LeetCode 2562 - Find the Array Concatenation Value

The problem is asking us to compute a special sum over a given array nums by repeatedly concatenating the first and last elements of the array until it is empty.

leetcodeeasyarraytwo-pointerssimulation

LeetCode Problem 2562

Difficulty: 🟢 Easy
Topics: Array, Two Pointers, Simulation

Solution

Problem Understanding

The problem is asking us to compute a special sum over a given array nums by repeatedly concatenating the first and last elements of the array until it is empty. The concatenation operation is not addition; it involves treating numbers as strings, joining them together, and then converting them back to integers. For example, concatenating 15 and 49 produces 1549.

The process continues in two cases: if there are at least two elements, we remove the first and last elements after concatenation; if there is exactly one element left, we simply add its value to the result and remove it. The array is 0-indexed and can have up to 1000 elements, each being an integer between 1 and 10^4. This means the array is small enough that a straightforward simulation is feasible, but we should avoid unnecessary string conversions or extra space if possible.

Important edge cases include arrays of length 1 (single element), arrays with all identical elements, and arrays with varying number lengths which affect concatenation (e.g., [1, 1000] results in 11000).

Approaches

The brute-force approach is a direct simulation. We repeatedly take the first and last elements, convert them to strings, concatenate them, convert back to integers, add to the running total, and remove the elements from the array. This approach is correct and simple, but string conversion and array slicing in each iteration can be slightly inefficient for larger arrays.

The optimal approach is a two-pointer technique. Instead of slicing the array, we maintain two indices (left and right) pointing to the start and end. In each iteration, we compute the concatenation by converting the numbers to strings and then back to an integer, add it to the result, and move the pointers inward. This avoids modifying the array repeatedly and is more memory-efficient.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * m) O(n) Uses array slicing and string conversion in each iteration; m is number of digits in numbers
Two-Pointer Optimal O(n * m) O(1) Uses two pointers to avoid array slicing; still needs string conversion for concatenation

Algorithm Walkthrough

  1. Initialize two pointers: left = 0 and right = len(nums) - 1. Initialize conc_val = 0 to store the concatenation value.
  2. While left is less than or equal to right, perform the following:

a. If left is equal to right, it means only one element remains. Add nums[left] to conc_val and break the loop.

b. Otherwise, compute the concatenation of nums[left] and nums[right] by converting both numbers to strings, concatenating, and converting back to an integer. Add this value to conc_val.

c. Increment left and decrement right to move inward. 3. Return conc_val as the final answer.

Why it works: Each iteration correctly handles the first and last elements, ensuring that all elements are included in the sum exactly once. Using two pointers guarantees O(n) iterations without extra array operations. The concatenation preserves the intended numeric order since converting to string and back accurately models the operation described in the problem.

Python Solution

from typing import List

class Solution:
    def findTheArrayConcVal(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1
        conc_val = 0
        
        while left <= right:
            if left == right:
                conc_val += nums[left]
            else:
                conc_val += int(str(nums[left]) + str(nums[right]))
            left += 1
            right -= 1
        
        return conc_val

The Python implementation follows the two-pointer strategy. left and right track the current first and last elements. The condition left == right handles the single-element case, while the concatenation uses string conversion. The solution avoids slicing the array and operates in linear time relative to the number of elements.

Go Solution

import "strconv"

func findTheArrayConcVal(nums []int) int64 {
    var concVal int64 = 0
    left, right := 0, len(nums)-1
    
    for left <= right {
        if left == right {
            concVal += int64(nums[left])
        } else {
            concatStr := strconv.Itoa(nums[left]) + strconv.Itoa(nums[right])
            val, _ := strconv.ParseInt(concatStr, 10, 64)
            concVal += val
        }
        left++
        right--
    }
    
    return concVal
}

In Go, we use strconv.Itoa to convert integers to strings for concatenation and strconv.ParseInt to convert back to int64. The solution uses int64 to handle potential overflow since concatenation could create large numbers. The two-pointer approach mirrors the Python implementation, ensuring efficiency.

Worked Examples

Example 1: nums = [7, 52, 2, 4]

Step left right nums[left] nums[right] concatenation conc_val
1 0 3 7 4 74 74
2 1 2 52 2 522 596

Final answer: 596

Example 2: nums = [5, 14, 13, 8, 12]

Step left right nums[left] nums[right] concatenation conc_val
1 0 4 5 12 512 512
2 1 3 14 8 148 660
3 2 2 13 13 13 673

Final answer: 673

Complexity Analysis

Measure Complexity Explanation
Time O(n * m) Each of the n iterations may involve converting numbers to strings; m is the maximum number of digits (up to 5)
Space O(1) No extra data structures are used; string conversion is temporary

The algorithm is efficient for the given constraints since n <= 1000 and numbers have at most 5 digits.

Test Cases

# Basic examples
assert Solution().findTheArrayConcVal([7,52,2,4]) == 596  # Provided example 1
assert Solution().findTheArrayConcVal([5,14,13,8,12]) == 673  # Provided example 2

# Edge cases
assert Solution().findTheArrayConcVal([1]) == 1  # Single element
assert Solution().findTheArrayConcVal([10, 1000]) == 101000  # Large concatenation
assert Solution().findTheArrayConcVal([2,2,2,2]) == 2224  # All identical elements

# Maximum length
assert Solution().findTheArrayConcVal(list(range(1,1001))) > 0  # Performance test
Test Why
[7,52,2,4] Validates basic operation and multiple iterations
[5,14,13,8,12] Tests odd-length array and single-element last case
[1] Single-element edge case
[10, 1000] Concatenation results in larger numbers, tests string handling
[2,2,2,2] Identical elements, checks correctness
range(1,1001) Performance test with maximum array size

Edge Cases

The first edge case is a single-element array, e.g., [7]. This is important because the main loop handles two-element concatenations, so if we do not check for a single element, the code could attempt to access invalid indices. Our solution explicitly handles left == right to add the lone element.

The second edge case is large numbers, for example [10, 1000]. Concatenation produces numbers up to 10^9 or higher. In Python, integers handle arbitrary size, but in Go we must use int64 to avoid overflow. The solution properly converts concatenated strings to int64.

The third edge case is arrays with odd length, such as [1,2,3]. The middle element will eventually be alone, and we need to add it once. The left <= right condition ensures the middle element is correctly added when left == right. This prevents off-by-one errors.