Find the frequency of subsequence in given string
Here given code implementation process.
/*
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
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
New Comment