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Split the string into N equal parts

Here given code implementation process.

/*
    Java Program for
    Split the string into N equal parts
*/
public class Splitting
{
	// Returns the substring of given start and end position
	public String subString(String text, int start, int end)
	{
		String ans = "";
		for (int i = start; i < end; ++i)
		{
			ans += text.charAt(i);
		}
		return ans;
	}
	public void splitNequalString(String text, int n)
	{
		int size = text.length();
		if (size == 0 || n > size)
		{
			// Invalid partition
			return;
		}
		System.out.println("Given text : " + text);
		if ((size % n) != 0)
		{
			System.out.println("Given " + n + 
                               " equal partition are not possible");
		}
		else
		{
			// Use to collect partition result
			String[] result = new String[n];
			// Get starting growth of partition
			int growth = size / n;
			int start = 0;
			int end = growth;
			// Collect partition result
			for (int i = 0; i < n; ++i)
			{
				result[i] = subString(text, start, end);
				// Get next partition position
				start = end;
				end += growth;
			}
			System.out.println("Partition of " + n + " equal size");
			// Display partition result
			for (int i = 0; i < n; ++i)
			{
				System.out.println(result[i]);
			}
		}
	}
	public static void main(String[] args)
	{
		Splitting task = new Splitting();
		// Given string numbers
		String num1 = "1234568902";
		String num2 = "ABCDEFGHIKLMNOP";
		String num3 = "xyz";
		// Test Case
		task.splitNequalString(num1, 2);
		task.splitNequalString(num2, 5);
		task.splitNequalString(num3, 3);
	}
}

Output

Given text : 1234568902
Partition of 2 equal size
12345
68902
Given text : ABCDEFGHIKLMNOP
Partition of 5 equal size
ABC
DEF
GHI
KLM
NOP
Given text : xyz
Partition of 3 equal size
x
y
z
// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
    C++ Program for
    Split the string into N equal parts
*/
class Splitting
{
	public:
		// Returns the substring of given start and end position
		string subString(string text, int start, int end)
		{
			string ans = "";
			for (int i = start; i < end; ++i)
			{
				ans += text[i];
			}
			return ans;
		}
	void splitNequalString(string text, int n)
	{
		int size = text.length();
		if (size == 0 || n > size)
		{
			// Invalid partition
			return;
		}
		cout << "Given text : " << text << endl;
		if ((size % n) != 0)
		{
			cout << "Given " << n 
          		 << " equal partition are not possible" << endl;
		}
		else
		{
			// Use to collect partition result
			string result[n];
			// Get starting growth of partition
			int growth = size / n;
			int start = 0;
			int end = growth;
			// Collect partition result
			for (int i = 0; i < n; ++i)
			{
				result[i] = this->subString(text, start, end);
				// Get next partition position
				start = end;
				end += growth;
			}
			cout << "Partition of " << n << " equal size" << endl;
			// Display partition result
			for (int i = 0; i < n; ++i)
			{
				cout << result[i] << endl;
			}
		}
	}
};
int main()
{
	Splitting *task = new Splitting();
	// Given string numbers
	string num1 = "1234568902";
	string num2 = "ABCDEFGHIKLMNOP";
	string num3 = "xyz";
	// Test Case
	task->splitNequalString(num1, 2);
	task->splitNequalString(num2, 5);
	task->splitNequalString(num3, 3);
	return 0;
}

