Introduction

In this blog post, we will explore how to solve the LeetCode problem "1764" 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 a 2D integer array groups of length n. You are also given an integer array nums. You are asked if you can choose n disjoint subarrays from the array nums such that the ith subarray is equal to groups[i] (0-indexed), and if i > 0, the (i-1)th subarray appears before the ith subarray in nums (i.e. the subarrays must be in the same order as groups). Return true if you can do this task, and false otherwise. Note that the subarrays are disjoint if and only if there is no index k such that nums[k] belongs to more than one subarray. A subarray is a contiguous sequence of elements within an array. Example 1: Input: groups = [[1,-1,-1],[3,-2,0]], nums = [1,-1,0,1,-1,-1,3,-2,0] Output: true Explanation: You can choose the 0th subarray as [1,-1,0,1,-1,-1,3,-2,0] and the 1st one as [1,-1,0,1,-1,-1,3,-2,0]. These subarrays are disjoint as they share no common nums[k] element. Example 2: Input: groups = [[10,-2],[1,2,3,4]], nums = [1,2,3,4,10,-2] Output: false Explanation: Note that choosing the subarrays [1,2,3,4,10,-2] and [1,2,3,4,10,-2] is incorrect because they are not in the same order as in groups. [10,-2] must come before [1,2,3,4]. Example 3: Input: groups = [[1,2,3],[3,4]], nums = [7,7,1,2,3,4,7,7] Output: false Explanation: Note that choosing the subarrays [7,7,1,2,3,4,7,7] and [7,7,1,2,3,4,7,7] is invalid because they are not disjoint. They share a common elements nums[4] (0-indexed). Constraints: groups.length == n 1 <= n <= 103 1 <= groups[i].length, sum(groups[i].length) <= 103 1 <= nums.length <= 103 -107 <= groups[i][j], nums[k] <= 107

Explanation

Here's the breakdown of the solution:

• Greedy Matching: Iterate through the groups array, trying to find the first occurrence of each group as a subarray within nums, starting from the end position of the previously matched group.
• Subarray Search: For each group, search for a matching subarray in nums. If a match is found, update the starting position for the next search to ensure disjoint subarrays and maintain the correct order.

• Early Exit: If any group cannot be found as a subarray, return false immediately. Otherwise, if all groups are found in the correct order, return true.

• Time Complexity: O(n m k), where n is the number of groups, m is the length of nums, and k is the maximum length of a group. Space Complexity: O(1).

Code

    def canChoose(groups, nums):
    """
    Checks if disjoint subarrays from nums can form groups in the given order.

    Args:
        groups: A 2D integer array of subarrays.
        nums: An integer array.

    Returns:
        True if nums contains disjoint subarrays matching groups in order, False otherwise.
    """
    nums_index = 0
    for group in groups:
        found = False
        while nums_index <= len(nums) - len(group):
            if nums[nums_index:nums_index + len(group)] == group:
                nums_index += len(group)
                found = True
                break
            nums_index += 1
        if not found:
            return False
    return True