Given an n x n matrix .In the given matrix, you have to print the elements of the matrix in the snake pattern.
Examples:
Input :mat[][] = { {10, 20, 30, 40},
{15, 25, 35, 45},
{27, 29, 37, 48},
{32, 33, 39, 50}};
Output : 10 20 30 40 45 35 25 15 27 29
37 48 50 39 33 32
Input :mat[][] = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
Output : 1 2 3 6 5 4 7 8 9
Python Program to Print matrix in snake pattern
We traverse all rows. For every row, we check if it is even or odd. If even, we print from left to right else print from right to left.
Python3
# Python 3 program to print# matrix in snake orderM = 4N = 4def printf(mat): global M, N # Traverse through all rows for i in range(M): # If current row is # even, print from # left to right if i % 2 == 0: for j in range(N): print(str(mat[i][j]), end=" ") # If current row is # odd, print from # right to left else: for j in range(N - 1, -1, -1): print(str(mat[i][j]), end=" ")# Driver codemat = [[10, 20, 30, 40], [15, 25, 35, 45], [27, 29, 37, 48], [32, 33, 39, 50]]printf(mat)# This code is contributed# by ChitraNayal |
Output
10 20 30 40 45 35 25 15 27 29 37 48 50 39 33 32
Time complexity: O(n^m) ,Traversing over all the elements of the matrix, therefore N X M elements are there.
Auxiliary space : O(1)
Reversing Alternate rows Using Slicing
Python3
# Python 3 program to print# matrix in snake orderM = 4N = 4def printf(mat): global M, N # Traverse through all rows for i in range(M): if i % 2 != 0: mat[i] = mat[i][::-1] for i in range(M): for j in range(N): print(mat[i][j], end=' ')# Driver codemat = [[10, 20, 30, 40], [15, 25, 35, 45], [27, 29, 37, 48], [32, 33, 39, 50]]printf(mat)# This code is contributed# by vikkycirus |
Output
10 20 30 40 45 35 25 15 27 29 37 48 50 39 33 32
Time Complexity: O(N x M), Traversing over all the elements of the matrix, therefore N X M elements are there.
Auxiliary Space: O(1)
Please refer complete article on Print matrix in snake pattern for more details!
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