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Find the frequency of subsequence in given string

In computer science and string manipulation, a subsequence is a sequence of characters obtained by deleting zero or more characters from the original string, without changing the order of the remaining characters. Given a string and a target subsequence, the problem is to find the frequency of occurrence of the subsequence in the given string. In this article, we will explore a Java solution to this problem using dynamic programming.

Problem Statement

Given a string text and a target subsequence sequence, we want to determine how many times the subsequence sequence appears in the string text. The subsequence does not have to be contiguous, but the characters must appear in the same relative order as they do in the target subsequence.

For example, consider the following input:

text = aaabbb
-------------
subsequence = ab

    a a a b b b
➀   -     -    ->ab
    a a a b b b
➁   -       -   ->ab
    a a a b b b
➂   -         - ->ab
    a a a b b b
➃     -   -     ->ab
    a a a b b b
➄     -     -   ->ab
    a a a b b b
➅     -       - ->ab
    a a a b b b
➆       - -     ->ab
    a a a b b b
➇       -   -   ->ab 
    a a a b b b            
➈       -     - ->ab 
______________________
Result = 9

The subsequence "ab" appears 9 times in the given string "aaabbb". The goal is to find this count.

Approach: Dynamic Programming

To solve this problem efficiently, we can use a dynamic programming approach. We'll define a two-dimensional array dp to store intermediate results. Here, dp[i][j] represents the count of occurrences of the subsequence sequence[0...j-1] in the string text[0...i-1].

We'll initialize the first column of the dp array as 1 because any subsequence can be formed with an empty string. The first row, except for dp[0][0], will be initialized as 0 because an empty subsequence cannot be formed with a non-empty string.

Next, we'll iterate over the characters of the string text and the target subsequence sequence. If the characters match, we update the current dp element by adding the count of two elements: dp[i-1][j-1] (indicating that we use the current characters to form the subsequence) and dp[i-1][j] (indicating that we don't use the current character to form the subsequence). If the characters don't match, we take the value from the previous row.

Finally, the value in the bottom-right cell of the dp array, dp[n][m], represents the count of occurrences of the subsequence sequence in the string text, where n is the length of text, and m is the length of sequence.

Code Solution

/*
    Java program for
    Find the frequency of subsequence in given string
*/
public class Subsequence
{
	public void countSubSequence(String text, String sequence)
	{
		// Get the length of given parameters
		int n = text.length();
		int m = sequence.length();
		int[][] dp = new int[n + 1][m + 1];
		// Set initial value of first column
		for (int i = 1; i <= n; ++i)
		{
			dp[i][0] = 1;
		}
		// Set initial value of first row
		for (int i = 1; i <= m; ++i)
		{
			dp[0][i] = 0;
		}
		// Active first row first column
		dp[0][0] = 1;
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= m; ++j)
			{
				if (text.charAt(i - 1) == sequence.charAt(j - 1))
				{
					// Change element value by using sum of top two elements
					dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
				}
				else
				{
					dp[i][j] = dp[i - 1][j];
				}
			}
		}
		// Display calculated result
		System.out.println(dp[n][m]);
	}
	public static void main(String[] args)
	{
		Subsequence task = new Subsequence();
		String text = "aaabbb";
		String sequence = "ab";
		/*
		    text = aaabbb
		    -------------
		    subsequence = ab
		    
		        a a a b b b
		    ➀   -     -    ->ab
		        a a a b b b
		    ➁   -       -   ->ab
		        a a a b b b
		    ➂   -         - ->ab
		        a a a b b b
		    ➃     -   -     ->ab
		        a a a b b b
		    ➄     -     -   ->ab
		        a a a b b b
		    ➅     -       - ->ab
		        a a a b b b
		    ➆       - -     ->ab
		        a a a b b b
		    ➇       -   -   ->ab 
		        a a a b b b            
		    ➈       -     - ->ab 
		    ______________________
		    Result = 9
		*/
		task.countSubSequence(text, sequence);
	}
}

