# Print Concatenation of Zig-Zag String in ‘n’ Rows

Given a string and number of rows ‘n’. Print the string formed by concatenating n rows when input string is written in row-wise Zig-Zag fashion.

Examples:

```Input: str = "ABCDEFGH"
n = 2
Output: "ACEGBDFH"
Explanation: Let us write input string in Zig-Zag fashion
in 2 rows.
A   C   E   G
B   D   F   H
Now concatenate the two rows and ignore spaces
in every row. We get "ACEGBDFH"

Input: str = "GEEKSFORGEEKS"
n = 3
Output: GSGSEKFREKEOE
Explanation: Let us write input string in Zig-Zag fashion
in 3 rows.
G       S       G       S
E   K   F   R   E   K
E       O       E
Now concatenate the two rows and ignore spaces
in every row. We get "GSGSEKFREKEOE"```

## We strongly recommend that you click here and practice it, before moving on to the solution.

The idea is to traverse the input string. Every character has to go to one of the rows. One by one add all characters to different rows. Below is algorithm:

```1) Create an array of n strings, arr[n]
2) Initialize direction as "down" and row as 0. The
direction indicates whether we need to move up or
down in rows.
3) Traverse the input string, do following for every
character.
a) Append current character to string of current row.
b) If row number is n-1, then change direction to 'up'
c) If row number is 0, then change direction to 'down'
d) If direction is 'down', do row++.  Else do row--.
4) One by one print all strings of arr[]. ```

Below is C++ implementation of above idea.

```// C++ program to print string obtained by concatenation
// of different rows of Zig-Zag fashion
#include<bits/stdc++.h>
using namespace std;

// Prints concatenation of all rows of str's Zig-Zag fasion
void printZigZagConcat(string str, int n)
{
// Corner Case (Only one row)
if (n == 1)
{
cout << str;
return;
}

// Find length of string
int len = str.length();

// Create an array of strings for all n rows
string arr[n];

// Initialize index for array of strings arr[]
int row = 0;
bool down; // True if we are moving down in rows,
// else false

// Travers through given string
for (int i = 0; i < len; ++i)
{
// append current character to current row
arr[row].push_back(str[i]);

// If last row is reached, change direction to 'up'
if (row == n-1)
down = false;

// If 1st row is reached, change direction to 'down'
else if (row == 0)
down = true;

// If direction is down, increment, else decrement
(down)? (row++): (row--);
}

// Print concatenation of all rows
for (int i = 0; i < n; ++i)
cout << arr[i];
}

// Driver program
int main()
{
string str = "GEEKSFORGEEKS";
int n = 3;
printZigZagConcat(str, n);
return 0;
}
```

Output:

`GSGSEKFREKEOE`

Time Complexity: O(len) where len is length of input string.
Auxiliary Space: O(len)

Thanks to Gaurav Ahirwar for suggesting above solution.

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