
+    }
+  }
+
+  return result;
+};