Reversal spiral of diamond element in square matrix
The problem at hand is to find the reversal spiral view of the diamond elements in a square matrix. The diamond elements are the central elements forming a diamond shape in the matrix. The reversal spiral view means traversing these diamond elements in a spiral manner but in the reverse order.
Problem Statement and Example
Consider the following square matrix:
Idea to Solve the Problem
To find the reversal spiral view of the diamond elements in the square 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 as 'midRow' and 'midCol'.
 We divide the spiral traversal into four cases (A, B, C, and D) representing each side of the diamond shape.
 For each case, we start from the corresponding position and move in a spiral manner while storing the elements in the 'result' array.
 After each case, the function reduces the size of the diamond (by updating 'x', 'y', 'm', 'n', and 'k') and makes a recursive call to continue the traversal until all diamond elements are covered.
 The 'elementSize' function calculates the total number of elements in the diamond shape based on the number of columns.
Pseudocode
getSpiralView(matrix, result, x, y, m, n, k):
midCol = y + ((n  y) / 2)
midRow = midCol
// Case A (Top to Righttop)
for i from midCol to n and k > 0:
result[k  1] = matrix[x][i]
k
// Case B (Righttop to Bottomright)
for i from n to midCol and k > 0:
result[k  1] = matrix[midRow][i]
k
// Case C (Bottomright to Leftbottom)
for i from midCol  1 to y and k > 0:
result[k  1] = matrix[n][i]
k
// Case D (Leftbottom to Topleft)
for i from y + 1 to midCol and k > 0:
result[k  1] = matrix[midRow][i]
k
if x + 1 <= m  1 and k > 0:
// Recursive call
getSpiralView(matrix, result, x + 1, y + 1, m  1, n  1, k)
elementSize(col):
counter = 0
while col > 0:
counter += col
col
return counter * 4
diamondSpiral(matrix):
row = number of rows in matrix
col = number of columns in matrix
if row != col or col % 2 == 0:
print "Not a valid perfect odd square matrix"
return
size = elementSize(col / 2) + 1
result = new array of size 'size'
getSpiralView(matrix, result, 0, 0, row  1, col  1, size)
for i from 0 to size  1:
print result[i]
Algorithm Explanation
 The 'getSpiralView' function takes 'matrix', 'result', 'x', 'y', 'm', 'n', and 'k' as input parameters. 'matrix' is the given square matrix, 'result' is the array to store the diamond elements, 'x' and 'y' are the starting row and column, 'm' and 'n' 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 'midRow' and 'midCol'.
 It then divides the spiral traversal into four cases (A, B, C, and D) corresponding to each side of the diamond shape.
 For each case, it starts from the corresponding position and moves in a spiral manner while storing the elements in the 'result' array.
 After each case, the function reduces the size of the diamond (by updating 'x', 'y', 'm', 'n', and 'k') and makes a recursive call to continue the traversal until all diamond elements are covered.
 The 'elementSize' function calculates the total number of elements in the diamond shape based on the number of columns.
 The 'diamondSpiral' function checks if the matrix is a valid oddsized square matrix and then calls 'getSpiralView' to find the spiral view of diamond elements and prints the result.
Code Solution

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Output Explanation
The given code creates a square matrix and calls the 'diamondSpiral' function to find the reversal spiral view of diamond elements. The output displays the reversal spiral view of the diamond elements as follows:
25 24 32 26 18 17 23 31 39 33 27 19 11 10 16 22 30 38 46 40 34 28 20 12 4
Time Complexity
The time complexity of the provided solution is O(N), where N is the total number of elements in the square matrix. The function 'getSpiralView' recursively traverses the matrix while storing the diamond elements in the 'result' array. 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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