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>

class Solution:
    def countPairs(self, nums: List[int], target: int) -> int:
        # Binary Search
        nums.sort()  
        count = 0
    
        for i in range(len(nums) - 1):
            left, right = i + 1, len(nums) - 1
            
            # Binary search element that added to nums[i] is not less than target.
            while left <= right:
                mid = left + (right - left) // 2
                if nums[i] + nums[mid] < target:
                    left = mid + 1
                else:
                    right = mid - 1
            # All elements between i+1 and right are valid pairs
            count += right - i
        
        return count

        # O(n * log(n)) time.
        # O(n) auxiliary space.
        # O(n) total space.
16400