Find the Array Concatenation Value

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ArraysTwo Pointers

You are given a 0-indexed integer array nums.

The concatenation of two numbers is the number formed by concatenating their numerals.

For example, the concatenation of 15, 49 is 1549.

The concatenation value of nums is initially equal to 0. Perform this operation until nums becomes empty:

If nums has a size greater than one, add the value of the concatenation of the first and the last element to the concatenation value of nums, and remove those two elements from nums. For example, if the nums was [1, 2, 4, 5, 6], add 16 to the concatenation value.
If only one element exists in nums, add its value to the concatenation value of nums, then remove it.

Return the concatenation value of nums.

Example 1:

Input: nums = [7,52,2,4]
Output: 596
Explanation: Before performing any operation, nums is [7,52,2,4] and concatenation value is 0.
 - In the first operation:
We pick the first element, 7, and the last element, 4.
Their concatenation is 74, and we add it to the concatenation value, so it becomes equal to 74.
Then we delete them from nums, so nums becomes equal to [52,2].
 - In the second operation:
We pick the first element, 52, and the last element, 2.
Their concatenation is 522, and we add it to the concatenation value, so it becomes equal to 596.
Then we delete them from the nums, so nums becomes empty.
Since the concatenation value is 596 so the answer is 596.

Example 2:

Input: nums = [5,14,13,8,12]
Output: 673
Explanation: Before performing any operation, nums is [5,14,13,8,12] and concatenation value is 0.
 - In the first operation:
We pick the first element, 5, and the last element, 12.
Their concatenation is 512, and we add it to the concatenation value, so it becomes equal to 512.
Then we delete them from the nums, so nums becomes equal to [14,13,8].
 - In the second operation:
We pick the first element, 14, and the last element, 8.
Their concatenation is 148, and we add it to the concatenation value, so it becomes equal to 660.
Then we delete them from the nums, so nums becomes equal to [13].
 - In the third operation:
nums has only one element, so we pick 13 and add it to the concatenation value, so it becomes equal to 673.
Then we delete it from nums, so nums become empty.
Since the concatenation value is 673 so the answer is 673.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁴

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers within the input array?
Can the input array be empty, or contain a single element?
Should the concatenation be treated as string concatenation followed by parsing to an integer, or is there another interpretation?
Is the input array guaranteed to contain only integers, or could there be other data types?
For string concatenation, how should leading zeros be handled when converting the concatenated string back to an integer (if applicable)?

Brute Force Solution

Approach

To find the array concatenation value, we will combine numbers from the start and end of the list until it's empty. The brute force method involves directly calculating the concatenation value for each possible combination without clever tricks.

Here's how the algorithm would work step-by-step:

Take the first and last number from the list.
Join these two numbers together as if they were text to create a new number.
Add this newly created number to our running total.
Remove the first and last numbers from the list.
Repeat this process with the new first and last numbers until there are no numbers left in the list.
If at any point there is only one number remaining in the list, add this number to our running total.
The running total is the final answer.

Code Implementation

def find_array_concatenation_value(numbers):
    total_concatenation_value = 0
    left_index = 0
    right_index = len(numbers) - 1

    # Iterate until the indices cross each other
    while left_index <= right_index:
        # Handle the case where there's only one number left
        if left_index == right_index:
            total_concatenation_value += numbers[left_index]
            
        else:
            # Concatenate the numbers at the left and right indices
            concatenated_number = int(str(numbers[left_index]) + str(numbers[right_index]))
            total_concatenation_value += concatenated_number

        # Move the indices inwards
        left_index += 1
        right_index -= 1

    return total_concatenation_value

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the array by repeatedly taking the first and last elements. In each iteration, it concatenates them (which is O(1) since it's a constant time string operation) and adds the resulting value to a running total. The first and last elements are then removed. This process continues until the array is empty or has a single element, meaning we process at most n/2 pairs (or n/2 iterations), and in the case of a single remaining element, it is simply added, which is O(1). Therefore, the time complexity is driven by the number of pairs processed, which is proportional to n. Thus the algorithm has a time complexity of O(n).

Space Complexity

O(1) – The provided solution primarily involves manipulating the input array in place and does not create any auxiliary data structures that scale with the input size N (where N is the number of elements in the input array). The algorithm uses a constant number of variables to track the first and last element and the running total. Therefore, the space complexity is constant, independent of the input size.

Optimal Solution

Approach

The goal is to pair numbers from the beginning and end of a list, creating larger numbers by joining them. We then simply add these joined numbers together to get the final result. It's all about pairing and summing.

Here's how the algorithm would work step-by-step:

Take the first and last numbers from the list.
Combine these two numbers into a single number (treat them like strings and put them together).
Add the combined number to your running total.
Remove the first and last numbers you just used.
Repeat this process (taking the new first and last numbers, combining them, and adding to the total) until there are no numbers left or only one number remains.
If there's a single number left in the middle, add it directly to your running total.
The final running total is the answer.

Code Implementation

def find_array_concatenation_value(numbers):
    left_index = 0
    right_index = len(numbers) - 1
    concatenation_sum = 0

    while left_index < right_index:
        # Concatenate numbers from both ends and add to total
        combined_string = str(numbers[left_index]) + str(numbers[right_index])
        concatenation_sum += int(combined_string)

        left_index += 1
        right_index -= 1

    # If there's a single number left, add it to the total
    if left_index == right_index:
        concatenation_sum += numbers[left_index]

    return concatenation_sum

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n by pairing the first and last elements and removing them. In each iteration, it performs a constant number of operations: concatenating two numbers (which takes constant time as the number of digits is fixed based on problem constraints) and adding to the running total. Since it removes two elements in each step, the loop runs approximately n/2 times. Therefore, the time complexity is proportional to n/2, which simplifies to O(n).

Space Complexity

O(1) – The provided plain English explanation describes an iterative process that only involves manipulating the input list and accumulating a running total. No auxiliary data structures, such as temporary lists, hash maps, or recursion stacks, are created or utilized. The space required for variables like the running total and loop indices remains constant irrespective of the input list's size (N), making the auxiliary space complexity O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 if the input array is null or empty as there are no numbers to concatenate and sum.
Array with a single element	Return the single element itself since only one number remains.
Array with two elements	Concatenate the two elements into a string, convert it to an integer, and return it.
Array with large positive integers that, when concatenated, could cause integer overflow	Use a long data type for the sum to prevent integer overflow during concatenation and summation.
Array with negative integers	The concatenation and summation should still work correctly with negative numbers.
Array with mixed positive and negative integers including zeros	The concatenation and summation should still work correctly with zeros and both positive and negative integers.
Array with all zeros	The result will be zero after concatenation and summation of zeros.
Large array size	The solution should iterate through the array with linear time complexity which scales reasonably well with larger input sizes.

Null or empty input array

How to Handle:

Return 0 if the input array is null or empty as there are no numbers to concatenate and sum.

Array with a single element

How to Handle:

Return the single element itself since only one number remains.

Array with two elements

How to Handle:

Concatenate the two elements into a string, convert it to an integer, and return it.

Array with large positive integers that, when concatenated, could cause integer overflow

How to Handle:

Use a long data type for the sum to prevent integer overflow during concatenation and summation.

Array with negative integers

How to Handle:

The concatenation and summation should still work correctly with negative numbers.

Array with mixed positive and negative integers including zeros

How to Handle:

The concatenation and summation should still work correctly with zeros and both positive and negative integers.

Array with all zeros

How to Handle:

The result will be zero after concatenation and summation of zeros.

Large array size

How to Handle:

The solution should iterate through the array with linear time complexity which scales reasonably well with larger input sizes.