How do you solve Compact Object on LeetCode?

Compact Object (LeetCode #2705) 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: Understanding how to handle nested structures is crucial for this problem.

What is the brute force solution for Compact Object?

The brute force approach for Compact Object is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves recursively traversing the object or array and checking each value. If the value is falsy, we simply skip it; otherwise, we keep it in the result. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Compact Object?

Explain your thought process clearly as you write the code. Consider edge cases, such as empty objects or arrays. Practice writing recursive functions, as they are common in data structure problems.

What are common mistakes when solving Compact Object?

Not handling nested objects or arrays correctly. Forgetting to check for both null and false values.

#2705

Compact Object

Medium

RecursionObject manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses a single pass to build the compact object or array. We utilize recursion effectively to handle nested structures while maintaining a clean and efficient traversal.

⚙️

Algorithm

6 steps

1Step 1: Create a function that takes the input object or array.
2Step 2: Initialize an empty result object or array.
3Step 3: Iterate through each key-value pair in the object or each element in the array.
4Step 4: For each value, check if it is falsy. If it is not, add it to the result.
5Step 5: If the value is an object or array, recursively call the function on it.
6Step 6: Return the result object or array.

solution.py7 lines

1def compact(obj):
2    if isinstance(obj, dict):
3        return {k: compact(v) for k, v in obj.items() if v}
4    elif isinstance(obj, list):
5        return [compact(v) for v in obj if v]
6    return obj
7

ℹ

Complexity note: This complexity is due to the need to store the results in a new object or array, where n is the total number of elements in the input.

1Understanding how to handle nested structures is crucial for this problem.
2Recognizing the difference between truthy and falsy values helps in filtering the data correctly.

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