Form Array by Concatenating Subarrays of Another Array

Problem Description

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 i-th subarray is equal to groups[i], and if i > 0, the (i-1)th subarray appears before the ith subarray in nums. 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.

Key Insights

• The order of the subarrays in groups must be preserved in nums.
• The subarrays must be disjoint, meaning they cannot overlap in nums.
• We can use a two-pointer technique to traverse through nums and match each subarray in groups.

Space and Time Complexity

Time Complexity: O(m + n), where m is the length of nums and n is the total number of elements in groups.
Space Complexity: O(1), as we are using a constant amount of extra space.

Solution

To solve this problem, we can use a two-pointer approach. We will iterate through the nums array using a pointer and, for each group in groups, we will try to match the corresponding subarray by checking if the elements in nums match the elements in the current group. We will maintain a separate pointer for the current position in nums, and when we find a match, we will move the pointer accordingly to ensure that the next subarray starts after the current one, maintaining disjointness.

Code Solutions