Anticlockwise spiral view of diamond element in matrix
The problem at hand is to find the anticlockwise spiral view of diamond elements in a given matrix. The diamond elements are the central elements forming a diamond shape in the matrix. The anticlockwise spiral view means traversing these diamond elements in an anticlockwise spiral manner.
Problem Statement and Example
Consider the following matrix:
1 2 3 4 5
6 7 8 9 10
11 12 1 15 16
17 18 19 20 21
22 23 24 25 26
The diamond elements in this matrix are:
3
7 8 9
11 12 1 15 16
18 19 20
24
The anticlockwise spiral view of these elements will be: 3 7 11 18 24 20 16 9 8 12 19 15 1
Idea to Solve the Problem
To find the anticlockwise spiral view of diamond elements in the matrix, we can follow the following approach:
 First, we need to check if the given matrix is a valid oddsized square matrix. If it is not, we return as the problem cannot be solved with an invalid matrix.
 We then find the middle row and middle column of the matrix. For an oddsized square matrix, both middle row and middle column will be the same.
 We divide the spiral traversal into four cases, corresponding to each side of the diamond shape.
 For each case, we start from the middle row and middle column and traverse in an anticlockwise spiral manner while printing the elements.
 After each case, we reduce the size of the diamond and make a recursive call to continue the traversal until all diamond elements are covered.
Pseudocode
spiralView(matrix, x, y, p, q, k):
mid_col = y + ((q  y) / 2)
mid_row = mid_col
// Case A (Top to Leftbottom)
i = mid_col
j = 0
while i > y and k > 0:
print matrix[x + j][i]
i
k
j++
// Case B (Middle left to middle right bottom)
i = y
j = 0
while i <= mid_col and k > 0:
print matrix[mid_row + j][i]
i++
k
j++
// Case C (Middle bottom to middle right side)
i = mid_col + 1
j = 1
while i <= q and k > 0:
print matrix[p  j][i]
i++
k
j++
// Case D (Middle right side to top middle)
i = q  1
j = 1
while i > mid_col and k > 0:
print matrix[mid_row  j][i]
i
k
j++
if x + 1 <= p  1 and k > 0:
// Recursive call
spiralView(matrix, x + 1, y + 1, p  1, q  1, k)
Algorithm Explanation
 The 'spiralView' function takes 'matrix', 'x', 'y', 'p', 'q', and 'k' as input parameters. 'matrix' is the given matrix, 'x' and 'y' are the starting row and column, 'p' and 'q' are the ending row and column, and 'k' is the number of diamond elements to traverse.
 The function finds the middle row and middle column of the matrix as 'mid_row' and 'mid_col'.
 It then divides the spiral traversal into four cases (A, B, C, and D) representing each side of the diamond shape.
 For each case, it starts from the corresponding position and moves in an anticlockwise spiral manner while printing the elements.
 After each case, the function reduces the size of the diamond (by updating 'x', 'y', 'p', 'q', and 'k') and makes a recursive call to continue the traversal until all diamond elements are covered.
Code Solution

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Output Explanation
The given Java code creates a matrix and calls the 'diamondAnticlockwise' function to find the anticlockwise spiral view of diamond elements. The output displays the anticlockwise spiral view of the diamond elements as follows:
3 7 11 18 24 20 16 9 8 12 19 15 1
Time Complexity
The time complexity of the provided solution is O(N), where N is the total number of elements in the matrix. The function 'spiralView' recursively traverses the matrix while printing the diamond elements in an anticlockwise spiral manner. The function makes O(N) recursive calls, and each call performs constant time operations. Therefore, the overall time complexity is linear in the number of elements in the matrix.
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