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:

1. First, we need to check if the given matrix is a valid odd-sized square matrix. If it is not, we return as the problem cannot be solved with an invalid matrix.
2. We then find the middle row and middle column of the matrix. For an odd-sized square matrix, both middle row and middle column will be the same.
3. We divide the spiral traversal into four cases, corresponding to each side of the diamond shape.
4. For each case, we start from the middle row and middle column and traverse in an anticlockwise spiral manner while printing the elements.
5. 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 Left-bottom)
    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

1. 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.
2. The function finds the middle row and middle column of the matrix as 'mid_row' and 'mid_col'.
3. It then divides the spiral traversal into four cases (A, B, C, and D) representing each side of the diamond shape.
4. For each case, it starts from the corresponding position and moves in an anticlockwise spiral manner while printing the elements.
5. 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

1) Anticlockwise spiral view of diamond matrix elements in java 2) Anticlockwise spiral view of diamond matrix elements in c++ 3) Anticlockwise spiral view of diamond matrix elements in c 4) Anticlockwise spiral view of diamond matrix elements in c# 5) Anticlockwise spiral view of diamond matrix elements in php 6) Anticlockwise spiral view of diamond matrix elements in node js 7) Anticlockwise spiral view of diamond matrix elements in typescript 8) Anticlockwise spiral view of diamond matrix elements in python 9) Anticlockwise spiral view of diamond matrix elements in ruby 10) Anticlockwise spiral view of diamond matrix elements in scala 11) Anticlockwise spiral view of diamond matrix elements in swift 12) Anticlockwise spiral view of diamond matrix elements in kotlin 13) Anticlockwise spiral view of diamond matrix elements in golang 14) Anticlockwise spiral view of diamond matrix elements in vb.net

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.