2675. Array of Objects to Matrix

Problem Description

The task is to write a function that converts an array of objects arr into a two-dimensional array, or matrix, m. The input array, arr, can contain objects that are nested, meaning that an object can have properties which are objects or arrays themselves. Additionally, these objects can have properties of different types including numbers, strings, booleans, and null values.

The first row of the resulting matrix m should contain column names. For non-nested objects, these names are simply the keys of the properties in the objects. For nested objects, the column names should represent the path to the property, with each level of nesting separated by a period ".".

Each subsequent row in the matrix corresponds to an object from the input array arr. The values are placed in the matrix in alignment with the column names. If an object does not have a value for a specific column, that matrix cell should be an empty string "".

Finally, the columns are to be sorted in lexicographically ascending order, which means dictionary order, considering the column names as strings.

Intuition

The proposed solution involves two main steps: First, it flattens the structure of the input objects to easily map them into rows of the matrix. Second, it constructs the matrix using column names in lexicographically ascending order.

For the flattening part, a depth-first search (DFS) algorithm is employed. DFS traverses nested objects and arrays and builds a flat key-value representation of the underlying data where the keys are the paths leading to each value. The paths reflect the structure of the original object with nested keys joined by a period ".".

Once flattened, the complete list of column names is derived from the keys of flattened objects. These keys are then deduplicated and sorted to form the first row of the matrix - the column header row.

The next part involves creating each row of the matrix corresponding to the input objects. For each object's flattened version, the values are filled in the appropriate columns, considering the sorted column names. If a value for a specific column is not found in the object, an empty string is inserted in that place.

By iterating through the entire array of objects and creating each row according to the gathered column names, the function builds the complete matrix, adhering to the requirements of the problem.

Solution Approach

The implementation of the solution can be broken down into several steps that reflect the thought process for handling the problem.

1. Depth-First Search to Flatten Objects: To navigate through nested objects, a DFS algorithm is used. DFS is ideal for exploring all the paths through nested data as it goes as deep as possible down one path before backing up and trying another. Here, the dfs function is recursively called with a key (representing the current path) and obj, the object or value currently being explored.

   • It starts by checking if the obj is a leaf node, which in this context is a value that is not an object or an array – i.e., numbers, strings, booleans, or null. If it is, a single key-value pair is returned with the current key and the obj itself.

   • If the obj is another object or array, it iterates over the object entries and recursively calls dfs on each key-value pair, appending the current key to the new key for nested properties by newKey = key ? ${key}.${k}:${k};, where k is the current object's key. The recursion ensures that all levels of nesting are accounted for in the final keys.

   • It accumulates and flattens these results to produce a single key-value pair for each property, regardless of how deeply nested it is in the original object.

2. Creating a List of Unique Keys: After flattening each object into a collection of key-value pairs, all the keys are gathered into one array and then reduced to a unique set to eliminate duplicates. The unique keys are then sorted lexicographically using the .sort() method to define the column names for the matrix.

3. Building the Matrix: With the list of keys prepared, the actual matrix construction begins. The sorted list of keys forms the first row of the matrix.

   • For each object (after flattening), the algorithm iterates over the list of sorted keys and checks if the object contains a value for each key. This is performed by searching the flattened objects for a property with that key. If the key is present, its value is inserted into the row; if not, an empty string "" is used as a placeholder.

   • The find function: row.find(r => r.hasOwnProperty(key))?.[key] is used to retrieve the value for a given key from the set of key-value pairs obtained after flattening.

   • The push method appends either the found value or an empty string to the newRow array, preparing a complete row of the matrix.

   • Each newRow is then added to the ans array, which accumulatively builds up the final matrix.

4. Returning the Result: At the end of the iteration, the ans array contains the first row with the sorted column names and the subsequent rows representing objects' properties. The completed matrix is returned.

