How do you solve Split Array into Consecutive Subsequences on LeetCode?

Split Array into Consecutive Subsequences (LeetCode #659) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each subsequence must be consecutive and of length at least 3.

What is the time complexity of Split Array into Consecutive Subsequences?

The optimal solution for Split Array into Consecutive Subsequences runs in O(n) time and uses O(n) space. The complexity is O(n) because we traverse the array a constant number of times, and the space complexity is O(n) due to the HashMap used for counting.

What is the brute force solution for Split Array into Consecutive Subsequences?

The brute force approach for Split Array into Consecutive Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences and checking if they meet the criteria of being consecutive and having a length of at least 3. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Split Array into Consecutive Subsequences?

Split Array into Consecutive Subsequences uses the following concepts: Array, Hash Table, Greedy, Heap (Priority Queue). The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Split Array into Consecutive Subsequences?

Think about how you can use data structures like HashMaps to track counts. Consider edge cases where numbers may repeat or be missing. Practice breaking down the problem into smaller parts.

What are common mistakes when solving Split Array into Consecutive Subsequences?

Failing to check if there are enough elements left to form a valid subsequence. Not updating counts correctly after forming subsequences.

#659

Split Array into Consecutive Subsequences

Medium

ArrayHash TableGreedyHeap (Priority Queue)Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a greedy approach with a HashMap to track the counts of elements and to build subsequences efficiently. This ensures that we can quickly determine if we can form valid subsequences of length 3 or more.

⚙️

Algorithm

4 steps

1Step 1: Count the frequency of each number in the array using a HashMap.
2Step 2: For each number in the array, attempt to build a subsequence starting from that number.
3Step 3: If a number can be added to an existing subsequence, do so; otherwise, create a new subsequence if possible.
4Step 4: Ensure that each subsequence has at least 3 elements.

solution.py22 lines

1# Full working Python code
2from collections import Counter
3
4def can_split(nums):
5    count = Counter(nums)
6    end = Counter()
7
8    for num in nums:
9        if count[num] == 0:
10            continue
11        count[num] -= 1
12
13        if end[num - 1] > 0:
14            end[num - 1] -= 1
15            end[num] += 1
16        elif count[num + 1] > 0 and count[num + 2] > 0:
17            count[num + 1] -= 1
18            count[num + 2] -= 1
19            end[num + 2] += 1
20        else:
21            return False
22    return True

ℹ

Complexity note: The complexity is O(n) because we traverse the array a constant number of times, and the space complexity is O(n) due to the HashMap used for counting.

1Each subsequence must be consecutive and of length at least 3.
2Using a greedy approach with a HashMap allows efficient tracking of subsequences.

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