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, 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

    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