Output

9
// Include header file
#include <iostream>

#include <string>

using namespace std;
/*
    C++ program for
    Find the frequency of subsequence in given string
*/
class Subsequence
{
	public: void countSubSequence(string text, string sequence)
	{
		// Get the length of given parameters
		int n = text.length();
		int m = sequence.length();
		int dp[n + 1][m + 1];
		// Set initial value of first column
		for (int i = 1; i <= n; ++i)
		{
			dp[i][0] = 1;
		}
		// Set initial value of first row
		for (int i = 1; i <= m; ++i)
		{
			dp[0][i] = 0;
		}
		// Active first row first column
		dp[0][0] = 1;
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= m; ++j)
			{
				if (text[i - 1] == sequence[j - 1])
				{
					// Change element value by using sum of top two elements
					dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
				}
				else
				{
					dp[i][j] = dp[i - 1][j];
				}
			}
		}
		// Display calculated result
		cout << dp[n][m] << endl;
	}
};
int main()
{
	Subsequence *task = new Subsequence();
	string text = "aaabbb";
	string sequence = "ab";
	/*
	    text = aaabbb
	    -------------
	    subsequence = ab
	    
	        a a a b b b
	    ➀   -     -    ->ab
	        a a a b b b
	    ➁   -       -   ->ab
	        a a a b b b
	    ➂   -         - ->ab
	        a a a b b b
	    ➃     -   -     ->ab
	        a a a b b b
	    ➄     -     -   ->ab
	        a a a b b b
	    ➅     -       - ->ab
	        a a a b b b
	    ➆       - -     ->ab
	        a a a b b b
	    ➇       -   -   ->ab 
	        a a a b b b            
	    ➈       -     - ->ab 
	    ______________________
	    Result = 9
	*/
	task->countSubSequence(text, sequence);
	return 0;
}

Output

9
// Include namespace system
using System;
/*
    Csharp program for
    Find the frequency of subsequence in given string
*/
public class Subsequence
{
	public void countSubSequence(String text, String sequence)
	{
		// Get the length of given parameters
		int n = text.Length;
		int m = sequence.Length;
		int[,] dp = new int[n + 1,m + 1];
		// Set initial value of first column
		for (int i = 1; i <= n; ++i)
		{
			dp[i,0] = 1;
		}
		// Set initial value of first row
		for (int i = 1; i <= m; ++i)
		{
			dp[0,i] = 0;
		}
		// Active first row first column
		dp[0,0] = 1;
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= m; ++j)
			{
				if (text[i - 1] == sequence[j - 1])
				{
					// Change element value by using sum of top two elements
					dp[i,j] = dp[i - 1,j - 1] + dp[i - 1,j];
				}
				else
				{
					dp[i,j] = dp[i - 1,j];
				}
			}
		}
		// Display calculated result
		Console.WriteLine(dp[n,m]);
	}
	public static void Main(String[] args)
	{
		Subsequence task = new Subsequence();
		String text = "aaabbb";
		String sequence = "ab";
		/*
		    text = aaabbb
		    -------------
		    subsequence = ab
		    
		        a a a b b b
		    ➀   -     -    ->ab
		        a a a b b b
		    ➁   -       -   ->ab
		        a a a b b b
		    ➂   -         - ->ab
		        a a a b b b
		    ➃     -   -     ->ab
		        a a a b b b
		    ➄     -     -   ->ab
		        a a a b b b
		    ➅     -       - ->ab
		        a a a b b b
		    ➆       - -     ->ab
		        a a a b b b
		    ➇       -   -   ->ab 
		        a a a b b b            
		    ➈       -     - ->ab 
		    ______________________
		    Result = 9
		*/
		task.countSubSequence(text, sequence);
	}
}

