Leetcode 1929: Concatenation of Array

Input: The input consists of a single array nums of integers.

Example: nums = [3, 4, 5]

Constraints:

• 1 <= n <= 1000

• 1 <= nums[i] <= 1000

Output: Return the array where nums is concatenated with itself.

Example: [3, 4, 5, 3, 4, 5]

Constraints:

Goal: Concatenate the array nums with itself.

Steps:

• Create a new array ans with a length twice as long as nums.

• Copy the elements of nums into the first half of ans.

• Copy the elements of nums into the second half of ans.

Goal: Ensure that the result array is formed by concatenating nums with itself, and handle up to 1000 elements in the array.

Steps:

• The length of nums is at most 1000.

• Each element of nums is a positive integer.

Assumptions:

• nums contains positive integers.

• The result array will not exceed the constraints set by the problem.

• Input: nums = [3, 4, 5]

• Explanation: The array ans is formed by concatenating nums with itself: [3, 4, 5, 3, 4, 5].

Approach: The problem can be solved by creating a new array with twice the length of nums, then copying the elements of nums twice.

Observations:

• The array needs to be expanded to twice its size.

• The solution can be efficiently achieved with a time complexity of O(n).

• Concatenation can be done in a single pass using a for loop or array operations.

Steps:

• Create an array of size 2n.

• Copy elements from nums to the first half of the new array.

• Copy elements from nums to the second half of the new array.

Empty Inputs:

• If nums is empty, the output should also be an empty array.

Large Inputs:

• The solution should efficiently handle large arrays with the maximum length of 1000.

Special Values:

• Ensure that values in nums are handled properly when they are at the upper constraint of 1000.

Constraints:

• The time complexity of O(n) is sufficient to handle arrays with the maximum length.

vector<int> getConcatenation(vector<int>& nums) {
    int n=nums.size();
    for(int i=0;i<n;i++)
    {
        nums.push_back(nums[i]);
    }
    return nums;
}

1 : Function Header

vector<int> getConcatenation(vector<int>& nums) {

Defines the function `getConcatenation`, which takes a vector `nums` by reference and returns a vector of integers.

2 : Variable Declaration

    int n=nums.size();

Calculates and stores the size of the input vector `nums` in the variable `n`.

3 : Loop Initialization

    for(int i=0;i<n;i++)

Starts a loop that iterates through each element of the vector `nums` using the index `i`.

4 : Array Modification

        nums.push_back(nums[i]);

Appends each element from the original vector `nums` to the end of the same vector, effectively duplicating its contents.

5 : Return Statement

    return nums;

Returns the modified vector `nums`, which now contains its original contents followed by a concatenation of those contents.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n), where n is the length of the input array nums, because we iterate through nums twice.

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is O(n), as we create a new array of size 2n.

Leetcode 1929: Concatenation of Array

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