Add solution #2616

Author: JoshCrozier

README.md

@@ -1943,6 +1943,7 @@
 2609|[Find the Longest Balanced Substring of a Binary String](./solutions/2609-find-the-longest-balanced-substring-of-a-binary-string.js)|Easy|
 2610|[Convert an Array Into a 2D Array With Conditions](./solutions/2610-convert-an-array-into-a-2d-array-with-conditions.js)|Medium|
 2615|[Sum of Distances](./solutions/2615-sum-of-distances.js)|Medium|
+2616|[Minimize the Maximum Difference of Pairs](./solutions/2616-minimize-the-maximum-difference-of-pairs.js)|Medium|
 2618|[Check if Object Instance of Class](./solutions/2618-check-if-object-instance-of-class.js)|Medium|
 2619|[Array Prototype Last](./solutions/2619-array-prototype-last.js)|Easy|
 2620|[Counter](./solutions/2620-counter.js)|Easy|

solutions/2616-minimize-the-maximum-difference-of-pairs.js

@@ -0,0 +1,53 @@
+/**
+ * 2616. Minimize the Maximum Difference of Pairs
+ * https://leetcode.com/problems/minimize-the-maximum-difference-of-pairs/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums and an integer p. Find p pairs of indices
+ * of nums such that the maximum difference amongst all the pairs is minimized. Also, ensure
+ * no index appears more than once amongst the p pairs.
+ *
+ * Note that for a pair of elements at the index i and j, the difference of this pair is
+ * |nums[i] - nums[j]|, where |x| represents the absolute value of x.
+ *
+ * Return the minimum maximum difference among all p pairs. We define the maximum of an empty
+ * set to be zero.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} p
+ * @return {number}
+ */
+var minimizeMax = function(nums, p) {
+  nums.sort((a, b) => a - b);
+  const n = nums.length;
+
+  let left = 0;
+  let right = nums[n - 1] - nums[0];
+  let result = 0;
+
+  while (left <= right) {
+    const mid = Math.floor((left + right) / 2);
+    if (canFormPairs(mid)) {
+      result = mid;
+      right = mid - 1;
+    } else {
+      left = mid + 1;
+    }
+  }
+
+  return result;
+
+  function canFormPairs(maxDiff) {
+    let pairs = 0;
+    for (let i = 0; i < n - 1 && pairs < p; i += 2) {
+      if (nums[i + 1] - nums[i] <= maxDiff) {
+        pairs++;
+      } else {
+        i--;
+      }
+    }
+    return pairs >= p;
+  }
+};