Rotate each ring of matrix anticlockwise by K elements - GeeksforGeeks
Given a matrix of order M*N and a value K, the task is to rotate each ring of the matrix anticlockwise by K elements. If in any ring elements are less than and equal K then don't rotate it.
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Given a matrix of order M*N and a value K, the task is to rotate each ring of the matrix anticlockwise by K elements. If in any ring elements are less than and equal K then don't rotate it.
Input : k = 3 mat[4][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}} Output: 4 8 12 16 3 10 6 15 2 11 7 14 1 5 9 13 Input : k = 2 mat[3][4] = {{1, 2, 3, 4}, {10, 11, 12, 5}, {9, 8, 7, 6}} Output: 3 4 5 6 2 11 12 7 1 10 9 8
The idea is to traverse matrix in spiral form. Here is the algorithm to solve this problem :
- Make an auxiliary array temp[] of size M*N.
- Start traversing matrix in spiral form and store elements of current ring in temp[] array. While storing the elements in temp, keep track of starting and ending positions of current ring.
- For every ring that is being stored in temp[], rotate that subarray temp[]
- Repeat this process for each ring of matrix.
- In last traverse matrix again spirally and copy elements of temp[] array to matrix.
// Fills temp array into mat[][] using spiral order
// traveral.
void
fillSpiral(
int
mat[][MAX],
int
m,
int
n,
int
temp[])
{
int
i, k = 0, l = 0;
/* k - starting row index
m - ending row index
l - starting column index
n - ending column index */
int
tIdx = 0;
// Index in temp array
while
(k < m && l < n)
{
/* first row from the remaining rows */
for
(
int
i = l; i < n; ++i)
mat[k][i] = temp[tIdx++];
k++;
/* last column from the remaining columns */
for
(
int
i = k; i < m; ++i)
mat[i][n-1] = temp[tIdx++];
n--;
/* last row from the remaining rows */
if
(k < m)
{
for
(
int
i = n-1; i >= l; --i)
mat[m-1][i] = temp[tIdx++];
m--;
}
/* first column from the remaining columns */
if
(l < n)
{
for
(
int
i = m-1; i >= k; --i)
mat[i][l] = temp[tIdx++];
l++;
}
}
}
// Function to spirally traverse matrix and
// rotate each ring of matrix by K elements
// mat[][] --> matrix of elements
// M --> number of rows
// N --> number of columns
void
spiralRotate(
int
mat[][MAX],
int
M,
int
N,
int
k)
{
// Create a temporary array to store the result
int
temp[M*N];
/* s - starting row index
m - ending row index
l - starting column index
n - ending column index; */
int
m = M, n = N, s = 0, l = 0;
int
*start = temp;
// Start position of current ring
int
tIdx = 0;
// Index in temp
while
(s < m && l < n)
{
// Initialize end position of current ring
int
*end = start;
// copy the first row from the remaining rows
for
(
int
i = l; i < n; ++i)
{
temp[tIdx++] = mat[s][i];
end++;
}
s++;
// copy the last column from the remaining columns
for
(
int
i = s; i < m; ++i)
{
temp[tIdx++] = mat[i][n-1];
end++;
}
n--;
// copy the last row from the remaining rows
if
(s < m)
{
for
(
int
i = n-1; i >= l; --i)
{
temp[tIdx++] = mat[m-1][i];
end++;
}
m--;
}
/* copy the first column from the remaining columns */
if
(l < n)
{
for
(
int
i = m-1; i >= s; --i)
{
temp[tIdx++] = mat[i][l];
end++;
}
l++;
}
// if elements in current ring greater than
// k then rotate elements of current ring
if
(end-start > k)
{
// Rotate current ring using revarsal
// algorithm for rotation
reverse(start, start+k);
reverse(start+k, end);
reverse(start, end);
// Reset start for next ring
start = end;
}
else
// There are less than k elements in ring
break
;
}
// Fill tenp array in original matrix.
fillSpiral(mat, M, N, temp);
}