Easy Solution with Proper Explanation || c++ ||Everything Explained properly :)
Intuition\n Describe your first thoughts on how to solve this problem. \nThe code leverages the properties of integer division and the remainder operation to ex
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code leverages the properties of integer division and the remainder operation to extract the separate digits from each number. Reversing the vector helps ensure correct digit extraction, and reversing the result vector restores the original order of the digits.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n The overall approach used in the code is as follows:\n\nThe input vector nums is reversed using reverse(nums.begin(), nums.end()). This step allows the digits to be extracted from least to most significant.\n\nThe code then iterates over each number in the reversed nums vector.\n\nInside the loop, the current number k is assigned to a temporary variable.\n\nA while loop is used to extract the digits from the current number. It continues until k becomes 0.\n\nInside the while loop, the least significant digit is extracted by calculating the remainder of k divided by 10. This digit is assigned to the variable c.\n\nThe extracted digit c is appended to the ans vector using ans.push_back(c).\n\nThe least significant digit is removed from the current number by dividing k by 10 (k = k / 10).\n\nThe process continues until k becomes 0, indicating that all digits of the current number have been extracted.\n\nFinally, the ans vector is reversed using reverse(ans.begin(), ans.end()) to restore the original order of the digits.\n\nThe ans vector is returned as the result of the separateDigits function.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe overall time complexity of the code is O(n * log10(k)), where n is the size of the vector and k is the maximum value among all the numbers in the vector.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is O(d), where d is the total number of digits in all the numbers in the vector.\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> separateDigits(vector<int>& nums) {\n        int n = nums.size();\n        int c;\n        vector<int> ans;\n        \n        // Reverse the input vector to extract digits from least to most significant\n        reverse(nums.begin(), nums.end());\n        \n        // Iterate over each number in the input vector\n        for (int i = 0; i < n; i++) {\n            int k = nums[i];\n            \n            // Extract digits from the current number\n            while (k) {\n                // Extract the least significant digit (remainder of division by 10)\n                c = k % 10;\n                \n                // Append the extracted digit to the answer vector\n                ans.push_back(c);\n                \n                // Remove the least significant digit from the current number\n                k = k / 10;\n            }\n        }\n        \n        // Reverse the answer vector to restore the original digit order\n        reverse(ans.begin(), ans.end());\n        \n        // Return the vector containing the separate digits\n        return ans;\n    }\n};\n\n```
['C++']