#2562

Easy

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Find the Array Concatenation Value

Compute the array concatenation value efficiently using two-pointer scanning while maintaining a running total of concatenations.

array two pointers simulation

Problem Statement

You are given a 0-indexed integer array nums. The concatenation of two numbers is defined as the number formed by joining their digits sequentially. Your task is to calculate the concatenation value of the array following a specific operation.

The operation proceeds as follows: while nums is not empty, pick the first and last elements, concatenate them as described, and add the result to a running total. Remove these elements from nums. If only one element remains, add it directly to the total. Return the final concatenation value after all elements are processed.

Examples

Example 1

Input: nums = [7,52,2,4]

Output: 596

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

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

Solution Approach

Two-Pointer Simulation

Initialize two pointers at the start and end of the array. Iterate while the start pointer is less than or equal to the end pointer, concatenate the numbers at both pointers, add to the total, and move pointers inward. This directly implements the two-pointer scanning pattern for accurate concatenation.

Single Element Handling

If the array has an odd number of elements, ensure the middle element is added once to the total when start equals end. This prevents double counting and aligns with the invariant that all elements contribute exactly once to the concatenation value.

Efficient String Concatenation

Convert numbers to strings for concatenation, then back to integers to add to the running total. This approach avoids arithmetic errors and respects the problem's simulation nature, ensuring correctness while iterating linearly through the array.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) since each element is processed exactly once with two-pointer scanning. Space complexity is O(1) additional space beyond input storage if string conversion is done in-place for concatenation.

What Interviewers Usually Probe

Notice the need to process elements from both ends simultaneously for correct concatenation order.
Check for odd-length arrays to avoid double counting the middle element.
Simulate the concatenation process step by step rather than attempting a formulaic shortcut.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle the single middle element in odd-length arrays, which leads to an incorrect total.
Attempting numeric concatenation without string conversion, which fails for multi-digit numbers.
Modifying the array incorrectly, which breaks the invariant of elements being counted exactly once.

Follow-up variants

Compute the concatenation value for arrays where numbers can have leading zeros, affecting string concatenation.
Return the intermediate total after each operation instead of only the final value.
Extend the problem to k-pair concatenation instead of just first and last elements, increasing pointer complexity.

FAQ

What is the best approach to solve Find the Array Concatenation Value?

A two-pointer simulation from both ends of the array ensures each concatenation is correctly calculated and added to the total.

How do I handle an odd-length array in this problem?

When the start and end pointers meet at a single element, add it once to the total to prevent double counting.

Can I concatenate numbers numerically instead of as strings?

Direct numeric concatenation may fail for multi-digit numbers. Convert numbers to strings first to ensure correct concatenation.

What is the time complexity for this two-pointer method?

Each element is processed once, so the time complexity is O(n) with linear traversal of the array.

Why is two-pointer scanning the key pattern here?

It preserves the order of concatenation from both ends, maintaining the problem's invariant and ensuring accurate total calculation.

terminal

Solution

Solution 1: Simulation

Starting from both ends of the array, we take out one element at a time, concatenate it with another element, and then add the concatenated result to the answer. We repeat this process until the array is empty.

1

2

3

4

5

6

7

8

9

10

class Solution:
    def findTheArrayConcVal(self, nums: List[int]) -> int:
        ans = 0
        i, j = 0, len(nums) - 1
        while i < j:
            ans += int(str(nums[i]) + str(nums[j]))
            i, j = i + 1, j - 1
        if i == j:
            ans += nums[i]
        return ans

Solution 2

#### Rust

1

2

3

4

5

6

7

8

9

10

class Solution:
    def findTheArrayConcVal(self, nums: List[int]) -> int:
        ans = 0
        i, j = 0, len(nums) - 1
        while i < j:
            ans += int(str(nums[i]) + str(nums[j]))
            i, j = i + 1, j - 1
        if i == j:
            ans += nums[i]
        return ans

Continue Practicing

#2462 total cost to hire k workers #2460 apply operations to an array #2161 partition array according to given pivot