Output

Given text : 1234568902
Partition of 2 equal size
12345
68902
Given text : ABCDEFGHIKLMNOP
Partition of 5 equal size
ABC
DEF
GHI
KLM
NOP
Given text : xyz
Partition of 3 equal size
x
y
z
// Include namespace system
using System;
/*
    Csharp Program for
    Split the string into N equal parts
*/
public class Splitting
{
	// Returns the substring of given start and end position
	public String subString(String text, int start, int end)
	{
		String ans = "";
		for (int i = start; i < end; ++i)
		{
			ans += text[i];
		}
		return ans;
	}
	public void splitNequalString(String text, int n)
	{
		int size = text.Length;
		if (size == 0 || n > size)
		{
			// Invalid partition
			return;
		}
		Console.WriteLine("Given text : " + text);
		if ((size % n) != 0)
		{
			Console.WriteLine("Given " + n + 
                              " equal partition are not possible");
		}
		else
		{
			// Use to collect partition result
			String[] result = new String[n];
			// Get starting growth of partition
			int growth = size / n;
			int start = 0;
			int end = growth;
			// Collect partition result
			for (int i = 0; i < n; ++i)
			{
				result[i] = this.subString(text, start, end);
				// Get next partition position
				start = end;
				end += growth;
			}
			Console.WriteLine("Partition of " + n + " equal size");
			// Display partition result
			for (int i = 0; i < n; ++i)
			{
				Console.WriteLine(result[i]);
			}
		}
	}
	public static void Main(String[] args)
	{
		Splitting task = new Splitting();
		// Given string numbers
		String num1 = "1234568902";
		String num2 = "ABCDEFGHIKLMNOP";
		String num3 = "xyz";
		// Test Case
		task.splitNequalString(num1, 2);
		task.splitNequalString(num2, 5);
		task.splitNequalString(num3, 3);
	}
}

Output

Given text : 1234568902
Partition of 2 equal size
12345
68902
Given text : ABCDEFGHIKLMNOP
Partition of 5 equal size
ABC
DEF
GHI
KLM
NOP
Given text : xyz
Partition of 3 equal size
x
y
z
package main

import "fmt"
/*
    Go Program for
    Split the string into N equal parts
*/

// Returns the substring of given start and end position
func subString(text string, start int, end int) string {
	var ans string = ""
	for i := start ; i < end ; i++ {
		ans += string(text[i])
	}
	return ans
}
func splitNequalString(text string, n int) {
	var size int = len(text)
	if size == 0 || n > size {
		// Invalid partition
		return
	}
	fmt.Println("Given text : ", text)
	if (size % n) != 0 {
		fmt.Println("Given ", n, " equal partition are not possible")
	} else {
		// Use to collect partition result
		var result = make([]string,n)
		// Get starting growth of partition
		var growth int = size / n
		var start int = 0
		var end int = growth
		// Collect partition result
		for i := 0 ; i < n ; i++ {
			result[i] = subString(text, start, end)
			// Get next partition position
			start = end
			end += growth
		}
		fmt.Println("Partition of ", n, " equal size")
		// Display partition result
		for i := 0 ; i < n ; i++ {
			fmt.Println(result[i])
		}
	}
}
func main() {
	
	// Given string numbers
	var num1 string = "1234568902"
	var num2 string = "ABCDEFGHIKLMNOP"
	var num3 string = "xyz"
	// Test Case
	splitNequalString(num1, 2)
	splitNequalString(num2, 5)
	splitNequalString(num3, 3)
}

Output

Given text :  1234568902
Partition of  2  equal size
12345
68902
Given text :  ABCDEFGHIKLMNOP
Partition of  5  equal size
ABC
DEF
GHI
KLM
NOP
Given text :  xyz
Partition of  3  equal size
x
y
z
<?php
/*
    Php Program for
    Split the string into N equal parts
*/
class Splitting
{
	// Returns the substring of given start and end position
	public	function subString($text, $start, $end)
	{
		$ans = "";
		for ($i = $start; $i < $end; ++$i)
		{
			$ans .= $text[$i];
		}
		return $ans;
	}
	public	function splitNequalString($text, $n)
	{
		$size = strlen($text);
		if ($size == 0 || $n > $size)
		{
			// Invalid partition
			return;
		}
		echo("Given text : ".$text."\n");
		if (($size % $n) != 0)
		{
			echo("Given ".$n.
				" equal partition are not possible\n");
		}
		else
		{
			// Use to collect partition result
			$result = array_fill(0, $n, "");
			// Get starting growth of partition
			$growth = (int)($size / $n);
			$start = 0;
			$end = $growth;
			// Collect partition result
			for ($i = 0; $i < $n; ++$i)
			{
				$result[$i] = $this->subString($text, $start, $end);
				// Get next partition position
				$start = $end;
				$end += $growth;
			}
			echo("Partition of ".$n." equal size\n");
			// Display partition result
			for ($i = 0; $i < $n; ++$i)
			{
				echo($result[$i]."\n");
			}
		}
	}
}