Output

9
package main
import "fmt"
/*
    Go program for
    Find the frequency of subsequence in given string
*/

func countSubSequence(text, sequence string) {
	// Get the length of given parameters
	var n int = len(text)
	var m int = len(sequence)
	var dp = make([][] int, n + 1)
	for i:= 0; i <= n ; i++{
		dp[i] = make([]int, m + 1)
	}
	// Set initial value of first column
	for i := 1 ; i <= n ; i++ {
		dp[i][0] = 1
	}
	// Set initial value of first row
	for i := 1 ; i <= m ; i++ {
		dp[0][i] = 0
	}
	// Active first row first column
	dp[0][0] = 1
	for i := 1 ; i <= n ; i++ {
		for j := 1 ; j <= m ; j++ {
			if text[i - 1] == sequence[j - 1] {
				// Change element value by using sum of top two elements
				dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
			} else {
				dp[i][j] = dp[i - 1][j]
			}
		}
	}
	// Display calculated result
	fmt.Println(dp[n][m])
}
func main() {

	var text string = "aaabbb"
	var sequence string = "ab"
	/*
	    text = aaabbb
	    -------------
	    subsequence = ab
	    
	        a a a b b b
	    ➀   -     -    ->ab
	        a a a b b b
	    ➁   -       -   ->ab
	        a a a b b b
	    ➂   -         - ->ab
	        a a a b b b
	    ➃     -   -     ->ab
	        a a a b b b
	    ➄     -     -   ->ab
	        a a a b b b
	    ➅     -       - ->ab
	        a a a b b b
	    ➆       - -     ->ab
	        a a a b b b
	    ➇       -   -   ->ab 
	        a a a b b b            
	    ➈       -     - ->ab 
	    ______________________
	    Result = 9
	*/
	countSubSequence(text, sequence)
}

Output

9
<?php
/*
    Php program for
    Find the frequency of subsequence in given string
*/
class Subsequence
{
	public	function countSubSequence($text, $sequence)
	{
		// Get the length of given parameters
		$n = strlen($text);
		$m = strlen($sequence);
		$dp = array_fill(0, $n + 1, array_fill(0, $m + 1, 0));
		// Set initial value of first column
		for ($i = 1; $i <= $n; ++$i)
		{
			$dp[$i][0] = 1;
		}
		// Set initial value of first row
		for ($i = 1; $i <= $m; ++$i)
		{
			$dp[0][$i] = 0;
		}
		// Active first row first column
		$dp[0][0] = 1;
		for ($i = 1; $i <= $n; ++$i)
		{
			for ($j = 1; $j <= $m; ++$j)
			{
				if ($text[$i - 1] == $sequence[$j - 1])
				{
					// Change element value by using sum of top two elements
					$dp[$i][$j] = $dp[$i - 1][$j - 1] + $dp[$i - 1][$j];
				}
				else
				{
					$dp[$i][$j] = $dp[$i - 1][$j];
				}
			}
		}
		// Display calculated result
		echo($dp[$n][$m]."\n");
	}
}

function main()
{
	$task = new Subsequence();
	$text = "aaabbb";
	$sequence = "ab";
	/*
	    text = aaabbb
	    -------------
	    subsequence = ab
	    
	        a a a b b b
	    ➀   -     -    ->ab
	        a a a b b b
	    ➁   -       -   ->ab
	        a a a b b b
	    ➂   -         - ->ab
	        a a a b b b
	    ➃     -   -     ->ab
	        a a a b b b
	    ➄     -     -   ->ab
	        a a a b b b
	    ➅     -       - ->ab
	        a a a b b b
	    ➆       - -     ->ab
	        a a a b b b
	    ➇       -   -   ->ab 
	        a a a b b b            
	    ➈       -     - ->ab 
	    ______________________
	    Result = 9
	*/
	$task->countSubSequence($text, $sequence);
}
main();

