c++ solution --> very simple-->O(n)
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nNo need to traverse all array when found one value simply break, as only 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nNo need to traverse all array when found one value simply break, as only minimum value is required.If value is not find ans intialized with -1 will be returned.\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n    int smallestEqual(vector<int>& nums) {\n        int ans=-1;\n        for(int i=0;i<nums.size();i++){\n         \n            if(i%10==nums[i]){\n                ans=i;\n                break;\n            }\n        }\n  return ans;  }\n};\n```
['C++']