Sync LeetCode submission Runtime - 95 ms (100.00%), Memory - 30.6 MB (91.73%)

Author: jimit105

2699-count-the-number-of-fair-pairs/README.md

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+<p>Given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>n</code> and two integers <code>lower</code> and <code>upper</code>, return <em>the number of fair pairs</em>.</p>
+
+<p>A pair <code>(i, j)</code> is <b>fair </b>if:</p>
+
+<ul>
+	<li><code>0 &lt;= i &lt; j &lt; n</code>, and</li>
+	<li><code>lower &lt;= nums[i] + nums[j] &lt;= upper</code></li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [0,1,7,4,4,5], lower = 3, upper = 6
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> There are 6 fair pairs: (0,3), (0,4), (0,5), (1,3), (1,4), and (1,5).
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,7,9,2,5], lower = 11, upper = 11
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> There is a single fair pair: (2,3).
+</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>nums.length == n</code></li>
+	<li><code><font face="monospace">-10<sup>9</sup></font>&nbsp;&lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code><font face="monospace">-10<sup>9</sup>&nbsp;&lt;= lower &lt;= upper &lt;= 10<sup>9</sup></font></code></li>
+</ul>

2699-count-the-number-of-fair-pairs/solution.py

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+# Approach 2: Two Pointer
+
+# Time: O(n log n)
+# Space: O(n)
+
+class Solution:
+    def countFairPairs(self, nums: List[int], lower: int, upper: int) -> int:
+        nums.sort()
+        return self.lower_bound(nums, upper + 1) - self.lower_bound(nums, lower)
+
+    # Calculate the number of pairs with sum less than `value`.
+    def lower_bound(self, nums, value):
+        left, right = 0, len(nums) - 1
+        result = 0
+
+        while left < right:
+            if nums[left] + nums[right] < value:
+                # If sum is less than value, add the size of window to result and move to the next index
+                result += right - left
+                left += 1
+            else:
+                # Otherwise, shift the right pointer backwards, until we get a valid window.
+                right -= 1
+
+        return result
+