Easy C++  solution W/o Sorting || 99% faster || commented fully
\nclass Solution {\npublic:\n    vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) {\n        \n        vector<bool> ans(
```\nclass Solution {\npublic:\n    vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) {\n        \n        vector<bool> ans(l.size(),false);\n        for(int i=0;i<l.size();i++)\n        {\n            //if the sub array size is equal to 2\n            if(r[i]-l[i]+1==2)\n                ans[i]=true;\n            else if(isArthemetic(nums,l[i],r[i]))\n                ans[i]=true;\n        }\n        return ans;\n    }\n    \n    //return true if the sub array can be rearranged in a arthemetic sequence\n    bool isArthemetic(vector<int>& nums,int start,int end)\n    {\n        //get the maximum and min elements \n        int mini=INT_MAX;\n        int maxi=INT_MIN;\n        for(int i=start;i<=end;i++)\n        {\n            mini=min(nums[i],mini);\n            maxi=max(nums[i],maxi);\n        }\n        \n        //when mini==maxi it means all the elements are same in the sub array\n        if(mini==maxi)\n            return true; \n        \n        //we cant have same common difference betweeen  two adjacent  elements\n        //when we arrange in arthemetic sequence\n        if((maxi-mini)%(end-start)!=0)\n            return false;\n        \n        \n        //the diff between every two integers when we rearrange sub array\n        int diff=(maxi-mini)/(end-start);\n        \n        //to check  if the duplicate elemnts are present\n        //ex- [2,4,6,6]\n        //6 is repeating two times \n        vector<bool> present(end-start+1,false);\n        for(int i=start;i<=end;i++)\n        {\n            \n            //we cant set a index of nums[i]\n            if((nums[i]-mini)%diff!=0)\n                return false;\n            \n            int ind=(nums[i]-mini)/diff;\n            \n            // same element is alreeady repeated ( 6 in the above  example)\n            if(present[ind])\n                return false;\n            //mark it presence\n            present[ind]=true;\n            \n        }\n        return true;\n        \n    }\n};\n```\n**Upvote if this helps you :)**
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