✅cpp easy with intuition || O(N) Time, O(1) space
```\nINTUITION\n1. since we have to find the minimum deletions, we dont have to \n\tactually delete the elements , we just have to count those elements.\n2. Now
```\nINTUITION\n1. since we have to find the minimum deletions, we dont have to \n\tactually delete the elements , we just have to count those elements.\n2. Now if we delete the element and shift all the elements towards left , it will\n\tcause time limit exceeded.\n3. To handle above case we can  observe one thing that, if we delete some element\n\tat a certain position then the indices of all the elements towards the right \n\twill get inverted means the odd index will become even and the even will become odd.\n4. In the end checking if the vector size if even or odd, if it is even simply return or else if it \n\tis odd then decrement result by 1 since we have to remove one element to have the vector size even.\nclass Solution {\npublic:\n    int minDeletion(vector<int>& nums) {\n        int res=0,n=nums.size(),i;\n        int flag=1;\n            for(i=0;i<nums.size()-1;i++){\n                    if(i%2==0 and nums[i]==nums[i+1]  and flag){\n                        res++;\n                        flag=0;\n                    }\n                    else if(i%2==1 and nums[i]==nums[i+1] and flag==0){\n                        res++;\n                        flag=1;\n                    }\n                }\n        int x=n-res;\n        if(x%2==0){\n            return res;\n        }\n        return res+1;\n    }\n};
['Greedy', 'C']