function main()
{
	$task = new Splitting();
	// Given string numbers
	$num1 = "1234568902";
	$num2 = "ABCDEFGHIKLMNOP";
	$num3 = "xyz";
	// Test Case
	$task->splitNequalString($num1, 2);
	$task->splitNequalString($num2, 5);
	$task->splitNequalString($num3, 3);
}
main();

Output

Given text : 1234568902
Partition of 2 equal size
12345
68902
Given text : ABCDEFGHIKLMNOP
Partition of 5 equal size
ABC
DEF
GHI
KLM
NOP
Given text : xyz
Partition of 3 equal size
x
y
z
/*
    Node JS Program for
    Split the string into N equal parts
*/
class Splitting
{
	// Returns the substring of given start and end position
	subString(text, start, end)
	{
		var ans = "";
		for (var i = start; i < end; ++i)
		{
			ans += text.charAt(i);
		}
		return ans;
	}
	splitNequalString(text, n)
	{
		var size = text.length;
		if (size == 0 || n > size)
		{
			// Invalid partition
			return;
		}
		console.log("Given text : " + text);
		if ((size % n) != 0)
		{
			console.log("Given " + n + " equal partition are not possible");
		}
		else
		{
			// Use to collect partition result
			var result = Array(n).fill(null);
			// Get starting growth of partition
			var growth = parseInt(size / n);
			var start = 0;
			var end = growth;
			// Collect partition result
			for (var i = 0; i < n; ++i)
			{
				result[i] = this.subString(text, start, end);
				// Get next partition position
				start = end;
				end += growth;
			}
			console.log("Partition of " + n + " equal size");
			// Display partition result
			for (var i = 0; i < n; ++i)
			{
				console.log(result[i]);
			}
		}
	}
}

function main()
{
	var task = new Splitting();
	// Given string numbers
	var num1 = "1234568902";
	var num2 = "ABCDEFGHIKLMNOP";
	var num3 = "xyz";
	// Test Case
	task.splitNequalString(num1, 2);
	task.splitNequalString(num2, 5);
	task.splitNequalString(num3, 3);
}
main();

Output

Given text : 1234568902
Partition of 2 equal size
12345
68902
Given text : ABCDEFGHIKLMNOP
Partition of 5 equal size
ABC
DEF
GHI
KLM
NOP
Given text : xyz
Partition of 3 equal size
x
y
z
#    Python 3 Program for
#    Split the string into N equal parts
class Splitting :
	#  Returns the substring of given start and end position
	def subString(self, text, start, ends) :
		ans = ""
		i = start
		while (i < ends) :
			ans += text[i]
			i += 1
		
		return ans
	
	def splitNequalString(self, text, n) :
		size = len(text)
		if (size == 0 or n > size) :
			#  Invalid partition
			return
		
		print("Given text : ", text)
		if ((size % n) != 0) :
			print("Given ", n ," equal partition are not possible")
		else :
			#  Use to collect partition result
			result = [None] * (n)
			#  Get starting growth of partition
			growth = int(size / n)
			start = 0
			ends = growth
			i = 0
			#  Collect partition result
			while (i < n) :
				result[i] = self.subString(text, start, ends)
				#  Get next partition position
				start = ends
				ends += growth
				i += 1
			
			print("Partition of ", n ," equal size")
			i = 0
			#  Display partition result
			while (i < n) :
				print(result[i])
				i += 1
			
