c++
if there exist any two sum with same sum set will overide the last one resulting into shorter set length\n.\n.\n.\nbool findSubarrays(vector& nums) {\n        s
*if there exist any two sum with same sum set will overide the last one resulting into shorter set length*\n.\n.\n.\nbool findSubarrays(vector<int>& nums) {\n        set<int> res;\n        for(int i=1;i<nums.size();i++){\n            res.insert(nums[i]+nums[i-1]);\n        }\n        return !(res.size()==nums.size()-1);\n    }
['Ordered Set']