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Javascript Program to Print a given matrix in reverse spiral form

Last Updated : 10 Sep, 2024
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Given a 2D array, print it in reverse spiral form. We have already discussed Print a given matrix in spiral form. This article discusses how to do the reverse printing. See the following examples. 

Input:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Output:
10 11 7 6 5 9 13 14 15 16 12 8 4 3 2 1

Input:
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
Output:
11 10 9 8 7 13 14 15 16 17 18 12 6 5 4 3 2 1
JavaScript
// This is a modified code of
// https://www.geeksforgeeks.org/
// print-a-given-matrix-in-spiral-form/

let R = 3;
let C = 6;

// Function that print matrix in 
// reverse spiral form.
function ReversespiralPrint(m, n, a) {
    // Large array to initialize it
    // with elements of matrix
    let b = new Array(100);

    /* k - starting row index
    l - starting column index*/
    let i, k = 0, l = 0;

    // Counter for single dimension array
    //in which elements will be stored
    let z = 0;

    // Total elements in matrix
    let size = m * n;

    while (k < m && l < n) {
        // Variable to store value of matrix.
        let val;

        /* Print the first row from
           the remaining rows */
        for (i = l; i < n; ++i) {
            // printf("%d ", a[k][i]);
            val = a[k][i];
            b[z] = val;
            ++z;
        }
        k++;

        /* Print the last column from 
           the remaining columns */
        for (i = k; i < m; ++i) {
            // printf("%d ", a[i][n-1]);
            val = a[i][n - 1];
            b[z] = val;
            ++z;
        }
        n--;

        /* Print the last row from the 
           remaining rows */
        if (k < m) {
            for (i = n - 1; i >= l; --i) {
                // printf("%d ", a[m-1][i]);
                val = a[m - 1][i];
                b[z] = val;
                ++z;
            }
            m--;
        }

        /* Print the first column from the 
           remaining columns */
        if (l < n) {
            for (i = m - 1; i >= k; --i) {
                // printf("%d ", a[i][l]);
                val = a[i][l];
                b[z] = val;
                ++z;
            }
            l++;
        }
    }
    let output='';
    for (let i = size - 1; i >= 0; --i) {
        output += b[i]+" "
    }
    console.log(output.trim() + " ");
}

/* Driver program to test above functions */
let a = [[1, 2, 3, 4, 5, 6],
[7, 8, 9, 10, 11, 12],
[13, 14, 15, 16, 17, 18]];
ReversespiralPrint(R, C, a);

Output
11 10 9 8 7 13 14 15 16 17 18 12 6 5 4 3 2 1 

Complexity Analysis:

  • Time complexity: O(R*C) where R and C are no of rows and columns in given matrix
  • Auxiliary Space: O(1) because using constant space

Please refer complete article on Print a given matrix in reverse spiral form for more details!



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