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Formatted question description: https://leetcode.ca/all/2562.html
2562. Find the Array Concatenation Value
Description
You are given a 0indexed integer array nums
.
The concatenation of two numbers is the number formed by concatenating their numerals.
 For example, the concatenation of
15
,49
is1549
.
The concatenation value of nums
is initially equal to 0
. Perform this operation until nums
becomes empty:
 If there exists more than one number in
nums
, pick the first element and last element innums
respectively and add the value of their concatenation to the concatenation value ofnums
, then delete the first and last element fromnums
.  If one element exists, add its value to the concatenation value of
nums
, then delete it.
Return the concatenation value of the 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] <= 10^{4}
Solutions
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.
The time complexity is $O(n \times \log M)$, and the space complexity is $O(\log M)$. Here, $n$ and $M$ are the length of the array and the maximum value in the array, respectively.

class Solution { public long findTheArrayConcVal(int[] nums) { long ans = 0; int i = 0, j = nums.length  1; for (; i < j; ++i, j) { ans += Integer.parseInt(nums[i] + "" + nums[j]); } if (i == j) { ans += nums[i]; } return ans; } }

class Solution { public: long long findTheArrayConcVal(vector<int>& nums) { long long ans = 0; int i = 0, j = nums.size()  1; for (; i < j; ++i, j) { ans += stoi(to_string(nums[i]) + to_string(nums[j])); } if (i == j) { ans += nums[i]; } return ans; } };

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

func findTheArrayConcVal(nums []int) (ans int64) { i, j := 0, len(nums)1 for ; i < j; i, j = i+1, j1 { x, _ := strconv.Atoi(strconv.Itoa(nums[i]) + strconv.Itoa(nums[j])) ans += int64(x) } if i == j { ans += int64(nums[i]) } return }

function findTheArrayConcVal(nums: number[]): number { const n = nums.length; let ans = 0; let i = 0; let j = n  1; while (i < j) { ans += Number(`${nums[i]}${nums[j]}`); i++; j; } if (i === j) { ans += nums[i]; } return ans; }

impl Solution { pub fn find_the_array_conc_val(nums: Vec<i32>) > i64 { let n = nums.len(); let mut ans = 0; let mut i = 0; let mut j = n  1; while i < j { ans += format!("{}{}", nums[i], nums[j]).parse::<i64>().unwrap(); i += 1; j = 1; } if i == j { ans += nums[i] as i64; } ans } }