This approach efficiently handles the conversion of an array of potentially nested objects into a cleanly structured matrix, taking care to accommodate missing keys and ensuring the final columns are in the required lexicographic order.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Suppose we have the following array of objects:

1const arr = [
2  { name: 'Alice', age: 30, address: { city: 'Wonderland', zip: '12345' } },
3  { name: 'Bob', age: 22, contact: { email: '[email protected]', phone: '555-4132' } }
4];

Following our solution approach, we'd perform these steps:

1. Depth-First Search to Flatten Objects:

   We use DFS to flatten each object in arr. For Alice, we would obtain:

   1{ 'name': 'Alice', 'age': 30, 'address.city': 'Wonderland', 'address.zip': '12345' }

   For Bob, similarly:

   1{ 'name': 'Bob', 'age': 22, 'contact.email': '[email protected]', 'contact.phone': '555-4132' }

2. Creating a List of Unique Keys:

   Collect all keys from the flattened objects:

   1['name', 'age', 'address.city', 'address.zip', 'contact.email', 'contact.phone']

   Then, we remove duplicates and sort them lexicographically to get the column names for our matrix:

   1['address.city', 'address.zip', 'age', 'contact.email', 'contact.phone', 'name']

3. Building the Matrix:

   With our sorted keys, we start building the matrix. The first row will be the column names:

   1[['address.city', 'address.zip', 'age', 'contact.email', 'contact.phone', 'name']]

   Then, we add rows for Alice and Bob. Since Alice does not have contact info, and Bob does not have an address, we put empty strings for those values:

   Alice's row:

   1['Wonderland', '12345', 30, '', '', 'Alice']

   Bob's row:

   1['', '', 22, '[email protected]', '555-4132', 'Bob']

   Adding these rows to our matrix results in:

   1[
   2  ['address.city', 'address.zip', 'age', 'contact.email', 'contact.phone', 'name'],
   3  ['Wonderland', '12345', 30, '', '', 'Alice'],
   4  ['', '', 22, '[email protected]', '555-4132', 'Bob']
   5]

4. Returning the Result:

   Our final output is the completed matrix:

   1[
   2  ['address.city', 'address.zip', 'age', 'contact.email', 'contact.phone', 'name'],
   3  ['Wonderland', '12345', 30, '', '', 'Alice'],
   4  ['', '', 22, '[email protected]', '555-4132', 'Bob']
   5]

And this is how the solution approach breaks down a complex data structure into a structured matrix with sorted columns, dealing properly with nested objects and missing values.

Solution Implementation

Time and Space Complexity

Time Complexity

To analyze the time complexity of the jsonToMatrix function, let's examine the steps involved:

1. dfs function - The dfs (Depth-First Search) function is a recursive function which is called for each property in the object.

   • For each property that is a primitive value (number, string, boolean, or null), it performs a constant amount of work.
   • For nested objects, it iterates through all keys and calls itself recursively.
   • Assuming that the average depth of the objects is d and the average number of keys at each level is k, the time complexity of the dfs function is approximately O(k^d).

2. arr.map call - The dfs function is called once for each item in the input array arr, with the length of the array being n, resulting in a time complexity of O(n * k^d) for this step.

3. Generation of keys array - This involves flattening the array of key-value pairs and extracting keys. The flattening operation has a complexity of O(n * k^d) since it involves concatenating arrays returned by the dfs function. The creation of the Set to filter out unique keys is O(u) where u is the number of unique keys, which is less than or equal to n * k^d.

4. The ans array - The nested loop involves iterating over each row and then each key. Assuming there are m unique keys, the complexity of this step is O(n * m).

Combining all these steps, the overall time complexity will be dominated by the arr.map call and the operations inside it. Therefore, the overall time complexity of the jsonToMatrix function is O(n * k^d + n * m) where:

• n is the length of the input array arr,
• k is the average number of keys in an object,
• d is the average depth of nested objects,
• m is the number of unique keys across all objects.

Space Complexity

The space complexity considers the memory used throughout the execution of the function:

1. Recursive dfs calls - Each recursive call creates a new execution context, and the maximum depth of the recursive call stack will be d, so the space complexity due to recursion is O(d).

2. kv and ans arrays - These arrays are used to store intermediate and final results. Since ans includes a row for each input object plus one row for the keys, its size is O(n * m). The kv array contains the results of the dfs function calls and has a similar size.

3. keys array - Storing the unique keys requires a space of O(u), where u is the number of unique keys, which can be as much as O(n * k^d) in the worst case.

The main contributing factor to space complexity is the output array ans and the set of unique keys. Therefore, the overall space complexity of the jsonToMatrix function is O(n * m + n * k^d + d). Simplified, assuming m is bounded by n * k^d, the final space complexity is O(n * k^d + d) where:

• d is the average depth of nested objects,
• n is the length of the input array arr,
• k is the average number of keys in an object.