Easy approach BEATS 100% using two pointers
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int sumOfSquares(int[] nums) {\n        int n=nums.length;\n        int prod=1,sum=0;\n       for(int i=1;i<=nums.length;i++)\n       {\n           if(n%i==0)\n           {\n               prod=nums[i-1]*nums[i-1];\n               sum+=prod;\n           }\n       } \n       return sum;\n    }\n}\n```
['Two Pointers', 'Binary Search', 'C', 'C++', 'Java']