Introduction

In this blog post, we will explore how to solve the LeetCode problem "2023" 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 an array of digit strings nums and a digit string target, return the number of pairs of indices (i, j) (where i != j) such that the concatenation of nums[i] + nums[j] equals target. Example 1: Input: nums = ["777","7","77","77"], target = "7777" Output: 4 Explanation: Valid pairs are: - (0, 1): "777" + "7" - (1, 0): "7" + "777" - (2, 3): "77" + "77" - (3, 2): "77" + "77" Example 2: Input: nums = ["123","4","12","34"], target = "1234" Output: 2 Explanation: Valid pairs are: - (0, 1): "123" + "4" - (2, 3): "12" + "34" Example 3: Input: nums = ["1","1","1"], target = "11" Output: 6 Explanation: Valid pairs are: - (0, 1): "1" + "1" - (1, 0): "1" + "1" - (0, 2): "1" + "1" - (2, 0): "1" + "1" - (1, 2): "1" + "1" - (2, 1): "1" + "1" Constraints: 2 <= nums.length <= 100 1 <= nums[i].length <= 100 2 <= target.length <= 100 nums[i] and target consist of digits. nums[i] and target do not have leading zeros.

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• Approach:

  • Iterate through all possible pairs of indices (i, j) in the nums array where i != j.
  • For each pair, concatenate nums[i] and nums[j] and check if the result equals the target string.
  • Increment a counter if the concatenation equals the target.

• Complexity:

  • Runtime: O(n^2 * k), where n is the length of nums and k is the average length of strings.
  • Storage: O(1).

Code

    def num_of_pairs(nums, target):
    """
    Counts the number of pairs (i, j) where i != j and nums[i] + nums[j] == target.

    Args:
        nums: A list of digit strings.
        target: A digit string.

    Returns:
        The number of pairs that satisfy the condition.
    """
    count = 0
    n = len(nums)
    for i in range(n):
        for j in range(n):
            if i != j:
                if nums[i] + nums[j] == target:
                    count += 1
    return count