Problem Description

You are given a 0-indexed integer array nums. Your task is to calculate the concatenation value of this array through a specific process.

Concatenation means joining two numbers together as strings. For example, concatenating 15 and 49 gives 1549 (not 64, which would be addition).

Starting with a concatenation value of 0, you need to repeatedly perform the following operation until the array becomes empty:

1. If the array has more than one element: Take the first and last elements, concatenate them (join them as strings to form a new number), add this concatenated number to your running total, and remove both elements from the array.

   • Example: If nums = [1, 2, 4, 5, 6], you would concatenate 1 and 6 to get 16, add 16 to your total, and remove both 1 and 6 from the array.

2. If the array has exactly one element: Simply add that element's value to your running total and remove it from the array.

The process continues until all elements are removed from the array. Return the final concatenation value.

Example walkthrough:

• For nums = [1, 2, 4, 5, 6]:
  • First iteration: Concatenate 1 and 6 → 16, add to total (total = 16)
  • Second iteration: Concatenate 2 and 5 → 25, add to total (total = 41)
  • Third iteration: Only 4 remains, add 4 to total (total = 45)
  • Return 45

Intuition

The problem asks us to repeatedly take elements from both ends of the array and concatenate them. This immediately suggests using a two-pointer approach - one pointer starting from the beginning and another from the end of the array.

Think about how we would manually solve this problem: we'd look at the first element, look at the last element, combine them, and then move inward. This is exactly what two pointers can do efficiently.

The key insight is that we don't actually need to modify the array by removing elements. Instead, we can simulate this process by:

• Maintaining a left pointer i starting at index 0
• Maintaining a right pointer j starting at the last index
• Moving both pointers toward each other after each concatenation

For concatenation, we need to join two numbers as strings. The straightforward way is to convert both numbers to strings using str(), concatenate them with +, and then convert back to an integer using int(). So for numbers a and b, the concatenation would be int(str(a) + str(b)).

The process continues while i < j, meaning we still have at least two elements to process. When the pointers meet (i == j), we have exactly one element left in the middle, which we simply add to our result without any concatenation.

This approach is elegant because:

1. We avoid the costly operation of actually removing elements from the array
2. We process the array in a single pass with O(n) time complexity
3. The logic naturally handles both even and odd-length arrays

Learn more about Two Pointers patterns.

Solution Approach

We implement the solution using the two-pointer technique to simulate the process of taking elements from both ends of the array.

Step-by-step implementation:

1. Initialize variables:

   • ans = 0 to store the concatenation value
   • i = 0 as the left pointer starting from the beginning
   • j = len(nums) - 1 as the right pointer starting from the end

2. Process pairs of elements:

   • While i < j, we have at least two elements to process
   • Convert both nums[i] and nums[j] to strings using str()
   • Concatenate them: str(nums[i]) + str(nums[j])
   • Convert the concatenated string back to integer: int(str(nums[i]) + str(nums[j]))
   • Add this value to ans
   • Move the pointers inward: i = i + 1 and j = j - 1

3. Handle the middle element (if exists):

   • After the loop, if i == j, there's exactly one element left in the middle
   • Simply add nums[i] to ans without any concatenation

4. Return the result:

   • Return ans as the final concatenation value

Example walkthrough with nums = [7, 52, 2, 4]:

• Initial: i = 0, j = 3, ans = 0
• First iteration: Concatenate 7 and 4 → 74, ans = 74, move to i = 1, j = 2
• Second iteration: Concatenate 52 and 2 → 522, ans = 74 + 522 = 596, move to i = 2, j = 1
• Loop ends since i >= j
• No middle element since i > j
• Return 596

Time Complexity: O(n) where n is the length of the array, as we visit each element exactly once.

Space Complexity: O(1) as we only use a constant amount of extra space for the pointers and the answer variable.

Example Walkthrough

Let's walk through the solution with nums = [3, 8, 15, 9]:

Initial Setup:

• Two pointers: i = 0 (pointing to 3), j = 3 (pointing to 9)
• ans = 0 (our running total)

Step 1: Process first pair (i=0, j=3)

• Elements: nums[0] = 3 and nums[3] = 9
• Concatenate: str(3) + str(9) = "3" + "9" = "39"
• Convert to integer: int("39") = 39
• Update total: ans = 0 + 39 = 39
• Move pointers inward: i = 1, j = 2

Step 2: Process second pair (i=1, j=2)

• Elements: nums[1] = 8 and nums[2] = 15
• Concatenate: str(8) + str(15) = "8" + "15" = "815"
• Convert to integer: int("815") = 815
• Update total: ans = 39 + 815 = 854
• Move pointers inward: i = 2, j = 1

Step 3: Check for middle element

• Now i = 2 and j = 1, so i > j
• No middle element exists (array had even length)
• Process complete

Final Result: Return ans = 854

The two-pointer approach efficiently processes the array from both ends without actually removing elements, completing in just n/2 iterations for an array of length n.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log M)

The algorithm iterates through the array using two pointers that meet in the middle, performing n/2 iterations where n is the length of the array. In each iteration, the main operation is concatenating two numbers by converting them to strings and back to an integer. Converting a number to a string takes O(log M) time where M is the maximum value in the array (since the number of digits in a number is proportional to log M). The string concatenation and conversion back to integer also take O(log M) time. Therefore, the overall time complexity is O(n/2 × log M) = O(n × log M).

Space Complexity: O(log M)

The space complexity comes from the temporary string objects created during the concatenation operation. When converting numbers to strings and concatenating them, the maximum space needed is proportional to the number of digits in the largest number, which is O(log M). The algorithm uses only a constant amount of additional variables (ans, i, j), and the string concatenation creates temporary strings whose maximum length is bounded by 2 × log M = O(log M).

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

Common Pitfalls

1. Incorrect String Concatenation Logic

A common mistake is trying to concatenate numbers mathematically instead of as strings. For example, attempting to concatenate 15 and 49 by calculating 15 * 100 + 49 = 1549. While this might work for some cases, it fails when dealing with numbers that have leading zeros in their decimal representation or when the second number has varying digits.

Wrong approach:

# This fails for nums[j] = 0 or when nums[j] has varying digit counts
concatenated = nums[left] * (10 ** len(str(nums[right]))) + nums[right]

Correct approach:

concatenated = int(str(nums[left]) + str(nums[right]))

2. Forgetting to Handle the Middle Element

When the array has an odd length, there will be one element left in the middle after processing all pairs. Forgetting to add this element to the total sum will result in an incorrect answer.

Incomplete solution:

while left < right:
    total += int(str(nums[left]) + str(nums[right]))
    left += 1
    right -= 1
# Missing: if left == right: total += nums[left]
return total

3. Off-by-One Error in Pointer Initialization

Initializing the right pointer incorrectly (e.g., right = len(nums) instead of len(nums) - 1) will cause an index out of bounds error.

Wrong initialization:

right_index = len(nums)  # This will cause IndexError

Correct initialization:

right_index = len(nums) - 1

4. Using Wrong Loop Condition

Using <= instead of < in the while loop condition would process the middle element twice in arrays with odd length, leading to incorrect results.

Wrong condition:

while left_index <= right_index:  # This processes middle element in the loop
    concatenated = int(str(nums[left_index]) + str(nums[right_index]))
    # When left_index == right_index, this concatenates the same element with itself

Correct condition:

while left_index < right_index:  # Stop when pointers meet or cross