# 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)
{
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
*/
}
}``````

#### 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()
{
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
*/
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)
{
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
*/
}
}``````

#### 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()
{
\$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
*/
}
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 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
*/
}
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() :
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

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()
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
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
*/
}
}``````

#### 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 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
*/
}
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 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
*/
}``````

#### 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).

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