# Solving Leetcode Interviews in Seconds with AI: Count Pairs Whose Sum is Less than Target


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2824" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given a 0-indexed integer array nums of length n and an integer target, return the number of pairs (i, j) where 0 <= i < j < n and nums[i] + nums[j] < target.   Example 1:  Input: nums = [-1,1,2,3,1], target = 2 Output: 3 Explanation: There are 3 pairs of indices that satisfy the conditions in the statement: - (0, 1) since 0 < 1 and nums[0] + nums[1] = 0 < target - (0, 2) since 0 < 2 and nums[0] + nums[2] = 1 < target  - (0, 4) since 0 < 4 and nums[0] + nums[4] = 0 < target Note that (0, 3) is not counted since nums[0] + nums[3] is not strictly less than the target.  Example 2:  Input: nums = [-6,2,5,-2,-7,-1,3], target = -2 Output: 10 Explanation: There are 10 pairs of indices that satisfy the conditions in the statement: - (0, 1) since 0 < 1 and nums[0] + nums[1] = -4 < target - (0, 3) since 0 < 3 and nums[0] + nums[3] = -8 < target - (0, 4) since 0 < 4 and nums[0] + nums[4] = -13 < target - (0, 5) since 0 < 5 and nums[0] + nums[5] = -7 < target - (0, 6) since 0 < 6 and nums[0] + nums[6] = -3 < target - (1, 4) since 1 < 4 and nums[1] + nums[4] = -5 < target - (3, 4) since 3 < 4 and nums[3] + nums[4] = -9 < target - (3, 5) since 3 < 5 and nums[3] + nums[5] = -3 < target - (4, 5) since 4 < 5 and nums[4] + nums[5] = -8 < target - (4, 6) since 4 < 6 and nums[4] + nums[6] = -4 < target    Constraints:  1 <= nums.length == n <= 50 -50 <= nums[i], target <= 50  

	# Explanation
	Here's a breakdown of the solution, followed by the Python code:

*   **High-Level Approach:**
    *   Sort the input array `nums`. This enables efficient searching for pairs that satisfy the condition.
    *   Iterate through the sorted array. For each element `nums[i]`, use a two-pointer approach (or binary search) to find the index `j` of the largest element such that `nums[i] + nums[j] < target`.
    *   The number of valid pairs involving `nums[i]` is `j - i`. Accumulate this count for each `i`.

*   **Complexity:**
    *   Runtime Complexity: O(n log n) - Sorting dominates. The two-pointer approach inside the loop takes O(n) in the worst case, but since the array is already sorted it can use binary search and become O(log n) so overall O(n log n).
    *   Storage Complexity: O(1) - Sorting can be done in-place for many implementations.

	
	# Code
	```python
	def countPairs(nums, target):
    """
    Counts the number of pairs (i, j) where 0 <= i < j < n and nums[i] + nums[j] < target.

    Args:
        nums: A list of integers.
        target: An integer.

    Returns:
        The number of pairs that satisfy the condition.
    """
    nums.sort()
    n = len(nums)
    count = 0

    for i in range(n):
        l, r = i + 1, n - 1
        j = i  # Initialize j to i, so if no element satisfies, the count doesn't change
        while l <= r:
            mid = (l + r) // 2
            if nums[i] + nums[mid] < target:
                j = mid
                l = mid + 1
            else:
                r = mid - 1
        count += (j - i)

    return count
	```