Surprise from leetcode
\nclass Solution {\npublic:\n    int smallestEqual(vector<int>& nums) {\n        int i=0, ans=-1;\n        for(i=0;i<nums.size();i++){\n            if(i%10==num
```\nclass Solution {\npublic:\n    int smallestEqual(vector<int>& nums) {\n        int i=0, ans=-1;\n        for(i=0;i<nums.size();i++){\n            if(i%10==nums[i]){\n                ans=i;\n                break;}\n        }\n        return ans;\n    }\n};\n```\n\n##### I know it is a easy question but there is some thing we can explore more like\n##### Since modullo operator is a very heavy operator , so instead of doing modullo every time ,\n##### We can just skip the elements who are greater than 9 and the code\'s execution time comes down\n##### from 16ms to 8ms . \n\n\t  for(int i=0 ; i< nums.size() ; i++)\n\t  {\n\t\t  if(nums[i] >= 10 ) continue;\n\t\t  if(nums[i] == i%10)\n\t\t\t  return i ;\n\t  }\n\t  return -1 ;\n\t  \n  Please **Upvote**
['C', 'C++']