Program to print the Fish Pattern
Last Updated :
23 Nov, 2021
Given an integer N, the task is to print a pattern of fish over 2N+1 rows.
Example:
Input: N=3
Output:
*
*** *
***** **
**********
***** **
*** *
*
Input: N=5
Output:
*
*** *
***** **
******* ***
********* ****
****************
********* ****
******* ***
***** **
*** *
*
Approach: The fish consists of three parts:
- Upper Part: Over N rows.
- Middle Part: Singe row in the middle
- Lower Part: Over N rows
Now, let's try to understand the pattern using an example:
For N=3, the fish is:
Now, to solve this question, follow the steps below:
- First, create the upper part:
- Run a loop for i=0 to i<N, and in each iteration of the loop:
- As depicted in the above image that first a string, say spaces1 having M (initially M=N) spaces appear, then a layer of stars, let's say stars1 having only 1 star, then the string spaces1 appears 2 times (having 2*M spaces) and then another layer of stars, say stars2 having 0 stars initially.
- Now in each iteration, spaces1 got reduced by a space, stars1 got increased by 2 stars and stars2 by 1 star.
- For the middle part, just print both stars1 and stars2, as no spaces appear in this row.
- Now, to get the lower part, reverse the algorithm for the upper part.
- After the loop ends, the pattern of fish will be created.
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 the pattern of a fish
// over N rows
void printFish(int N)
{
string spaces1 = "", spaces2 = "";
string stars1 = "*", stars2 = "";
for (int i = 0; i < N; ++i) {
spaces1 += ' ';
}
spaces2 = spaces1;
for (int i = 0; i < 2 * N + 1; ++i) {
// For upper part
if (i < N) {
cout << spaces1 << stars1
<< spaces1 << spaces1
<< stars2 << endl;
spaces1.pop_back();
stars1 += "**";
stars2 += "*";
}
// For middle part
if (i == N) {
cout << spaces1 << stars1
<< spaces1 << spaces1
<< stars2 << endl;
}
// For lower part
if (i > N) {
spaces1 += ' ';
stars1.pop_back();
stars1.pop_back();
stars2.pop_back();
cout << spaces1 << stars1
<< spaces1 << spaces1
<< stars2 << endl;
}
}
}
// Driver Code
int main()
{
int N = 5;
printFish(N);
}
Java
// Java program for the above approach
class GFG {
// Function to print the pattern of a fish
// over N rows
public static void printFish(int N) {
String spaces1 = "";
String stars1 = "*", stars2 = "";
for (int i = 0; i < N; ++i) {
spaces1 += ' ';
}
for (int i = 0; i < 2 * N + 1; ++i)
{
// For upper part
if (i < N) {
System.out.print(spaces1 + stars1 + spaces1 + spaces1);
System.out.println(stars2);
spaces1 = spaces1.substring(0, spaces1.length() - 1);
stars1 += "**";
stars2 += "*";
}
// For middle part
if (i == N) {
System.out.print(spaces1 + stars1 + spaces1 + spaces1);
System.out.println(stars2);
}
// For lower part
if (i > N) {
spaces1 += ' ';
stars1 = stars1.substring(0, stars1.length() - 1);
stars1 = stars1.substring(0, stars1.length() - 1);
stars2 = stars2.substring(0, stars2.length() - 1);
System.out.print(spaces1 + stars1 + spaces1 + spaces1);
System.out.println(stars2);
}
}
}
// Driver Code
public static void main(String args[]) {
int N = 5;
printFish(N);
}
}
// This code is contributed by gfgking.
Python3
# Python3 program for the above approach
# Function to print the pattern of a fish
# over N rows
def printFish(N) :
spaces1 = ""; spaces2 = "";
stars1 = "*"; stars2 = "";
for i in range(N) :
spaces1 += ' ';
spaces2 = spaces1;
for i in range( 2 * N + 1) :
# For upper part
if (i < N) :
print(spaces1,end="");
print(stars1,end="");
print(spaces1,end="");
print(spaces1,end="");
print(stars2);
spaces1 = spaces1[:-1]
stars1 += "**";
stars2 += "*";
# For middle part
if (i == N) :
print(spaces1,end="");
print(stars1,end="");
print(spaces1,end="");
print(spaces1,end="");
print(stars2);
# For lower part
if (i > N) :
spaces1 += ' ';
stars1 = stars1[:-1];
stars1 = stars1[:-1];
stars2 = stars2[:-1];
print(spaces1,end="");
print(stars1,end="")
print(spaces1,end="");
print(spaces1,end="");
print(stars2);
# Driver Code
if __name__ == "__main__" :
N = 5;
printFish(N);
# This code is contributed by AnkThon
C#
// C# program for the above approach
using System;
class GFG {
// Function to print the pattern of a fish
// over N rows
static void printFish(int N) {
string spaces1 = "";
string stars1 = "*", stars2 = "";
for (int i = 0; i < N; ++i) {
spaces1 += ' ';
}
for (int i = 0; i < 2 * N + 1; ++i)
{
// For upper part
if (i < N) {
Console.Write(spaces1 + stars1 + spaces1 + spaces1);
Console.Write(stars2 + "\n");
spaces1 = spaces1.Substring(0, spaces1.Length - 1);
stars1 += "**";
stars2 += "*";
}
// For middle part
if (i == N) {
Console.Write(spaces1 + stars1 + spaces1 + spaces1);
Console.Write(stars2 + "\n");
}
// For lower part
if (i > N) {
spaces1 += ' ';
stars1 = stars1.Substring(0, stars1.Length - 1);
stars1 = stars1.Substring(0, stars1.Length - 1);
stars2 = stars2.Substring(0, stars2.Length - 1);
Console.Write(spaces1 + stars1 + spaces1 + spaces1);
Console.Write(stars2 + "\n");
}
}
}
// Driver Code
public static void Main() {
int N = 5;
printFish(N);
}
}
// This code is contributed by Samim Hossain Mondal.
JavaScript
<script>
// JavaScript program for the above approach
// Function to print the pattern of a fish
// over N rows
const printFish = (N) => {
let spaces1 = "", spaces2 = "";
let stars1 = "*", stars2 = "";
for (let i = 0; i < N; ++i) {
spaces1 += "  ";
}
spaces2 = spaces1;
for (let i = 0; i < 2 * N + 1; ++i) {
// For upper part
if (i < N) {
document.write(`${spaces1}${stars1}${spaces1}${spaces1}${stars2}<br/>`);
spaces1 = spaces1.substr(0, spaces1.length - 10);
stars1 += "**";
stars2 += "*";
}
// For middle part
if (i == N) {
document.write(`${spaces1}${stars1}${spaces1}${spaces1}${stars2}<br/>`);
}
// For lower part
if (i > N) {
spaces1 += "  ";
stars1 = stars1.substr(0, stars1.length - 2);
stars2 = stars2.substr(0, stars2.length - 1);
document.write(`${spaces1}${stars1}${spaces1}${spaces1}${stars2}<br/>`);
}
}
}
// Driver Code
let N = 5;
printFish(N);
// This code is contributed by rakeshsahni
</script>
Output *
*** *
***** **
******* ***
********* ****
****************
********* ****
******* ***
***** **
*** *
*
Time Complexity: O(N)
Auxiliary Space: O(N)