Introduction

In this blog post, we will explore how to solve the LeetCode problem "646" 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

You are given an array of n pairs pairs where pairs[i] = [lefti, righti] and lefti < righti. A pair p2 = [c, d] follows a pair p1 = [a, b] if b < c. A chain of pairs can be formed in this fashion. Return the length longest chain which can be formed. You do not need to use up all the given intervals. You can select pairs in any order. Example 1: Input: pairs = [[1,2],[2,3],[3,4]] Output: 2 Explanation: The longest chain is [1,2] -> [3,4]. Example 2: Input: pairs = [[1,2],[7,8],[4,5]] Output: 3 Explanation: The longest chain is [1,2] -> [4,5] -> [7,8]. Constraints: n == pairs.length 1 <= n <= 1000 -1000 <= lefti < righti <= 1000

Explanation

Here's a solution to find the longest chain of pairs:

• Sort by End Time: The key idea is to sort the pairs based on their end times (righti). This greedy approach allows us to efficiently select the next pair in the chain.

• Greedy Selection: Iterate through the sorted pairs. If a pair's start time (lefti) is greater than the end time of the previously selected pair, we include it in the chain and update the end time.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(1) excluding the input.

Code

    def findLongestChain(pairs):
    """
    Finds the length of the longest chain of pairs.

    Args:
        pairs: A list of pairs, where each pair is a list [left, right].

    Returns:
        The length of the longest chain.
    """

    pairs.sort(key=lambda x: x[1])  # Sort by end times
    current_end = float('-inf')
    chain_length = 0

    for pair in pairs:
        if pair[0] > current_end:
            chain_length += 1
            current_end = pair[1]

    return chain_length