Easy C++ solution || commented
\nclass Solution {\npublic:\n    int minElements(vector<int>& nums, int limit, int goal) {\n     \n        //just calculate how many limit values to be added to
```\nclass Solution {\npublic:\n    int minElements(vector<int>& nums, int limit, int goal) {\n     \n        //just calculate how many limit values to be added to the array to make our currSum close to the goal\n        long int currSum=0;\n        for(auto ele:nums)\n            currSum+=ele;\n        long int remaining=abs(currSum-goal);\n        if(remaining==0)\n            return 0;\n        long int res=(remaining/limit);\n        //if remaining exists still after adding limit k times\n        //ex - remaining - 7 limit -3 we have to add a 1 after adding two times of limit values (6)\n        res+= (remaining%limit!=0) ? 1 : 0;\n        return res;\n        \n    }\n};\n```\n**Do upvote if this helps you :)**
['Greedy', 'C']