3069. Distribute Elements Into Two Arrays I

EasyArraySimulation
Leetcode Link

Problem Description

You have an array of distinct integers, numbered from 1 and following the sequence of natural numbers. Your goal is to separate all the elements of this array into two new arrays while following a specific set of rules during distribution.

In the beginning, you'll place the first element of the given array into the first new array (let's call it arr1), and the second element into the second new array (arr2). For each subsequent element, you'll make a decision based on the comparison between the last elements of arr1 and arr2. If the last element of arr1 is greater than the last element of arr2, you'll add the next element into arr1. Otherwise, you put it into arr2.

After you've gone through all the elements, you'll combine both arr1 and arr2 into a single array called result, where arr1 is followed by arr2. The task is to return this concatenated result array. Notably, since integers are distinct and the distribution is influenced by the comparison between the latest numbers in each array, the sequence of elements in the final array will reflect the sequence of decision-making during the distribution process.

Intuition

The fundamental insight for approaching this problem is that the distribution process is sequential and decision-based, depending entirely on the comparison between the trailing elements of arr1 and arr2. This suggests a straightforward, step-by-step simulation where you iterate through the list and distribute elements based on the stated rules.

By starting with the initial conditions (first element in arr1 and second in arr2), you merely need to keep checking the last elements of both arrays to decide the placement of the next item. Since the elements are distinct, there will be no ties, and the decision will always be clear-cut.

After placing all elements using these rules, combining arr1 and arr2 is trivial: one can simply concatenate one to the other. The essence of the solution is to map the rules as directly as possible into code, ensuring that the logic is simple and the implementation is a direct mirror of the rules outlined in the problem. This approach not only streamlines the flow of the algorithm but also makes the code easy to understand and verify.

Solution Approach

The implementation of the solution is a direct simulation of the process described in the problem definition. We translate the distinct steps of decision-making into code structure. The Python programming language offers a simple and expressive way to achieve this in just a few lines.

Initially, we create two arrays (arr1 and arr2) to represent the two groups where elements of the input array (nums) will be distributed. The first element (nums[0]) is put in arr1, and the second (nums[1]) goes into arr2. This sets up the initial state of the arrays as per the problem's rules.

We then go into a loop starting at the third element (index 2, since arrays are 1-indexed in the description) of nums. For each element x in the array from index 2 onwards, we must decide whether x will be added to arr1 or arr2. The decision is based on a simple comparison:

  • If the last element of arr1 (retrieved with arr1[-1]) is greater than the last element of arr2 (arr2[-1]), x is appended to arr1.
  • Otherwise, x goes into arr2.

This process relies on two key aspects:

  • Array indexing and manipulation: By using negative indexes (like -1), we can easily access the last elements of the arrays in Python, which is a convenient feature of the language's data structure functionality.
  • Conditional logic: The decision-making process used for distribution is implemented through a simple if-else statement, a fundamental control structure in programming that allows us to execute different actions based on certain conditions.

After all elements have been placed into arr1 or arr2, the two arrays are concatenated using the + operator, which in Python can merge lists. This last step creates the final result array as specified.

The overall design pattern follows a linear, iterative approach, meaning the elements of nums are processed in order. No complex data structures, backtracking, or optimization techniques are needed because the problem statement guarantees that there will be no ambiguity in the decision-making process; thus, each step depends only on the state from the previous step.

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Example Walkthrough

Let's consider an example with an array of distinct integers nums = [1, 2, 3, 4, 5]. We will walk through the application of the solution approach to this array.

  1. We start by creating two new arrays, arr1 and arr2.
  2. According to the rules, we must place the first element of nums in arr1, and the second in arr2. Thus, after this step, arr1 = [1] and arr2 = [2].
  3. Next, we begin iterating from the third element of nums. The third element is 3.
    • We compare the last elements of both arrays: arr1[-1] = 1 and arr2[-1] = 2. Since 1 is not greater than 2, we add 3 to arr2.
    • Now, arr1 = [1] and arr2 = [2, 3].
  4. Move to the fourth element, which is 4.
    • We compare again: arr1[-1] = 1 and arr2[-1] = 3. 1 is less than 3, so we add 4 to arr2.
    • arr1 = [1] and arr2 = [2, 3, 4].
  5. Finally, we take the fifth element, 5.
    • Comparing arr1[-1] = 1 with arr2[-1] = 4, 1 is less than 4, so we add 5 to arr2.
    • arr1 = [1] and arr2 = [2, 3, 4, 5].
  6. After the iteration is done, we concatenate arr1 and arr2 to form the result array.
    • result = arr1 + arr2 = [1] + [2, 3, 4, 5] = [1, 2, 3, 4, 5].

Thus, the result array is [1, 2, 3, 4, 5], which reflects the distribution decision made at each step. However, in this particular example, after placing the first two initial elements, all elements went into the second array due to the initial placement and specific order of nums. Different input arrays might yield a more varied distribution between arr1 and arr2.

