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2917-count-pairs-whose-sum-is-less-than-target/README.md

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+Given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, return <em>the number of pairs</em> <code>(i, j)</code> <em>where</em> <code>0 &lt;= i &lt; j &lt; n</code> <em>and</em> <code>nums[i] + nums[j] &lt; target</code>.
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [-1,1,2,3,1], target = 2
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> There are 3 pairs of indices that satisfy the conditions in the statement:
+- (0, 1) since 0 &lt; 1 and nums[0] + nums[1] = 0 &lt; target
+- (0, 2) since 0 &lt; 2 and nums[0] + nums[2] = 1 &lt; target 
+- (0, 4) since 0 &lt; 4 and nums[0] + nums[4] = 0 &lt; target
+Note that (0, 3) is not counted since nums[0] + nums[3] is not strictly less than the target.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [-6,2,5,-2,-7,-1,3], target = -2
+<strong>Output:</strong> 10
+<strong>Explanation:</strong> There are 10 pairs of indices that satisfy the conditions in the statement:
+- (0, 1) since 0 &lt; 1 and nums[0] + nums[1] = -4 &lt; target
+- (0, 3) since 0 &lt; 3 and nums[0] + nums[3] = -8 &lt; target
+- (0, 4) since 0 &lt; 4 and nums[0] + nums[4] = -13 &lt; target
+- (0, 5) since 0 &lt; 5 and nums[0] + nums[5] = -7 &lt; target
+- (0, 6) since 0 &lt; 6 and nums[0] + nums[6] = -3 &lt; target
+- (1, 4) since 1 &lt; 4 and nums[1] + nums[4] = -5 &lt; target
+- (3, 4) since 3 &lt; 4 and nums[3] + nums[4] = -9 &lt; target
+- (3, 5) since 3 &lt; 5 and nums[3] + nums[5] = -3 &lt; target
+- (4, 5) since 4 &lt; 5 and nums[4] + nums[5] = -8 &lt; target
+- (4, 6) since 4 &lt; 6 and nums[4] + nums[6] = -4 &lt; target
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length == n &lt;= 50</code></li>
+	<li><code>-50 &lt;= nums[i], target &lt;= 50</code></li>
+</ul>

2917-count-pairs-whose-sum-is-less-than-target/solution.py

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+# Approach: Two Pointer
+
+# Time: O(n log n)
+# Space: O(n)
+
+class Solution:
+    def countPairs(self, nums: List[int], target: int) -> int:
+        nums.sort()
+        count = 0
+        left, right = 0, len(nums) - 1
+
+        while left < right:
+            if (nums[left] + nums[right]) < target:
+                count += right - left
+                left += 1
+            else:
+                right -= 1
+
+        return count
+