🔥🚀Simplest Solution🚀||🔥Full Explanation||🔥C++🔥|| Python3 || Java
Consider\uD83D\uDC4D\n\n                    Please Upvote If You Find It Helpful\n\n# Intuition\nWe are taking k as carry.\nWe start from the last or lowest dig
# Consider\uD83D\uDC4D\n```\n                    Please Upvote If You Find It Helpful\n```\n# Intuition\nWe are taking `k` as carry.\nWe start from the last or lowest digit in array `num` add `k`.\nThen **update** `k` and move untill the highest digit.\nAfter traversing array if carry is  **>** `0` then we add it to begining of `num`.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n    Example: `num` = [2,1,5], `k` = 806\n    At index 2 num = [2, 1, 811] \n    So, `k` = 81 and `num` = [2, 1, 1]\n\n    At index 1 num = [2, 82, 1]\n    So, `k` = 8 and `num` = [2, 2, 1]\n\n    At index 0 num = [10, 2, 1]\n    So, `k` = 1 and `num` = [0, 2, 1]\n\n    Now `k` > 0\n    So, we add at the beginning of num\n    `num` = [1, 0, 2, 1]\n\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```C++ []\nclass Solution {\npublic:\n    vector<int> addToArrayForm(vector<int>& num, int k) {\n        for(int i=num.size()-1;i>=0;i--){\n            num[i] += k;\n            k = num[i]/10;\n            num[i] %= 10;\n        }\n        while(k > 0){\n            num.insert(num.begin(), k%10);\n            k /= 10;\n        }\n        return num;\n    }\n};\n```\n```python []\nclass Solution:\n    def addToArrayForm(self, num: List[int], k: int) -> List[int]:\n        for i in range(len(num) - 1, -1, -1):\n            k, num[i] = divmod(num[i] + k, 10)\n        while k:\n            k, a = divmod(k, 10)\n            num = [a] + num\n        return num\n```\n```Java []\npublic List<Integer> addToArrayForm(int[] num, int K) {\n    List<Integer> res = new LinkedList<>();\n    for (int i = num.length - 1; i >= 0; --i) {\n        res.add(0, (num[i] + K) % 10);\n        K = (num[i] + K) / 10;\n    }\n    while (K > 0) {\n        res.add(0, K % 10);\n        K /= 10;\n    }\n    return res;\n}\n```\n\n```\n                    Give a \uD83D\uDC4D. It motivates me alot\n```\nLet\'s Connect On [Linkedin](https://www.linkedin.com/in/naman-agarwal-0551aa1aa/)
['Array', 'Python', 'C++', 'Java', 'Python3']