		
	

def main() :
	task = Splitting()
	#  Given string numbers
	num1 = "1234568902"
	num2 = "ABCDEFGHIKLMNOP"
	num3 = "xyz"
	#  Test Case
	task.splitNequalString(num1, 2)
	task.splitNequalString(num2, 5)
	task.splitNequalString(num3, 3)

if __name__ == "__main__": main()

Output

Given text :  1234568902
Partition of  2  equal size
12345
68902
Given text :  ABCDEFGHIKLMNOP
Partition of  5  equal size
ABC
DEF
GHI
KLM
NOP
Given text :  xyz
Partition of  3  equal size
x
y
z
#    Ruby Program for
#    Split the string into N equal parts
class Splitting 
	#  Returns the substring of given start and end position
	def subString(text, start, ends) 
		ans = ""
		i = start
		while (i < ends) 
			ans += text[i]
			i += 1
		end

		return ans
	end

	def splitNequalString(text, n) 
		size = text.length
		if (size == 0 || n > size) 
			#  Invalid partition
			return
		end

		print("Given text : ", text, "\n")
		if ((size % n) != 0) 
			print("Given ", n ," equal partition are not possible", "\n")
		else
 
			#  Use to collect partition result
			result = Array.new(n) {""}
			#  Get starting growth of partition
			growth = size / n
			start = 0
			ends = growth
			i = 0
			#  Collect partition result
			while (i < n) 
				result[i] = self.subString(text, start, ends)
				#  Get next partition position
				start = ends
				ends += growth
				i += 1
			end

			print("Partition of ", n ," equal size", "\n")
			i = 0
			#  Display partition result
			while (i < n) 
				print(result[i], "\n")
				i += 1
			end

		end

	end

end

def main() 
	task = Splitting.new()
	#  Given string numbers
	num1 = "1234568902"
	num2 = "ABCDEFGHIKLMNOP"
	num3 = "xyz"
	#  Test Case
	task.splitNequalString(num1, 2)
	task.splitNequalString(num2, 5)
	task.splitNequalString(num3, 3)
end

main()

Output

Given text : 1234568902
Partition of 2 equal size
12345
68902
Given text : ABCDEFGHIKLMNOP
Partition of 5 equal size
ABC
DEF
GHI
KLM
NOP
Given text : xyz
Partition of 3 equal size
x
y
z

/*
    Scala Program for
    Split the string into N equal parts
*/
class Splitting()
{
	// Returns the substring of given start and end position
	def subString(text: String, start: Int, end: Int): String = {
		var ans: String = "";
		var i: Int = start;
		while (i < end)
		{
			ans += text.charAt(i);
			i += 1;
		}
		return ans;
	}
	def splitNequalString(text: String, n: Int): Unit = {
		var size: Int = text.length();
		if (size == 0 || n > size)
		{
			// Invalid partition
			return;
		}
		println("Given text : " + text);
		if ((size % n) != 0)
		{
			println("Given " + n + " equal partition are not possible");
		}
		else
		{
			// Use to collect partition result
			var result: Array[String] = Array.fill[String](n)("");
			// Get starting growth of partition
			var growth: Int = size / n;
			var start: Int = 0;
			var end: Int = growth;
			var i: Int = 0;
			// Collect partition result
			while (i < n)
			{
				result(i) = subString(text, start, end);
				// Get next partition position
				start = end;
				end += growth;
				i += 1;
			}
			println("Partition of " + n + " equal size");
			i = 0;
			// Display partition result
			while (i < n)
			{
				println(result(i));
				i += 1;
			}
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Splitting = new Splitting();
		// Given string numbers
		var num1: String = "1234568902";
		var num2: String = "ABCDEFGHIKLMNOP";
		var num3: String = "xyz";
		// Test Case
		task.splitNequalString(num1, 2);
		task.splitNequalString(num2, 5);
		task.splitNequalString(num3, 3);
	}
}

Output

Given text : 1234568902
Partition of 2 equal size
12345
68902
Given text : ABCDEFGHIKLMNOP
Partition of 5 equal size
ABC
DEF
GHI
KLM
NOP
Given text : xyz
Partition of 3 equal size
x
y
z
import Foundation;
/*
    Swift 4 Program for
    Split the string into N equal parts
*/
class Splitting
{
	// Returns the substring of given start and end position
	func subString(_ text: [Character], _ start: Int, _ end: Int) -> String
	{
		var ans: String = "";
		var i: Int = start;
		while (i < end)
		{
			ans += String(text[i]);
			i += 1;
		}
		return ans;
	}
	func splitNequalString(_ data: String, _ n: Int)
	{
      	let text = Array(data);
		let size: Int = text.count;
		if (size == 0 || n > size)
		{
			// Invalid partition
			return;
		}
		print("Given text : ", data);
		if ((size % n)  != 0)
		{
			print("Given ", n ," equal partition are not possible");
		}
		else
		{
			// Use to collect partition result
			var result: [String] = Array(repeating: "", count: n);
			// Get starting growth of partition
			let growth: Int = size / n;
			var start: Int = 0;
			var end: Int = growth;
			var i: Int = 0;
			// Collect partition result
			while (i < n)
			{
				result[i] = self.subString(text, start, end);
				// Get next partition position
				start = end;
				end += growth;
				i += 1;
			}
			print("Partition of ", n ," equal size");
			i = 0;
			// Display partition result
			while (i < n)
			{
				print(result[i]);
				i += 1;
			}
		}
	}
}
func main()
{
	let task: Splitting = Splitting();
	// Given string numbers
	let num1: String = "1234568902";
	let num2: String = "ABCDEFGHIKLMNOP";
	let num3: String = "xyz";
	// Test Case
	task.splitNequalString(num1, 2);
	task.splitNequalString(num2, 5);
	task.splitNequalString(num3, 3);
}
main();

Output

Given text :  1234568902
Partition of  2  equal size
12345
68902
Given text :  ABCDEFGHIKLMNOP
Partition of  5  equal size
ABC
DEF
GHI
KLM
NOP
Given text :  xyz
Partition of  3  equal size
x
y
z
/*
    Kotlin Program for
    Split the string into N equal parts
*/
class Splitting
{
	// Returns the substring of given start and end position
	fun subString(text: String, start: Int, end: Int): String 
	{
		var ans: String = "";
		var i: Int = start;
		while (i < end)
		{
			ans += text.get(i);
			i += 1;
		}
		return ans;
	}
	fun splitNequalString(text: String, n: Int): Unit
	{
		val size: Int = text.length;
		if (size == 0 || n > size)
		{
			// Invalid partition
			return;
		}
		println("Given text : " + text);
		if ((size % n) != 0)
		{
			println("Given " + n + " equal partition are not possible");
		}
		else
		{
			// Use to collect partition result
			val result: Array < String > = Array(n)
			{
				""
			};
			// Get starting growth of partition
			val growth: Int = size / n;
			var start: Int = 0;
			var end: Int = growth;
			var i: Int = 0;
			// Collect partition result
			while (i < n)
			{
				result[i] = this.subString(text, start, end);
				// Get next partition position
				start = end;
				end += growth;
				i += 1;
			}
			println("Partition of " + n + " equal size");
			i = 0;
			// Display partition result
			while (i < n)
			{
				println(result[i]);
				i += 1;
			}
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Splitting = Splitting();
	// Given string numbers
	val num1: String = "1234568902";
	val num2: String = "ABCDEFGHIKLMNOP";
	val num3: String = "xyz";
	// Test Case
	task.splitNequalString(num1, 2);
	task.splitNequalString(num2, 5);
	task.splitNequalString(num3, 3);
}

Output

Given text : 1234568902
Partition of 2 equal size
12345
68902
Given text : ABCDEFGHIKLMNOP
Partition of 5 equal size
ABC
DEF
GHI
KLM
NOP
Given text : xyz
Partition of 3 equal size
x
y
z
Last updated on by Kalkicode




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