Output

9
/*
    Node JS program for
    Find the frequency of subsequence in given string
*/
class Subsequence
{
	countSubSequence(text, sequence)
	{
		// Get the length of given parameters
		var n = text.length;
		var m = sequence.length;
		var dp = Array(n + 1).fill(0).map(() => new Array(m + 1).fill(0));
		// Set initial value of first column
		for (var i = 1; i <= n; ++i)
		{
			dp[i][0] = 1;
		}
		// Set initial value of first row
		for (var i = 1; i <= m; ++i)
		{
			dp[0][i] = 0;
		}
		// Active first row first column
		dp[0][0] = 1;
		for (var i = 1; i <= n; ++i)
		{
			for (var j = 1; j <= m; ++j)
			{
				if (text.charAt(i - 1) == sequence.charAt(j - 1))
				{
					// Change element value by using sum of top two elements
					dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
				}
				else
				{
					dp[i][j] = dp[i - 1][j];
				}
			}
		}
		// Display calculated result
		console.log(dp[n][m]);
	}
}

function main()
{
	var task = new Subsequence();
	var text = "aaabbb";
	var sequence = "ab";
	/*
	    text = aaabbb
	    -------------
	    subsequence = ab
	    
	        a a a b b b
	    ➀   -     -    ->ab
	        a a a b b b
	    ➁   -       -   ->ab
	        a a a b b b
	    ➂   -         - ->ab
	        a a a b b b
	    ➃     -   -     ->ab
	        a a a b b b
	    ➄     -     -   ->ab
	        a a a b b b
	    ➅     -       - ->ab
	        a a a b b b
	    ➆       - -     ->ab
	        a a a b b b
	    ➇       -   -   ->ab 
	        a a a b b b            
	    ➈       -     - ->ab 
	    ______________________
	    Result = 9
	*/
	task.countSubSequence(text, sequence);
}
main();

Output

9
#    Python 3 program for
#    Find the frequency of subsequence in given string
class Subsequence :
	def countSubSequence(self, text, sequence) :
		#  Get the length of given parameters
		n = len(text)
		m = len(sequence)
		dp = [[0] * (m + 1) for _ in range(n + 1) ]
		i = 1
		#  Set initial value of first column
		while (i <= n) :
			dp[i][0] = 1
			i += 1
		
		i = 1
		#  Set initial value of first row
		while (i <= m) :
			dp[0][i] = 0
			i += 1
		
		#  Active first row first column
		dp[0][0] = 1
		i = 1
		while (i <= n) :
			j = 1
			while (j <= m) :
				if (text[i - 1] == sequence[j - 1]) :
					#  Change element value by using sum of top two elements
					dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
				else :
					dp[i][j] = dp[i - 1][j]
				
				j += 1
			
			i += 1
		
		#  Display calculated result
		print(dp[n][m])
	

def main() :
	task = Subsequence()
	text = "aaabbb"
	sequence = "ab"
	#    text = aaabbb
	#    -------------
	#    subsequence = ab
	#        a a a b b b
	#    ➀   -     -    ->ab
	#        a a a b b b
	#    ➁   -       -   ->ab
	#        a a a b b b
	#    ➂   -         - ->ab
	#        a a a b b b
	#    ➃     -   -     ->ab
	#        a a a b b b
	#    ➄     -     -   ->ab
	#        a a a b b b
	#    ➅     -       - ->ab
	#        a a a b b b
	#    ➆       - -     ->ab
	#        a a a b b b
	#    ➇       -   -   ->ab 
	#        a a a b b b            
	#    ➈       -     - ->ab 
	#    ______________________
	#    Result = 9
	task.countSubSequence(text, sequence)

if __name__ == "__main__": main()

Output

9
#    Ruby program for
#    Find the frequency of subsequence in given string
class Subsequence 
	def countSubSequence(text, sequence) 
		#  Get the length of given parameters
		n = text.length
		m = sequence.length
		dp = Array.new(n + 1) {Array.new(m + 1) {0}}
		i = 1
		#  Set initial value of first column
		while (i <= n) 
			dp[i][0] = 1
			i += 1
		end

		i = 1
		#  Set initial value of first row
		while (i <= m) 
			dp[0][i] = 0
			i += 1
		end