Solution Implementation

1from typing import List
2
3class Solution:
4    def result_array(self, nums: List[int]) -> List[int]:
5        # Initialize the first array with the first element of nums
6        first_array = [nums[0]]
7      
8        # Initialize the second array with the second element of nums
9        second_array = [nums[1]]
10      
11        # Process the remaining elements, starting from the third
12        for number in nums[2:]:
13            # Compare the last elements of first_array and second_array
14            # Place the current number in the array with the smaller last element
15            if first_array[-1] > second_array[-1]:
16                first_array.append(number)
17            else:
18                second_array.append(number)
19      
20        # Combine the two arrays and return the result
21        return first_array + second_array
22
23# Example usage:
24# solution = Solution()
25# result = solution.result_array([1, 2, 3, 4, 5])
26# print(result) # This will print: [1, 2, 3, 4, 5]
27
1class Solution {
2    // method to construct a result array based on certain rules
3    public int[] resultArray(int[] nums) {
4        int n = nums.length; // length of the input array
5        int[] sortedHalf1 = new int[n]; // first half of the sorted array
6        int[] sortedHalf2 = new int[n]; // second half of the sorted array
7      
8        // Initialize the first element of each sorted half with the first two numbers
9        sortedHalf1[0] = nums[0];
10        sortedHalf2[0] = nums[1];
11      
12        // Index to track the last filled positions in sortedHalf1 and sortedHalf2
13        int i = 0, j = 0; 
14
15        // Iterate over the rest of nums to separate into two sorted halves
16        for (int k = 2; k < n; ++k) {
17            // If the current last number in sortedHalf1 is greater than that of sortedHalf2
18            if (sortedHalf1[i] > sortedHalf2[j]) {
19                // Place the current number in the next spot in sortedHalf1
20                sortedHalf1[++i] = nums[k];
21            } else {
22                // Otherwise, place it in the next spot in sortedHalf2
23                sortedHalf2[++j] = nums[k];
24            }
25        }
26
27        // Append the contents of sortedHalf2 to the end of sortedHalf1
28        for (int k = 0; k <= j; ++k) {
29            sortedHalf1[++i] = sortedHalf2[k];
30        }
31      
32        // Return the combined array
33        return sortedHalf1;
34    }
35}
36
1#include <vector>
2using namespace std;
3
4class Solution {
5public:
6    // Function to create a result array from the input array 'nums'
7    vector<int> resultArray(vector<int>& nums) {
8        int n = nums.size();  // Size of the input array
9      
10        // Initial array containing the first element
11        vector<int> array1 = {nums[0]};
12      
13        // Check for edge cases where nums might contain less than 2 elements
14        if (n < 2) {
15            return array1;  // If less than 2, return array1 as the result
16        }
17      
18        // Second array containing the second element
19        vector<int> array2 = {nums[1]};
20      
21        // Loop through the rest of elements starting from the third element
22        for (int k = 2; k < n; ++k) {
23            // Decide which array to append the current element (nums[k]) based on the last elements of array1 and array2
24            if (array1.back() > array2.back()) {
25                array1.push_back(nums[k]);
26            } else {
27                array2.push_back(nums[k]);
28            }
29        }
30      
31        // Merge array2 into array1, thus array1 will contain elements of both arrays
32        array1.insert(array1.end(), array2.begin(), array2.end());
33      
34        return array1;  // Return the merged array as the result
35    }
36};
37
1/**
2 * Takes an array of numbers and returns a new array.
3 * The new array is a concatenation of two subarrays:
4 * - The first subarray contains elements from the original array at even indices.
5 * - The second subarray contains elements from the original array at odd indices.
6 * Each subarray builds from the second element by comparing the last elements
7 * and pushing the next value to the subarray with the smaller last value.
8 * @param {number[]} nums - The input array of numbers.
9 * @returns {number[]} The resulting concatenated array from two subarrays.
10 */
11function resultArray(nums: number[]): number[] {
12    // Initialize first subarray with the first element of the input array
13    const subArrayEven: number[] = [nums[0]];
14
15    // Initialize second subarray with the second element of the input array
16    const subArrayOdd: number[] = [nums[1]];
17
18    // Iterate over the rest of the elements starting from the third element
19    for (const num of nums.slice(2)) {
20        // Compare the last elements of both subarrays and push the current
21        // number (num) to the subarray with the smaller last item
22        if (subArrayEven[subArrayEven.length - 1] > subArrayOdd[subArrayOdd.length - 1]) {
23            // If the last item of subArrayEven is greater, push to subArrayOdd
24            subArrayOdd.push(num);
25        } else {
26            // Otherwise, push to subArrayEven
27            subArrayEven.push(num);
28        }
29    }
30
31    // Concatenate the two subarrays and return the result
32    return subArrayEven.concat(subArrayOdd);
33}
34

Time and Space Complexity

The time complexity of the provided code is O(n). This is because there is a single loop that iterates through the nums array starting from the third element, which runs in O(n-2) time. However, since constant factors are disregarded in Big O notation, it simplifies to O(n).

The space complexity is also O(n). Two new arrays, arr1 and arr2, are created and potentially, all elements of nums could be added to these arrays. In the worst case, both arr1 and arr2 together would contain all elements of the input nums, making the space complexity linear with respect to the size of the input.

Learn more about how to find time and space complexity quickly using problem constraints.


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