2634. Filter Elements from Array


Problem Description

In this problem, you are provided with two main elements: an integer array arr and a function fn which acts as a filter. The goal is to construct a new array, filteredArr, that contains only certain elements from the original array arr. To decide which elements make it into filteredArr, you use the filter function fn.

The function fn can accept one or two parameters. The first parameter is an element from the array (arr[i]), and the second parameter is the index of that element within the array (i). The role of fn is to evaluate each element (and its index) and return a truthy value if the element should be included in the filtered array or a falsy value if it should not. In JavaScript, a truthy value is any value that, when converted to a boolean, becomes true.

The requirement of the problem is to write your own filter logic without using the built-in Array.filter method provided by JavaScript. This is a fundamental programming task that reinforces your understanding of array iteration and conditional statements.

Intuition

The intuition behind the solution involves iterating through the given array, analyzing each element using the provided function fn, and selectively gathering items into a new array based on the result of that function. The steps to arrive at the solution are quite straightforward:

  1. Initialize a new array, let's call it ans, to store the elements that satisfy the filtering condition.
  2. Loop through each element of the original array arr using a for loop. As we loop through, we have access to both the current element arr[i] and its index i.
  3. On each iteration, apply the filter function fn to the current element and its index. If the function returns a truthy value, include this element in the ans array by adding (pushing) it.
  4. After going through all the elements in arr, end the loop and return the ans array as the filtered array.

By following these steps, you effectively replicate the behavior of the Array.filter method using basic iteration and conditional logic. This method ensures that the original array remains unchanged while producing a new array that meets the specified filtering criteria.

Solution Approach

The implementation of the solution uses a simple and effective algorithm that involves array traversal and conditional logic:

  1. Array traversal: We start by iterating through each element of the input array arr using a for loop. The loop runs from the beginning to the end of the array, granting us access to every element and its corresponding index.

  2. Apply the filter function: During each iteration of the loop, we invoke the function fn with two arguments: the current element arr[i] and its index i. The purpose of fn is to determine whether the element meets the filtering criteria specified.

  3. Conditional logic: If the function fn returns a truthy value, it signifies that the current element passes the filter condition and should be included in the resulting array. We check the result of fn(arr[i], i) with a simple if condition.

  4. Building the answer array: When the if condition is satisfied, we use the Array.prototype.push method to add the current element arr[i] to a new array called ans. This array is initialized beforehand to store all elements that meet the condition.

  5. Return the result: After the loop has finished processing all the elements in the input array, the ans array now contains all the elements for which fn(arr[i], i) returned a truthy value. We return ans as the filtered array.

By following these steps, the code effectively filters the input array based on the logic defined in the filter function fn without utilizing the built-in Array.filter method. This approach demonstrates fundamental programming concepts like loops, conditionals, and array manipulation in TypeScript, which is the language of the provided solution.

Here is the code snippet provided using the steps mentioned above:

function filter(arr: number[], fn: (n: number, i: number) => any): number[] {
    const ans: number[] = [];
    for (let i = 0; i < arr.length; ++i) { // Loop through array
        if (fn(arr[i], i)) { // Apply filter function and check for truthy
            ans.push(arr[i]); // Add element to answer array if condition is met
        }
    }
    return ans; // Return filtered array
}

Notice that no special data structures are needed for this task. An additional array ans is the only requirement to gather filtered results. The given solution is straightforward, adhering to basic algorithmic patterns—an essential practice for any programmer to master.

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Example Walkthrough

Let's walk through a simple example to illustrate the solution approach:

Assume we have an integer array arr and a filter function fn defined as follows:

const arr = [1, 2, 3, 4, 5];
const fn = (element: number, index: number) => (element % 2 === 0);

In this example, our function fn is designed to return true for even numbers and false for odd numbers.

Now, let's use these to walk through the filter function's steps:

  1. Array traversal: We start by looking at each element of the array arr. The array elements are [1, 2, 3, 4, 5].

  2. Apply the filter function: For each element, we call the function fn:

    • fn(1, 0) returns false because 1 is odd.
    • fn(2, 1) returns true because 2 is even.
    • fn(3, 2) returns false because 3 is odd.
    • fn(4, 3) returns true because 4 is even.
    • fn(5, 4) returns false because 5 is odd.
  3. Conditional logic: We check the result of the function for each occurrence:

    • Because the call for arr[0] returned false, we do not include 1 in the ans array.
    • The call for arr[1] returned true, so 2 is included in the ans array.
    • The call for arr[2] returned false, so 3 is not included.
    • Because arr[3] returned true, 4 is added to the ans array.
    • Finally, arr[4] returned false, so 5 is left out.
  4. Building the answer array: As we apply the filter function and evaluate the result, our ans array begins to build up with the even numbers [2, 4].

  5. Return the result: Once we've finished iterating through the entire arr array, the function ends, and the final ans array [2, 4] is returned, which contains only the elements from the array arr that are even.

In conclusion, following the above steps and example, the filter function replicates Array.filter behavior by creating a new array that only includes elements passing the filtering condition defined by fn. The final result is an array of even numbers from the original array, demonstrating how the algorithm effectively selects elements based on specified criteria.