		#  Active first row first column
		dp[0][0] = 1
		i = 1
		while (i <= n) 
			j = 1
			while (j <= m) 
				if (text[i - 1] == sequence[j - 1]) 
					#  Change element value by using sum of top two elements
					dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
				else
 
					dp[i][j] = dp[i - 1][j]
				end

				j += 1
			end

			i += 1
		end

		#  Display calculated result
		print(dp[n][m], "\n")
	end

end

def main() 
	task = Subsequence.new()
	text = "aaabbb"
	sequence = "ab"
	#    text = aaabbb
	#    -------------
	#    subsequence = ab
	#        a a a b b b
	#    ➀   -     -    ->ab
	#        a a a b b b
	#    ➁   -       -   ->ab
	#        a a a b b b
	#    ➂   -         - ->ab
	#        a a a b b b
	#    ➃     -   -     ->ab
	#        a a a b b b
	#    ➄     -     -   ->ab
	#        a a a b b b
	#    ➅     -       - ->ab
	#        a a a b b b
	#    ➆       - -     ->ab
	#        a a a b b b
	#    ➇       -   -   ->ab 
	#        a a a b b b            
	#    ➈       -     - ->ab 
	#    ______________________
	#    Result = 9
	task.countSubSequence(text, sequence)
end

main()

Output

9

import scala.collection.mutable._;
/*
    Scala program for
    Find the frequency of subsequence in given string
*/
class Subsequence()
{
	def countSubSequence(text: String, sequence: String): Unit = {
		// Get the length of given parameters
		var n: Int = text.length();
		var m: Int = sequence.length();
		var dp: Array[Array[Int]] = Array.fill[Int](n + 1, m + 1)(0);
		var i: Int = 1;
		// Set initial value of first column
		while (i <= n)
		{
			dp(i)(0) = 1;
			i += 1;
		}
		i = 1;
		// Set initial value of first row
		while (i <= m)
		{
			dp(0)(i) = 0;
			i += 1;
		}
		// Active first row first column
		dp(0)(0) = 1;
		i = 1;
		while (i <= n)
		{
			var j: Int = 1;
			while (j <= m)
			{
				if (text.charAt(i - 1) == sequence.charAt(j - 1))
				{
					// Change element value by using sum of top two elements
					dp(i)(j) = dp(i - 1)(j - 1) + dp(i - 1)(j);
				}
				else
				{
					dp(i)(j) = dp(i - 1)(j);
				}
				j += 1;
			}
			i += 1;
		}
		// Display calculated result
		println(dp(n)(m));
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Subsequence = new Subsequence();
		var text: String = "aaabbb";
		var sequence: String = "ab";
		/*
		    text = aaabbb
		    -------------
		    subsequence = ab
		    
		        a a a b b b
		    ➀   -     -    ->ab
		        a a a b b b
		    ➁   -       -   ->ab
		        a a a b b b
		    ➂   -         - ->ab
		        a a a b b b
		    ➃     -   -     ->ab
		        a a a b b b
		    ➄     -     -   ->ab
		        a a a b b b
		    ➅     -       - ->ab
		        a a a b b b
		    ➆       - -     ->ab
		        a a a b b b
		    ➇       -   -   ->ab 
		        a a a b b b            
		    ➈       -     - ->ab 
		    ______________________
		    Result = 9
		*/
		task.countSubSequence(text, sequence);
	}
}

