Sync LeetCode submission Runtime - 353 ms (78.14%), Memory - 32.7 MB (40.08%)

Author: jimit105

2720-minimize-the-maximum-difference-of-pairs/README.md

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+<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>

2720-minimize-the-maximum-difference-of-pairs/solution.py

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+class Solution:
+    def minimizeMax(self, nums: List[int], p: int) -> int:
+        nums.sort()
+        n = len(nums)
+
+        def countValidPairs(threshold):
+            idx, count = 0, 0
+            while idx < n - 1:
+                # If a valid pair is found, skip both numbers
+                if nums[idx + 1] - nums[idx] <= threshold:
+                    count += 1
+                    idx += 1
+                idx += 1
+
+            return count
+
+        left, right = 0, nums[-1] - nums[0]
+        while left < right:
+            mid = left + (right - left) // 2
+
+            # If there are enough pairs, look for a smaller threshold.
+            # Otherwise, look for a larger threshold.
+            if countValidPairs(mid) >= p:
+                right = mid
+            else:
+                left = mid + 1
+
+        return left
+