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

  • Difficulty Level : Medium
  • Last Updated : 13 Sep, 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 the 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 fashion
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
 
    // Traverse 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 fashion
    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
 
        // Traverse 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

Python3




# 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 fashion
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
     
    # Traverse 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 fashion
    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
 
    // Traverse 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

Javascript




<script>
// Javascript program to print string
// obtained by concatenation
// of different rows of
// Zig-Zag fashion
 
// Prints concatenation
    // of all rows of str's
    // Zig-Zag fashion
function printZigZagConcat(str,n)
{
    // Corner Case (Only one row)
        if (n == 1)
        {
            document.write(str);
            return;
        }
        let str1 = str.split("");
  
        // Find length of string
        let len = str.length;
  
        // Create an array of
        // strings for all n rows
        let arr = new Array(n);
        for(let i=0;i<n;i++)
        {
            arr[i]="";
        }
         
  
        // Initialize index for
        // array of strings arr[]
        let row = 0;
        let down = true; // True if we are moving
        // down in rows, else false
  
        // Traverse through
        // given string
        for (let 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 (let i = 0; i < n; ++i)
        {
            document.write(arr[i]);
        }
}
 
// Driver Code
let str = "GEEKSFORGEEKS";
let n = 3;
printZigZagConcat(str, n);
 
// This code is contributed by ab2127
</script>
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 through 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:

C++




// C++ Program for above approach
#include<bits/stdc++.h>
using namespace std;
// Function for zig-zag Concatenation
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
int main()
{
    string str = "GEEKSFORGEEKS";
    int n = 3;
    cout<< zigZagConcat(str, n);
}
 
// This code is contributed by Mayank Tyagi

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 contributed 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

Javascript




<script>
// Javascript Program for above approach
 
// Function for zig-zag Concatenation
function zigZagConcat(s,n)
{
 
    // Check is n is less
        // or equal to 1
        if (n <= 1)
        {
            return s;
        }
  
        let result = [];
  
        // Iterate rowNum from 0 to n - 1
        for (let rowNum = 0; rowNum < n; rowNum++)
        {
            let i = rowNum;
            let up = true;
            
            // Iterate i till s.length() - 1
            while (i < s.length)
            {
  
                result.push(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.join("");
}
 
// Driver Code
let str = "GEEKSFORGEEKS";
let n = 3;
document.write(zigZagConcat(str, n));
 
// This code is contributed by unknown2108
</script>
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

 

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