27 August "Problem of the Day" Answer

                                                                  GeeksforGeeks
                                                           
The Question is :- "Alternate positive and negative numbers"
Answer :-

class Solution{
public:

	void rearrange(int arr[], int n) {
	    // code here
	    
	    //check for the efficient soln.
	    
	    //wht i did here is
	    //As we have time and space complexity as N
	    //we can stroe them, in another arrays and do the normal answer.
	    
	    //so wht i did is 
	    //i stored both the positive and neg elenmnts in the queues, not vectors bczo
	    //if queue, thn we remove the elements at top, with out using the indices, we can just use the key word, pop
	    //and for pushing also, we just need to use push,
	    //but where as for the vectors, we need to use the indeices to acces the elements
	    //and also to delete thm, it will b difficult.
	    
	    
	    //so, after storing thm in the queues, wht i did is
	    
	    //first as it was mentioned, we need positive first,
	    //we will start with the pos elements only.
	    
	    //so, we will run the loop, n times only.
	    //this only better coz, we know, how many times, we hav to run.
	    
	    //thn each time, if i is even, thn and if pos not empty.
	    //we replace the arra element as pos.front
	    //and pop.
	    //if pos empty, thn we wcheck for the neg elemnt
	    //thn we will pop frm neg elemmt.
	    
	    //similarly, we do like this for the i whn they r not divided by 2.
	    //in tht case, alos
	    //if neg not emty, thn we pop.
	    //if empty, thn we pop frm the pos if thts not empty.
	    
	    //thts it.
	    
	    
	    queue<int> pos,neg;
	    
	    for(int i=0;i<n;i++)
	    {
	        if(arr[i] < 0)
	        {
	            neg.push(arr[i]);
	        }
	        else
	        {
	            pos.push(arr[i]);    
	        }
	    }
	    for(int i=0;i<n;i++)
	    {
	        if(i%2 == 0)
	        {
	            if(!pos.empty())
	            {
	                arr[i] = pos.front();
	                pos.pop();
	            }
	            else
	            {
	                arr[i] = neg.front();
	                neg.pop();
	            }
	        }
	        
	        else
	        {
	            if(!neg.empty())
	            {
	                arr[i] = neg.front();
	                neg.pop();
	            }
	            else
	            {
	                arr[i] = pos.front();
	                pos.pop();
	            }
	        }
	    }
	    
	}
};

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