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C++ Program to Print matrix in zag-zag fashion

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  • Last Updated : 27 Jan, 2022
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Given a matrix of 2D array of n rows and m columns. Print this matrix in ZIG-ZAG fashion as shown in figure. 
 

matrix_zag-zag

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!
 


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