feat: add solution 2616. Minimize the Maximum Difference of Pairs

Author: tzuyi0817

solutions/2616. Minimize the Maximum Difference of Pairs/README.md

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+# [2616. Minimize the Maximum Difference of Pairs](https://leetcode.com/problems/minimize-the-maximum-difference-of-pairs)
+
+## Description
+
+<div class="elfjS" data-track-load="description_content"><p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and an integer <code>p</code>. Find <code>p</code> pairs of indices of <code>nums</code> such that the <strong>maximum</strong> difference amongst all the pairs is <strong>minimized</strong>. Also, ensure no index appears more than once amongst the <code>p</code> pairs.</p>
+
+<p>Note that for a pair of elements at the index <code>i</code> and <code>j</code>, the difference of this pair is <code>|nums[i] - nums[j]|</code>, where <code>|x|</code> represents the <strong>absolute</strong> <strong>value</strong> of <code>x</code>.</p>
+
+<p>Return <em>the <strong>minimum</strong> <strong>maximum</strong> difference among all </em><code>p</code> <em>pairs.</em> We define the maximum of an empty set to be zero.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> nums = [10,1,2,7,1,3], p = 2
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. 
+The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> nums = [4,2,1,2], p = 1
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= p &lt;= (nums.length)/2</code></li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+## Solutions
+
+**Solution: `Binary Search + Greedy`**
+
+- Time complexity: <em>O(nlogn+nlog(maxDiff))</em>
+- Space complexity: <em>O(1)</em>
+
+<p>&nbsp;</p>
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} p
+ * @return {number}
+ */
+const minimizeMax = function (nums, p) {
+  const n = nums.length;
+
+  nums.sort((a, b) => a - b);
+
+  let left = 0;
+  let right = nums[n - 1] - nums[0];
+
+  const pairsCount = diff => {
+    let index = 1;
+    let count = 0;
+
+    while (index < n) {
+      const current = nums[index] - nums[index - 1];
+
+      if (current <= diff) {
+        count += 1;
+        index += 2;
+      } else {
+        index += 1;
+      }
+    }
+
+    return count;
+  };
+
+  while (left < right) {
+    const mid = Math.floor((left + right) / 2);
+
+    pairsCount(mid) >= p ? (right = mid) : (left = mid + 1);
+  }
+
+  return left;
+};
+```

solutions/2616. Minimize the Maximum Difference of Pairs/solution.js

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+/**
+ * @param {number[]} nums
+ * @param {number} p
+ * @return {number}
+ */
+const minimizeMax = function (nums, p) {
+  const n = nums.length;
+
+  nums.sort((a, b) => a - b);
+
+  let left = 0;
+  let right = nums[n - 1] - nums[0];
+
+  const pairsCount = diff => {
+    let index = 1;
+    let count = 0;
+
+    while (index < n) {
+      const current = nums[index] - nums[index - 1];
+
+      if (current <= diff) {
+        count += 1;
+        index += 2;
+      } else {
+        index += 1;
+      }
+    }
+
+    return count;
+  };
+
+  while (left < right) {
+    const mid = Math.floor((left + right) / 2);
+
+    pairsCount(mid) >= p ? (right = mid) : (left = mid + 1);
+  }
+
+  return left;
+};