✔️C++ SOLUTION✔️
PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT\n# CONNECT WITH ME\n### https://www.linkedin.com/in/pratay-nandy-9ba57b229/\n#### https://www.instagram.com/pratay_nand
# **PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT**\n# **CONNECT WITH ME**\n### **[https://www.linkedin.com/in/pratay-nandy-9ba57b229/]()**\n#### **[https://www.instagram.com/pratay_nandy/]()**\n# Approach\nThe provided C++ code is an updated version of the previous code and defines a class "Solution" with a public member function "addToArrayForm." This function takes a vector of integers `nums` and an integer `k` as input and aims to add the integer `k` to the number represented by the elements of the `nums` vector and return the result as a new vector. This updated code addresses memory usage by minimizing the need for an extra vector and optimization for better clarity.\n\nHere\'s the approach used in this code:\n\n1. Initialize an empty vector `ans` to store the result.\n\n2. Initialize two variables, `i` and `carry`. `i` is set to the index of the last element in the `nums` vector, and `carry` is initially set to 0.\n\n3. The code uses a while loop to perform the addition. It handles three cases:\n\n   a. **Both number and array of the same size:** While `i` is greater than or equal to 0 and `k` is greater than 0, the code extracts the least significant digit (rightmost digit) of `k` and the corresponding digit from the `nums` array, adds them together along with the carry from the previous operation, calculates the new carry and the sum. The sum is added to the `ans` vector, and `i` and `k` are decremented.\n\n   b. **When the array is bigger than the number:** If `i` is still greater than or equal to 0 after the first loop, it means the array is longer than the number. In this case, the code adds the elements of `nums` and the carry, calculates the new carry, and adds the result to the `ans` vector.\n\n   c. **When the number is bigger than the array:** If `k` is still greater than 0 after the first loop, it means the number is longer than the array. The code extracts the least significant digit of `k`, adds it to the carry, calculates the new carry, and adds the result to the `ans` vector.\n\n4. After all the additions, if there is still a carry, it is added as the most significant digit to the `ans` vector in a separate loop.\n\n5. Finally, the `ans` vector is reversed to ensure the least significant digit is at the beginning, and it is returned as the result.\n\nThe code effectively performs the addition of the integer `k` to the number represented by the `nums` vector and returns the result as a new vector, taking into account different scenarios involving array and number sizes. This updated code is designed to optimize memory usage.\n\n\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:0(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:0(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> addToArrayForm(vector<int>& nums, int k) {\n        vector<int>ans;\n        int i = nums.size()-1; \n        int carry = 0;\n        // both number and array of same size\n        while(i>=0 && k>0)\n        {\n            int left = k%10;\n            k = k/10;\n            int sum = nums[i] + left + carry;\n            carry = sum / 10;\n            sum=sum % 10;  \n            ans.push_back(sum);\n            i--; \n        }\n        // when array is big\n        while(i>=0)\n        {\n            int sum = nums[i] + carry;\n            carry = sum / 10;\n            sum=sum % 10;  \n            ans.push_back(sum);\n            i--; \n        }\n        // when number is big\n        while(k>0)\n        {\n            int left = k%10;\n            k = k/10;\n            int sum =  left + carry;\n            carry = sum / 10;\n            sum=sum % 10;  \n            ans.push_back(sum);\n        }\n        while(carry!=0)\n        {\n            int sum = carry;\n            carry = sum / 10;\n            sum=sum % 10;  \n            ans.push_back(sum);\n        }\n        reverse(ans.begin(), ans.end());\n        return ans;\n    }\n};\n```
['Array', 'Math', 'C++']