Concatenation of Zig-Zag String in N Rows

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P        A      H      N
  A   P   L  S   I   I    G
    Y        I       R

And then read line by line: PAHNAPLSIIGYIR.

Therefore, for given string str and an integer N, the task is to print the string formed by concatenating N rows when str is written in row-wise Zig-Zag fashion.

Example:

Input: str = "PAYPALISHIRING", N = 3
Output: PAHNAPLSIIGYIR

Input: str = "ABCDEFGH", N = 2
Output: ACEGBDFH
Explanation: The input string can be written in Zig-Zag fashion in 2 rows as follows:
A   C    E   G    
   B   D   F   H
Hence, upon reading the above pattern row-wise, the output string is "ACEGBDFH"

Approach: The given problem is an implementation based problem that can be solved by following the below steps

• Create an array of N strings, arr[N].
• Initialize direction as "down" and row as 0. The direction indicates whether the current pointer is moving up or down in rows.
• Traverse the input string, do the following for every character.
  • Append the current character to the string representing the current row.
  • If row number is N - 1, then change direction to 'up'
  • If row number is 0, then change direction to 'down'
  • If direction is 'down', do row++.  Else do row--.
• One by one print all strings of arr[].

Below is the implementation of the above approach:

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that Prints concatenation of
// all rows of str's Zig-Zag fashion
void printZigZagConcat(string str, int n)
{
    if (n == 1)
    {
        cout << str << endl;
    }
    string res = "";
    string arr[n] = {""};
    bool down;
    int row = 0; // helps in building individual blocks of strings

    for (int i = 0; i < str.size(); i++)
    {
        arr[row].push_back(str[i]);
        if (row == n - 1)
        {
            down = false;
        }
        if (row == 0)
        {
            down = true;
        }
        if (!down)
            row--;
        else
            row++;
    }

    for (int i = 0; i < n; i++)
    {
        cout << arr[i];
    }
}

int main()
{
    // Driver Code
    string str = "PAYPALISHIRING";
    int N = 3;
    printZigZagConcat(str, N);
    return 0;
}

// This code is contributed by Potta Lokesh

// Java code for the above approach
import java.util.*;

class GFG {

  // Function that Prints concatenation of
  // all rows of str's Zig-Zag fashion
  static void printZigZagConcat(String str, int n)
  {
    if (n == 1) {
      System.out.print(str + "\n");
    }
    String res = "";
    String[] arr = new String[n];
    for (int i = 0; i < n; i++)
      arr[i] = "";
    boolean down = false;
    int row = 0; // helps in building individual blocks
    // of Strings

    for (int i = 0; i < str.length(); i++) {
      if (row >= 0)
        arr[row] += (str.charAt(i));
      if (row == n - 1) {
        down = false;
      }
      if (row == 0) {
        down = true;
      }
      if (!down)
        row--;
      else
        row++;
    }

    for (int i = 0; i < n; i++) {
      System.out.print(arr[i]);
    }
  }

  public static void main(String[] args)
  {
    // Driver Code
    String str = "PAYPALISHIRING";
    int N = 3;
    printZigZagConcat(str, N);
  }
}

// This code is contributed by umadevi9616

# Python 3 program of the above approach

# Function that 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 = "PAYPALISHIRING"
N = 3
printZigZagConcat(str, N)

// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG
{

  // Function that Prints concatenation of
  // all rows of str's Zig-Zag fashion
  static void printZigZagConcat(String str, int n)
  {
    if (n == 1) {
      Console.WriteLine(str);
    }

    String[] arr = new String[n];
    for (int i = 0 ; i < n ; i++)
      arr[i] = "";

    bool down = false;
    int row = 0; // helps in building individual blocks
    // of Strings

    for (int i = 0 ; i < str.Length ; i++) {
      if (row >= 0)
        arr[row] += (str[i]);
      if (row == n - 1) {
        down = false;
      }
      if (row == 0) {
        down = true;
      }
      if (!down)
        row--;
      else
        row++;
    }

    for (int i = 0; i < n; i++) {
      Console.Write(arr[i]);
    }
  }


  // Driver code
  public static void Main(string[] args){

    // Driver Code
    String str = "PAYPALISHIRING";
    int N = 3;
    printZigZagConcat(str, N);

  }
}

// This code is contributed by subhamgoyal2014.

    <script>
        // JavaScript code for the above approach

        // Function that Prints concatenation of
        // all rows of str's Zig-Zag fashion
        const printZigZagConcat = (str, n) => {
            if (n == 1) {
                document.write(`${str}<br/>`);
            }
            let res = "";
            let arr = new Array(n).fill("");
            let down = false;
            let row = 0; // helps in building individual blocks of strings

            for (let i = 0; i < str.length; i++) {
                arr[row] += str[i];
                if (row == n - 1) {
                    down = false;
                }
                if (row == 0) {
                    down = true;
                }
                if (!down)
                    row--;
                else
                    row++;
            }

            for (let i = 0; i < n; i++) {
                document.write(arr[i]);
            }
        }

        // Driver Code
        let str = "PAYPALISHIRING";
        let N = 3;
        printZigZagConcat(str, N);

    // This code is contributed by rakeshsahni

    </script>

PAHNAPLSIIGYIR

Time Complexity: O(N)
Auxiliary Space: O(N)