Skip to content
Related Articles

Related Articles

Print Concatenation of Zig-Zag String in ‘n’ Rows
  • Difficulty Level : Medium
  • Last Updated : 05 Mar, 2021

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++




// 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;
}
Java



// 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
Python 3



# 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
C#



// 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
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.


Another Approach:
If we assume that we are iterating thorugh imaginary matrix of row count n one by one and printing chars
for first and last row , index will increment by a value of i+= 2*(n-1) 
for middle rows, if we are going in upward direction, index will increment as i+= 2*(n-rowNum-1), and for downward direction, index will increment as 2*rowNum. Following is the working implementation. in JAVA
Below is the implementation of the above approach:
Java



// Java Program for above approach
import java.lang.*;
 
class Solution
{
     
    // Function for zig-zag Concatenation
    private static String zigZagConcat(
                                String s, int n)
    {
 
        // Check is n is less
        // or equal to 1
        if (n <= 1)
        {
            return s;
        }
 
        StringBuilder result = new
                             StringBuilder();
 
        // Iterate rowNum from 0 to n - 1
        for (int rowNum = 0; rowNum < n; rowNum++)
        {
            int i = rowNum;
            boolean up = true;
           
            // Iterate i till s.length() - 1
            while (i < s.length())
            {
 
                result = result.append(s.charAt(i));
               
                // Check is rowNum is 0 or n - 1
                if (rowNum == 0 || rowNum == n - 1)
                {
                    i += (2 * n - 2);
                }
                else
                {
                    if (up)
                    {
                        i += (2 * (n - rowNum) - 2);
                    }
                    else
                    {
                        i += rowNum * 2;
                    }
                    up ^= true;
                }
            }
        }
        return result.toString();
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String str = "GEEKSFORGEEKS";
        int n = 3;
        System.out.println(zigZagConcat(str, n));
    }
}
 
// This code is conntributed by Sakshi Sachdeva
Python3



# Python Program for above approach
 
# Function for zig-zag Concatenation
def zigZagConcat(s, n):
     
    # Check is n is less 
    # or equal to 1
    if (n <= 1):
        return s
     
    result = ""
     
    # Iterate rowNum from 0 to n - 1
    for rowNum in range(n):
        i = rowNum
        up = True
         
        # Iterate i till s.length() - 1
        while(i < len(s)):
            result += s[i]
             
            # Check is rowNum is 0 or n - 1
            if (rowNum == 0 or rowNum == n - 1):
                i += (2 * n - 2)
            else:
                if(up):
                    i += (2 * (n - rowNum) - 2)
                else:
                    i += rowNum * 2
                up ^= True
     
    return result
 
# Driver Code
str = "GEEKSFORGEEKS"
n = 3
print(zigZagConcat(str, n))
 
# This code is contributed by rag2127
C#



// C# Program for above approach
using System;
public class GFG{
 
  // Function for zig-zag Concatenation
  static string zigZagConcat( string s, int n)
  {
    // Check is n is less 
    // or equal to 1
    if (n <= 1) 
    {
      return s;
    }
    string result="";
    // Iterate rowNum from 0 to n - 1
    for (int rowNum = 0; rowNum < n; rowNum++)
    {
      int i = rowNum;
      bool up = true;
      // Iterate i till s.length() - 1
      while (i < s.Length) 
      {
 
        result += s[i];
 
        // Check is rowNum is 0 or n - 1
        if (rowNum == 0 || rowNum == n - 1) 
        {
          i += (2 * n - 2);
        }
        else
        {
          if (up) 
          {
            i += (2 * (n - rowNum) - 2);
          }
          else
          {
            i += rowNum * 2;
          }
          up ^= true;
        }
      }
    }
    return result;
  }
 
 
  // Driver Code
  static public void Main (){
    string str = "GEEKSFORGEEKS";
    int n = 3;
    Console.WriteLine(zigZagConcat(str, n));
  }
}
 
// This code is contributed by avanitrachhadiya2155
 
 
Output
GSGSEKFREKEOE
 
Time Complexity: O(length)
 
Space Complexity: O(1)
 
 
Thanks to Sakshi Sachdeva for suggesting above solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
My Personal Notes
arrow_drop_up
Next
last_page
KMP Algorithm for Pattern Searching
Recommended Articles
Page :
Longest palindromic string formed by concatenation of prefix and suffix of a string
06, May 20
Check if given string is a substring of string formed by repeated concatenation of z to a
22, Jul 20
Check if a string is concatenation of another given string
06, Oct 20
Minimum characters to be replaced to make a string concatenation of a K-length palindromic string
19, Nov 20
Ropes Data Structure (Fast String Concatenation)
24, May 17
Longest palindromic String formed using concatenation of given strings in any order
18, Feb 20
Minimum flips required to convert given string into concatenation of equal substrings of length K
21, Jul 20
Count of pairs of strings whose concatenation forms a palindromic string
16, Jul 20
Minimize cost to convert given string into concatenation of equal substrings of length K
16, Oct 20
Lexicographical concatenation of all substrings of a string
18, Jan 18
Minimum removals to make a string concatenation of a substring of 0s followed by a substring of 1s
14, Dec 20
Rearrange characters of a string to make it a concatenation of palindromic substrings
17, Mar 21
Concatenation of two strings in PHP
26, Jan 18
Count of three non-overlapping sub-strings which on concatenation forms a palindrome
29, May 19
Caesar Concatenation
01, May 20
Pairs whose concatenation contain all digits
20, May 18
Count pair of strings whose concatenation of substrings form a palindrome
20, Jun 20
Number of pairs with Pandigital Concatenation
16, Mar 18
Find the number obtained after concatenation of binary representation of M and N
02, Jan 20
Check if concatenation of splitted substrings of two given strings forms a palindrome or not
02, Dec 20
Count pair of strings whose concatenation has every vowel
15, Jul 20
Print shortest path to print a string on screen
24, Jun 16
Check if all rows of a matrix are circular rotations of each other
14, Jan 17
Print the longest prefix of the given string which is also the suffix of the same string
15, Mar 19
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
GeeksforGeeks
Vote for difficulty
Current difficulty :
Medium





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox