Easy Solution | Faster using Two Pointers  and  Strings 3ms
Intuition\n1. We need first element & last element from array -> so we will use Two Pointer to solve.\n2. We need to Concate these two elements from array -> so
# Intuition\n1. We need first element & last element from array -> so we will use Two Pointer to solve.\n2. We need to Concate these two elements from array -> so we will use Strings.\n3. And then we need to process the ans. \n\n# Approach\nWe will get both elements from pointers (low,high) and then convert them into Strings , and then concate them, store the result into another string.\nNow we have to add the resultant string in the form of Integer so, convert the resultant string into Integer.\n\nNow we have to cases:\n  \n![WhatsApp Image 2023-02-12 at 12.55.16.jpeg](https://assets.leetcode.com/users/images/0308d2f4-c13c-4a26-8238-5ed0e9796b94_1676186919.783601.jpeg)\n\nCase 2:\nIf we run a loop (low <= high) then our middle element will get concatenated twice, & if not then our middle element will be remained from getting concatenated.\n\nTo avoid this if array length is odd then after loop directly add the middle element. \n\n# Complexity\n- Time complexity: O(n)\n\n# Code\n```\nclass Solution {\n    public long findTheArrayConcVal(int[] nums) {\n        \n        int low = 0;\n        int high = nums.length-1;\n        \n        long ans = 0;\n        \n        while(low < high){\n           \n            String s1 = Integer.toString(nums[low]);\n            String s2 = Integer.toString(nums[high]);\n            \n            String s3 = s1 + s2;\n            \n            ans += (Integer.parseInt(s3));\n            \n            low++;\n            high--;\n        }\n        \n        if(nums.length%2 != 0)\n            ans += nums[low];\n        \n        return ans;\n    }\n}\n```
['Two Pointers', 'String', 'Java']