48. Rotate Image

//🎯DAY 23 PROBLEM 1
//First transpose along the axis of top right to bottom left i.e.
// for I/P:   1 2 3      After transposing, we get :   8     9     6    3         9 6 3 
//            4 5 6                                      \     \     \            8 5 2
 //           7 8 9                                   7   4     5      2          7 4 1
//                                                                \
//                                                                  1
//


class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n=matrix.size();
      int i=0;
            for(int j=n-2;j>=0;j--)
            {
                int a=i;//we use a and b extra variables so as to make sure j dosen't changes as j is used for looping the first row and rest of the elements on right half of 
                        //diagonal is taken care by a++, b++
                int b=j;
                while(a+b<=(n-2))//SO THAT WE SWAP ONLY HALF WAY ALONG THE DIAGONAL i.e. we swap only 1,2,4 in the above matrix 1 2 3
                //                                                                                                              4 5 6
                //                                                                                                              7 8 9
                {
                   swap( matrix[a][b],matrix[(n-1)-b][(n-1)-a]);// to swap 2 i.e. cell at 0th row 1st col with  cell at (2-1) 1st row and (2-0) 0th col i.e. 6 
                    a++;
                    b++;
                }
            }
        int j=0;
        //starting from i=1 as 0 row 0 col element is already covered in above j for loop
        for( i=1;i<n-1;i++)
        {
            int a=i;
            int b=j;
                while(a+b<=(n-2))
                {
                   swap( matrix[a][b],matrix[(n-1)-b][(n-1)-a]);
                    a++;
                    b++;
                }
            
        }
//After the above process(transposing along the top right to bottom left axis), the given I/p matrix 
//1 2 3
//4 5 6
//7 8 9
//will be converted to:
//9 6 3
//8 5 2
//7 4 1
//Then reverse it:  i.e reverse of [[9,6,3],[8,5,2],[7,4,1]] is    [[7,4,1],[8,5,2],[9,6,3]]
//in matrix form, 7 4 1
//                8 5 2
//                9 6 3    is the anticlockwise image of given I/P
        reverse(matrix.begin(),matrix.end());
        
    }
};