Given a matrix of 2D array of n rows and m columns. Print this matrix in ZIG-ZAG fashion as shown in figure.
Example:
Input: 1 2 3 4 5 6 7 8 9 Output: 1 2 4 7 5 3 6 8 9
Approach of C++ code
The approach is simple. Just simply iterate over every diagonal elements one at a time and change the direction according to the previous match.
C++
/* C++ Program to print matrix in Zig-zag pattern*/ #include <iostream> using namespace std;
#define C 3 // Utility function to print matrix // in zig-zag form void zigZagMatrix( int arr[][C], int n, int m)
{ int row = 0, col = 0;
// Boolean variable that will true if we
// need to increment 'row' value otherwise
// false- if increment 'col' value
bool row_inc = 0;
// Print matrix of lower half zig-zag pattern
int mn = min(m, n);
for ( int len = 1; len <= mn; ++len) {
for ( int i = 0; i < len; ++i) {
cout << arr[row][col] << " " ;
if (i + 1 == len)
break ;
// If row_increment value is true
// increment row and decrement col
// else decrement row and increment
// col
if (row_inc)
++row, --col;
else
--row, ++col;
}
if (len == mn)
break ;
// Update row or col value according
// to the last increment
if (row_inc)
++row, row_inc = false ;
else
++col, row_inc = true ;
}
// Update the indexes of row and col variable
if (row == 0) {
if (col == m - 1)
++row;
else
++col;
row_inc = 1;
}
else {
if (row == n - 1)
++col;
else
++row;
row_inc = 0;
}
// Print the next half zig-zag pattern
int MAX = max(m, n) - 1;
for ( int len, diag = MAX; diag > 0; --diag) {
if (diag > mn)
len = mn;
else
len = diag;
for ( int i = 0; i < len; ++i) {
cout << arr[row][col] << " " ;
if (i + 1 == len)
break ;
// Update row or col value according
// to the last increment
if (row_inc)
++row, --col;
else
++col, --row;
}
// Update the indexes of row and col variable
if (row == 0 || col == m - 1) {
if (col == m - 1)
++row;
else
++col;
row_inc = true ;
}
else if (col == 0 || row == n - 1) {
if (row == n - 1)
++col;
else
++row;
row_inc = false ;
}
}
} // Driver code int main()
{ int matrix[][3] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
zigZagMatrix(matrix, 3, 3);
return 0;
} |
Output:
1 2 4 7 5 3 6 8 9
Time complexity: O(n*m)
Auxiliary space: O(1)
Please refer complete article on Print matrix in zig-zag fashion for more details!