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 implementation of above idea.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to print string 
// obtained by concatenation
// of different rows of 
// Zig-Zag fashion
import java.util.Arrays;
  
class GFG {
  
    // Prints concatenation 
    // of all rows of str's 
    // Zig-Zag fasion
    static void printZigZagConcat(String str,
            int n) 
    {
  
        // Corner Case (Only one row)
        if (n == 1
        {
            System.out.print(str);
            return;
        }
        char[] str1 = str.toCharArray();
  
        // Find length of string
        int len = str.length();
  
        // Create an array of
        // strings for all n rows
        String[] arr = new String[n];
        Arrays.fill(arr, "");
  
        // Initialize index for
        // array of strings arr[]
        int row = 0;
        boolean down = true; // 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] += (str1[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
            if (down)
            {
                row++;
            
            else 
            {
                row--;
            }
        }
  
        // Print concatenation
        // of all rows
        for (int i = 0; i < n; ++i) 
        {
            System.out.print(arr[i]);
        }
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        String str = "GEEKSFORGEEKS";
        int n = 3;
        printZigZagConcat(str, n);
    }
}
  
// This code is contributed by PrinciRaj1992

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 program to print
# string obtained by 
# concatenation of different
# rows of Zig-Zag fashion
  
# Prints concatenation of all 
# rows of str's Zig-Zag fasion
def printZigZagConcat(str, n):
      
    # Corner Case (Only one row)
    if n == 1:
        print(str)     
        return
  
    # Find length of string
    l = len(str)
  
    # Create an array of 
    # strings for all n rows
    arr=["" for x in range(l)]
  
    # Initialize index for 
    # array of strings arr[]
    row = 0
      
    # Travers through
    # given string
    for i in range(l):
          
        # append current character
        # to current row
        arr[row] += 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'
        elif row == 0:
            down = True
  
        # If direction is down, 
        # increment, else decrement
        if down:
            row += 1
        else:
            row -= 1
  
    # Print concatenation
    # of all rows
    for i in range(n):
        print(arr[i], end = "")
  
# Driver Code
str = "GEEKSFORGEEKS"
n = 3
printZigZagConcat(str, n)
  
# This code is contributed
# by ChitraNayal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to print string 
// obtained by concatenation
// of different rows of 
// Zig-Zag fashion
using System;
  
class GFG 
{
      
    // Prints concatenation 
    // of all rows of str's 
    // Zig-Zag fasion
    static void printZigZagConcat(string str,
                                  int n)
    {
          
    // Corner Case (Only one row)
    if (n == 1)
    {
        Console.Write(str);     
        return;
    
      
    char[] str1 = str.ToCharArray();
      
    // Find length of string
    int len = str.Length;
  
    // Create an array of
    // strings for all n rows
    string []arr = new string[n];
  
    // Initialize index for
    // array of strings arr[]
    int row = 0;
    bool down = true; // 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] += (str1[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
        if(down)
        row++;
        else
        row--;
    }
  
    // Print concatenation
    // of all rows
    for (int i = 0; i < n; ++i)
        Console.Write(arr[i]);
    }
  
    // Driver Code
    public static void Main()
    {
        String str = "GEEKSFORGEEKS";
        int n = 3;
        printZigZagConcat(str, n);
    }
}
  
// This code is contributed 
// by ChitraNayal

chevron_right



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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



My Personal Notes arrow_drop_up

Improved By : chitranayal, princiraj1992



Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.