Program to print a string in vertical zigzag manner
Last Updated :
06 Jul, 2021
Given a string, S of size N, and a number of rows R, the task is to print the given string in a vertical zigzag fashion with respect to the given number of rows as shown in the examples.
Examples:
Input: S = "123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz", R = 9
Output:
Input: S = "AttentionReaders!Don'tStopLearning!HappyLearning!", R = 12
Output:
Approach: In order to print the characters line by line, the idea is to find the interval between the major columns and step value for in-between columns for printing the spaces until the last character of the string is reached. Follow the steps below to solve this problem:
- Initialize a variable interval as 2*R-2 to store the gap between the major columns.
- Iterate in the range [0, R-1] using the variable i
- Initialize a variable step as interval-2*i to store step values for each row.
- Iterate in the range [i, N-1] using the variable j, incrementing j by interval in each iteration,
- Print the character, S[j].
- If the value of step lies in the range [1, interval-1] and step+j<N, then print (interval-R-i) number of spaces, then print s[j+step] and finally print (i-1) spaces.
- Else print (interval-R) number of spaces.
- Print newline after each iteration of the outer loop.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print any string
// in zigzag fashion
void zigzag(string s, int rows)
{
// Store the gap between the major columns
int interval = 2 * rows - 2;
// Traverse through rows
for (int i = 0; i < rows; i++) {
// Store the step value for each row
int step = interval - 2 * i;
// Iterate in the range [1, N-1]
for (int j = i; j < s.length(); j = j + interval) {
// Print the character
cout << s[j];
if (step > 0 && step < interval
&& step + j < s.length()) {
// Print the spaces before character
// s[j+step]
for (int k = 0; k < (interval - rows - i);
k++)
cout << " ";
// Print the character
cout << s[j + step];
// Print the spaces after character
// after s[j+step]
for (int k = 0; k < i - 1; k++)
cout << " ";
}
else {
// Print the spaces for first and last rows
for (int k = 0; k < (interval - rows); k++)
cout << " ";
}
}
cout << endl;
}
}
// Driver Code
int main()
{
// Given Input
string s = "123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefgh"
"ijklmnopqrstuvwxyz";
int rows = 9;
// Function Call
zigzag(s, rows);
}
// Java program for the above approach
public class GFG{
// Function to print any string
// in zigzag fashion
static void zigzag(String s, int rows)
{
// Store the gap between the major columns
int interval = 2 * rows - 2;
// Traverse through rows
for(int i = 0; i < rows; i++)
{
// Store the step value for each row
int step = interval - 2 * i;
// Iterate in the range [1, N-1]
for(int j = i; j < s.length(); j = j + interval)
{
// Print the character
System.out.print(s.charAt(j));
if (step > 0 && step < interval &&
step + j < s.length())
{
// Print the spaces before character
// s[j+step]
for(int k = 0; k < (interval - rows - i); k++)
System.out.print(" ");
// Print the character
System.out.print(s.charAt(j + step));
// Print the spaces after character
// after s[j+step]
for(int k = 0; k < i - 1; k++)
System.out.print(" ");
}
else
{
// Print the spaces for first and last rows
for(int k = 0; k < (interval - rows); k++)
System.out.print(" ");
}
}
System.out.println();
}
}
// Driver Code
public static void main(String args[])
{
// Given Input
String s = "123456789ABCDEFGHIJKLM" +
"NOPQRSTUVWXYZabcdefghi" +
"jklmnopqrstuvwxyz";
int rows = 9;
// Function Call
zigzag(s, rows);
}
}
// This code is contributed by SoumikMondal
# Python3 program for the above approach
# Function to print any string
# in zigzag fashion
def zigzag(s, rows):
# Store the gap between the major columns
interval = 2 * rows - 2
# Traverse through rows
for i in range(rows):
# Store the step value for each row
step = interval - 2 * i
# Iterate in the range [1, N-1]
for j in range(i, len(s), interval):
# Print the character
print(s[j], end = "")
if (step > 0 and step < interval and
step + j < len(s)):
# Print the spaces before character
# s[j+step]
for k in range((interval - rows - i)):
print(end = " ")
# Print the character
print(s[j + step], end = "")
# Print the spaces after character
# after s[j+step]
for k in range(i - 1):
print(end = " ")
else:
# Print the spaces for first and
# last rows
for k in range(interval - rows):
print(end = " ")
print()
# Driver Code
if __name__ == '__main__':
# Given Input
s = "123456789ABCDEFGHIJKL"\
"MNOPQRSTUVWXYZabcdefghi"\
"jklmnopqrstuvwxyz"
rows = 9
# Function Call
zigzag(s, rows)
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print any string
// in zigzag fashion
static void zigzag(string s, int rows)
{
// Store the gap between the major columns
int interval = 2 * rows - 2;
// Traverse through rows
for(int i = 0; i < rows; i++)
{
// Store the step value for each row
int step = interval - 2 * i;
// Iterate in the range [1, N-1]
for(int j = i; j < s.Length; j = j + interval)
{
// Print the character
Console.Write(s[j]);
if (step > 0 && step < interval &&
step + j < s.Length)
{
// Print the spaces before character
// s[j+step]
for(int k = 0; k < (interval - rows - i); k++)
Console.Write(" ");
// Print the character
Console.Write(s[j + step]);
// Print the spaces after character
// after s[j+step]
for(int k = 0; k < i - 1; k++)
Console.Write(" ");
}
else
{
// Print the spaces for first and last rows
for(int k = 0; k < (interval - rows); k++)
Console.Write(" ");
}
}
Console.WriteLine();
}
}
// Driver Code
public static void Main()
{
// Given Input
string s = "123456789ABCDEFGHIJKLM" +
"NOPQRSTUVWXYZabcdefghi" +
"jklmnopqrstuvwxyz";
int rows = 9;
// Function Call
zigzag(s, rows);
}
}
// This code is contributed by SURENDRA_GANGWAR
<script>
// JavaScript program for the above approach
// Function to print any string
// in zigzag fashion
function zigzag(s,rows)
{
// Store the gap between the major columns
let interval = 2 * rows - 2;
// Traverse through rows
for(let i = 0; i < rows; i++)
{
// Store the step value for each row
let step = interval - 2 * i;
// Iterate in the range [1, N-1]
for(let j = i; j < s.length; j = j + interval)
{
// Print the character
document.write(s[j]);
if (step > 0 && step < interval &&
step + j < s.length)
{
// Print the spaces before character
// s[j+step]
for(let k = 0; k < (interval - rows - i); k++)
document.write("  ");
// Print the character
document.write(s[j + step]);
// Print the spaces after character
// after s[j+step]
for(let k = 0; k < i - 1; k++)
document.write("  ");
}
else
{
// Print the spaces for first and last rows
for(let k = 0; k < (interval - rows); k++)
document.write("  ");
}
}
document.write("<br>");
}
}
// Driver Code
// Given Input
let s = "123456789ABCDEFGHIJKLM" +
"NOPQRSTUVWXYZabcdefghi" +
"jklmnopqrstuvwxyz";
let rows = 9;
// Function Call
zigzag(s, rows);
// This code is contributed by patel2127
</script>
Output1 H X n
2 GI WY mo
3 F J V Z l p
4 E K U a k q
5 D L T b j r z
6 C M S c i s y
7 B N R d h t x
8A OQ eg uw
9 P f v
Time Complexity: O(R2*N)
Auxiliary Space: O(1)