2705. Compact Object LeetCode Solution

In this guide, you will get 2705. Compact Object LeetCode Solution with the best time and space complexity. The solution to Compact Object problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Compact Object solution in C++
Compact Object solution in Java
Compact Object solution in Python
Additional Resources

2705. Compact Object LeetCode Solution image

Problem Statement of Compact Object

Given an object or array obj, return a compact object.
A compact object is the same as the original object, except with keys containing falsy values removed. This operation applies to the object and any nested objects. Arrays are considered objects where the indices are keys. A value is considered falsy when Boolean(value) returns false.
You may assume the obj is the output of JSON.parse. In other words, it is valid JSON.

Example 1:

Input: obj = [null, 0, false, 1]
Output: [1]
Explanation: All falsy values have been removed from the array.

Example 2:

Input: obj = {“a”: null, “b”: [false, 1]}
Output: {“b”: [1]}
Explanation: obj[“a”] and obj[“b”][0] had falsy values and were removed.
Example 3:

Input: obj = [null, 0, 5, [0], [false, 16]]
Output: [5, [], [16]]
Explanation: obj[0], obj[1], obj[3][0], and obj[4][0] were falsy and removed.

Constraints:

obj is a valid JSON object
2 <= JSON.stringify(obj).length <= 106

Complexity Analysis

Time Complexity: Google AdSense
Space Complexity: Google Analytics

2705. Compact Object LeetCode Solution in C++

type JSONValue =
  | null
  | boolean
  | number
  | string
  | JSONValue[]
  | { [key: string]: JSONValue };
type Obj = Record<string, JSONValue> | Array<JSONValue>;

function compactObject(obj: Obj): Obj {
  return dfs(obj) as Obj;
}

function dfs(value: JSONValue): JSONValue {
  if (value === null) {
    return null;
  }
  if (Array.isArray(value)) {
    return value.filter(Boolean).map(dfs);
  }
  if (typeof value === 'object') {
    for (const key of Object.keys(value)) {
      if (Boolean(value[key])) {
        value[key] = dfs(value[key]);
      } else {
        delete value[key];
      }
    }
  }
  return value;
}
/* code provided by PROGIEZ */