What is the time complexity of Form Array by Concatenating Subarrays of Another Array?

The optimal solution for Form Array by Concatenating Subarrays of Another Array runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only make a single pass through `nums`, checking each group in constant time relative to its size.

What is the brute force solution for Form Array by Concatenating Subarrays of Another Array?

The brute force approach for Form Array by Concatenating Subarrays of Another Array is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible combination of subarrays in `nums` to see if they match the groups in order. This approach is straightforward but inefficient because it may require checking many overlapping subarrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Form Array by Concatenating Subarrays of Another Array?

Form Array by Concatenating Subarrays of Another Array uses the following concepts: Array, Two Pointers, Greedy, String Matching. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Form Array by Concatenating Subarrays of Another Array?

Clarify the problem statement to ensure you understand the requirements, especially about order and disjoint conditions. Think about edge cases, such as when `groups` or `nums` are empty. Practice explaining your thought process as you code, as this helps interviewers understand your reasoning.

What are common mistakes when solving Form Array by Concatenating Subarrays of Another Array?

Failing to check the order of groups when matching subarrays. Not handling cases where groups have overlapping elements in `nums`.

#1764

Form Array by Concatenating Subarrays of Another Array

Medium

ArrayTwo PointersGreedyString MatchingTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

We can use a single pass through `nums` while checking for each group. By using a pointer to track our position in `nums`, we can efficiently find and validate each group without unnecessary checks.

⚙️

Algorithm

6 steps

1Step 1: Initialize a pointer `i` to 0 to track the current position in `nums`.
2Step 2: Iterate through each group in `groups`.
3Step 3: For each group, check if the subarray in `nums` starting from `i` matches the group.
4Step 4: If it matches, increment `i` by the length of the group to move the pointer forward.
5Step 5: If a match is not found, return false immediately.
6Step 6: If all groups are matched successfully, return true.

solution.py8 lines

1def canFormArray(groups, nums):
2    i = 0
3    for group in groups:
4        group_len = len(group)
5        if nums[i:i + group_len] != group:
6            return False
7        i += group_len
8    return True

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through `nums`, checking each group in constant time relative to its size.

1The order of groups matters, so we must ensure each group is checked sequentially.
2Disjoint subarrays mean that once we use elements from `nums`, they cannot be reused for subsequent groups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.