Solution Implementation

1def filter_by_function(array, filter_function):
2    """
3    Filters elements of an array based on a provided filter function.
4
5    :param array: The array to filter.
6    :param filter_function: The filter function used to determine if an element 
7                            should be included in the result array.
8                            It takes the element value and index as its arguments.
9    :return: A new list with elements that pass the filter function condition.
10    """
11    result = []  # Initialize an empty list to hold the filtered results
12    for index, value in enumerate(array):  # Loop over each item and its index in the array
13        if filter_function(value, index):  # If the filter function returns 'true', include the element in the result list
14            result.append(value)  # Append the current element to the result list
15    return result  # Return the filtered list
16
1import java.util.ArrayList;
2import java.util.List;
3import java.util.function.BiPredicate;
4
5/**
6 * Filters elements of an array based on a provided filter function.
7 * @param array The array to filter.
8 * @param filterFunction The filter function used to determine if an element should be included in the result array.
9 * It takes the element value and its index in the array as its arguments.
10 * @return A new list with elements that pass the filter function condition.
11 */
12public List<Integer> filter(int[] array, BiPredicate<Integer, Integer> filterFunction) {
13    List<Integer> result = new ArrayList<>(); // Initialize an empty list to hold the filtered results
14    for (int index = 0; index < array.length; index++) {
15        if (filterFunction.test(array[index], index)) {
16            // If the filter function returns 'true', include the element in the result list
17            result.add(array[index]);
18        }
19    }
20    return result; // Return the filtered list
21}
22
1#include <vector>
2#include <functional> // For std::function
3
4/**
5 * Filters elements of a vector based on a provided filter function.
6 * @param inputArray - The vector to filter.
7 * @param filterFunction - The filter function used to determine if an element should be included in the result vector.
8 *     It takes the element value and index as its arguments.
9 * @returns A new vector with elements that pass the filter function condition.
10 */
11std::vector<int> filter(const std::vector<int> &inputArray, const std::function<bool(int, int)> &filterFunction) {
12    std::vector<int> result; // Initialize an empty vector to hold the filtered results
13    for (int index = 0; index < inputArray.size(); ++index) {
14        if (filterFunction(inputArray[index], index)) {
15            // If the filter function returns 'true', include the element in the result vector
16            result.push_back(inputArray[index]);
17        }
18    }
19    return result; // Return the filtered vector
20}
21
1/**
2 * Filters elements of an array based on a provided filter function.
3 * @param {number[]} array - The array to filter.
4 * @param {Function} filterFunction - The filter function used to determine if an element should be included in the result array.
5 *    It takes the element value and index as its arguments.
6 * @returns {number[]} A new array with elements that pass the filter function condition.
7 */
8function filter(array: number[], filterFunction: (value: number, index: number) => boolean): number[] {
9    const result: number[] = []; // Initialize an empty array to hold the filtered results
10    for (let index = 0; index < array.length; ++index) {
11        if (filterFunction(array[index], index)) {
12            // If the filter function returns 'true', include the element in the result array
13            result.push(array[index]);
14        }
15    }
16    return result; // Return the filtered array
17}
18

Time and Space Complexity

The time complexity of the provided code is O(n), where n is the length of the array arr. This is because the function iterates through each element of the array exactly once.

The space complexity of the function is O(1), ignoring the space consumption of the ans array, since there are no additional data structures that grow with the size of the input. The variables ans and i use a fixed amount of space regardless of the input size.


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Question 1 out of 10

What does the following code do?

1def f(arr1, arr2):
2  i, j = 0, 0
3  new_arr = []
4  while i < len(arr1) and j < len(arr2):
5      if arr1[i] < arr2[j]:
6          new_arr.append(arr1[i])
7          i += 1
8      else:
9          new_arr.append(arr2[j])
10          j += 1
11  new_arr.extend(arr1[i:])
12  new_arr.extend(arr2[j:])
13  return new_arr
14
1public static List<Integer> f(int[] arr1, int[] arr2) {
2  int i = 0, j = 0;
3  List<Integer> newArr = new ArrayList<>();
4
5  while (i < arr1.length && j < arr2.length) {
6      if (arr1[i] < arr2[j]) {
7          newArr.add(arr1[i]);
8          i++;
9      } else {
10          newArr.add(arr2[j]);
11          j++;
12      }
13  }
14
15  while (i < arr1.length) {
16      newArr.add(arr1[i]);
17      i++;
18  }
19
20  while (j < arr2.length) {
21      newArr.add(arr2[j]);
22      j++;
23  }
24
25  return newArr;
26}
27
1function f(arr1, arr2) {
2  let i = 0, j = 0;
3  let newArr = [];
4  
5  while (i < arr1.length && j < arr2.length) {
6      if (arr1[i] < arr2[j]) {
7          newArr.push(arr1[i]);
8          i++;
9      } else {
10          newArr.push(arr2[j]);
11          j++;
12      }
13  }
14  
15  while (i < arr1.length) {
16      newArr.push(arr1[i]);
17      i++;
18  }
19  
20  while (j < arr2.length) {
21      newArr.push(arr2[j]);
22      j++;
23  }
24  
25  return newArr;
26}
27

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