✅ EASY & BETTER C++ Solution | EXPLAINED | IN-PLACE | O(n)
Algorithm:\n1. Traversing the vector array from last and continuously adding the last element of integer \'K\'\n\ta. Adding k to num[i]\n\tb. Updating k by num[
Algorithm:\n1. Traversing the vector array from last and continuously adding the last element of integer \'K\'\n\ta. Adding k to num[i]\n\tb. Updating k by num[i]/10\n\tb. Updating num[i] by num[i]%10  \n2. If any carry left, it will be inserted at start of the vector array num\n3. Return the array.\n\nTime Complexity: O(n)\nSpace Complexity: O(1)\n\nThanks for reading, If you like it, please upvote \u2B06\uFE0F and help me gain good reputation.\n\n```\nclass Solution {\npublic:\n    vector<int> addToArrayForm(vector<int>& num, int k) {\n        // Peforming element wise sum\n        for (int i=num.size()-1; i>=0 && k>0; i--) {\n            num[i]+=k;\n            k=num[i]/10;\n            num[i]%=10;\n        }\n        \n        // if carry left, insert it in vector in front\n        while(k>0) {\n            num.insert(num.begin(), k%10);\n            k/=10;\n        }\n        \n        return num;\n    }\n};\n```
['C']