Output

9
import Foundation;
/*
    Swift 4 program for
    Find the frequency of subsequence in given string
*/
class Subsequence
{
	func countSubSequence(_ t: String, _ s: String)
	{
      	let text = Array(t);
      	let sequence = Array(s);
		// Get the length of given parameters
		let n: Int = text.count;
		let m: Int = sequence.count;
		var dp: [
			[Int]
		] = Array(repeating: Array(repeating: 0, count: m + 1), count: n + 1);
		var i: Int = 1;
		// Set initial value of first column
		while (i <= n)
		{
			dp[i][0] = 1;
			i += 1;
		}
		i = 1;
		// Set initial value of first row
		while (i <= m)
		{
			dp[0][i] = 0;
			i += 1;
		}
		// Active first row first column
		dp[0][0] = 1;
		i = 1;
		while (i <= n)
		{
			var j: Int = 1;
			while (j <= m)
			{
				if (text[i - 1] == sequence[j - 1])
				{
					// Change element value by using sum of top two elements
					dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
				}
				else
				{
					dp[i][j] = dp[i - 1][j];
				}
				j += 1;
			}
			i += 1;
		}
		// Display calculated result
		print(dp[n][m]);
	}
}
func main()
{
	let task: Subsequence = Subsequence();
	let text: String = "aaabbb";
	let sequence: String = "ab";
	/*
	    text = aaabbb
	    -------------
	    subsequence = ab
	    
	        a a a b b b
	    ➀   -     -    ->ab
	        a a a b b b
	    ➁   -       -   ->ab
	        a a a b b b
	    ➂   -         - ->ab
	        a a a b b b
	    ➃     -   -     ->ab
	        a a a b b b
	    ➄     -     -   ->ab
	        a a a b b b
	    ➅     -       - ->ab
	        a a a b b b
	    ➆       - -     ->ab
	        a a a b b b
	    ➇       -   -   ->ab 
	        a a a b b b            
	    ➈       -     - ->ab 
	    ______________________
	    Result = 9
	*/
	task.countSubSequence(text, sequence);
}
main();

Output

9
/*
    Kotlin program for
    Find the frequency of subsequence in given string
*/
class Subsequence
{
	fun countSubSequence(text: String, sequence: String): Unit
	{
		// Get the length of given parameters
		val n: Int = text.length;
		val m: Int = sequence.length;
		var dp: Array < Array < Int >> = Array(n + 1)
		{
			Array(m + 1)
			{
				0
			}
		};
		var i: Int = 1;
		// Set initial value of first column
		while (i <= n)
		{
			dp[i][0] = 1;
			i += 1;
		}
		i = 1;
		// Set initial value of first row
		while (i <= m)
		{
			dp[0][i] = 0;
			i += 1;
		}
		// Active first row first column
		dp[0][0] = 1;
		i = 1;
		while (i <= n)
		{
			var j: Int = 1;
			while (j <= m)
			{
				if (text.get(i - 1) == sequence.get(j - 1))
				{
					// Change element value by using sum of top two elements
					dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
				}
				else
				{
					dp[i][j] = dp[i - 1][j];
				}
				j += 1;
			}
			i += 1;
		}
		// Display calculated result
		println(dp[n][m]);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Subsequence = Subsequence();
	val text: String = "aaabbb";
	val sequence: String = "ab";
	/*
	    text = aaabbb
	    -------------
	    subsequence = ab
	    
	        a a a b b b
	    ➀   -     -    ->ab
	        a a a b b b
	    ➁   -       -   ->ab
	        a a a b b b
	    ➂   -         - ->ab
	        a a a b b b
	    ➃     -   -     ->ab
	        a a a b b b
	    ➄     -     -   ->ab
	        a a a b b b
	    ➅     -       - ->ab
	        a a a b b b
	    ➆       - -     ->ab
	        a a a b b b
	    ➇       -   -   ->ab 
	        a a a b b b            
	    ➈       -     - ->ab 
	    ______________________
	    Result = 9
	*/
	task.countSubSequence(text, sequence);
}

Output

9

Output

The above code will output 9 because the subsequence "ab" appears 9 times in the string "aaabbb".

Time Complexity Analysis

The time complexity of this solution is determined by the nested loops that iterate over the characters of the input strings. Let n be the length of the text string and m be the length of the sequence string. The nested loops iterate from 1 to n and 1 to m respectively. Therefore, the time complexity of this solution is O(n * m).

Last updated on by Kalkicode




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