Matrix Rotations
You are given a square matrix A of dimensions NxN. You need to apply the below given 3 operations on the matrix A.
Rotation: It is represented as R S where S is an integer in {90, 180, 270, 360, 450, ...} which denotes the number of degrees to rotate. You need to rotate the matrix A by angle S in the clockwise direction. The angle of rotation(S) will always be in multiples of 90 degrees.
Update: It is represented as U X Y Z. In initial matrix A (as given in input), you need to update the element at row index X and column index Y with value Z.
After the update, all the previous rotation operations have to be applied to the updated initial matrix.
Querying: It is represented as Q K L. You need to print the value at row index K and column index L of the matrix A.
Input
The first line contains a single integer N.
Next N lines contain N space-separated integers Aij (i - index of the row, j - index of the column).
Next lines contain various operations on the array. Each operation on each line (Beginning either with R, U or Q).
-1 will represent the end of input.
Output
For each Query operation print the element present at row index K and colum index L of the matrix in its current state.
Explanation
For Input:
2
1 2
3 4
R 90
Q 0 0
Q 0 1
R 90
Q 0 0
U 0 0 6
Q 1 1
-1
Initial Matrix
1 2
3 4
For R 90, clockwise rotation by 90 degrees, the matrix will become
3 1
4 2
For Q 0 0, print the element at row index 0 and column index 0 of A, which is 3.
For Q 0 1, print the element at row index 0 and column index 1 of A, which is 1.
Again for R 90, clockwise rotation by 90 degrees, the matrix will become
4 3
2 1
For Q 0 0, print the element at row index 0 and column index 0 of A, which is 4.
For U 0 0 6, update the value at row index 0 and column index 1 in the initial matrix to 6. So the updated matrix will be,
6 2
3 4
After updating, we need to rotate the matrix by sum of all rotation angles applied till now(i.e. R 90 and R 90 => 90 + 90 => 180 degrees in clockwise direction).
After rotation the matrix will now become
4 3
2 6
Next for Q 1 1, print the element at row index 1 and column index 1 of A, which is 6.
output
3
1
4
6
Sample Input 1
2
1 2
3 4
R 90
Q 0 0
Q 0 1
R 90
Q 0 0
U 0 0 6
Q 1 1
-1
Sample Output 1
3
1
4
6
Sample Input 2
2
5 6
7 8
R 90
Q 0 1
R 270
Q 1 1
R 180
U 0 0 4
Q 0 0
-1
Sample Output 2
5
8
8
i want exact sample outputs sir
def rotateMatrix(mat):
if not len(mat):
return
""
top : starting row index
bottom : ending row index
left : starting column index
right : ending column index
""
top = 0
bottom = len(mat)-1
left = 0
right = len(mat[0])-1
while left < right and top < bottom:
# Store the first element of next row,
# this element will replace first element of
# current row
prev = mat[top+1][left]
# Move elements of top row one step right
for i in range(left, right+1):
curr = mat[top][i]
mat[top][i] = prev
prev = curr
top += 1
# Move elements of rightmost column one step downwards
for i in range(top, bottom+1):
curr = mat[i][right]
mat[i][right] = prev
prev = curr
right -= 1
# Move elements of bottom row one step left
for i in range(right, left-1, -1):
curr = mat[bottom][i]
mat[bottom][i] = prev
prev = curr
bottom -= 1
# Move elements of leftmost column one step upwards
for i in range(bottom, top-1, -1):
curr = mat[i][left]
mat[i][left] = prev
prev = curr
left += 1
return mat
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