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Shortest uncommon subsequence using dynamic programming

Dynamic programming is a powerful technique for solving optimization problems. In this article, we will explore the problem of finding the shortest uncommon subsequence between two strings using dynamic programming. We will discuss the problem statement, present an algorithm, provide a step-by-step solution, and analyze the time complexity of the code.

Problem Statement

The problem is to find the length of the shortest uncommon subsequence that exists in str1 but not in str2. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. The uncommon subsequence is the one that exists in str1 but is not a subsequence of str2.

Algorithm

We can solve this problem using dynamic programming with the following steps:

  1. Initialize a 2D array dp of size (m + 1) x (n + 1), where m and n are the lengths of str1 and str2 respectively.
  2. Set max to the maximum integer value. Reduce the size of max by the length of the longest substring between str1 and str2.
  3. Initialize the first row of dp to 1, as an empty string is always a subsequence of any string.
  4. Initialize the first column of dp to max, as it represents the case where str2 is an empty string and any non-empty subsequence of str1 would be uncommon.
  5. Iterate over the remaining cells of dp (excluding the first row and column) and calculate the values based on the following rules:
    • If the characters at the current positions in str1 and str2 are the same, set the current cell value to the minimum of the previous row value and the previous diagonal cell value plus one.
    • If the characters are different, find the last occurrence of the character at the current position in str1 in str2. If it exists, set the current cell value to the minimum of the previous row value and the value at the last occurrence position plus one. Otherwise, set it to 1.
  6. If the bottom-right cell of dp is not equal to max, it means there is an uncommon subsequence, and we assign its value to result.

Solution

/*
    Java program for
    Shortest uncommon subsequence using dynamic programming
*/
public class Subsequence
{
	public int minValue(int a, int b)
	{
		if (a < b)
		{
			return a;
		}
		return b;
	}
	// Find the length of shortest uncommon subsequence which is exist
	// In str1 but not in str2.
	public void uncommonSubsequence(
    	String str1, 
     		String str2)
	{
		// Get the length
		int m = str1.length();
		int n = str2.length();
      	// Auxiliary space
		int[][] dp = new int[m + 1][n + 1];
		int result = 0;
		int max = Integer.MAX_VALUE;
		// Reduce size of max by length of longest substring
		if (n > m)
		{
			max -= n;
		}
		else
		{
			max -= m;
		}
		for (int i = 0; i <= m; i++)
		{
			dp[i][0] = 1;
		}
		for (int i = 0; i <= n; i++)
		{
			dp[0][i] = max;
		}
		for (int i = 1; i <= m; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				int k = j - 1;
				while (k >= 0 && str2.charAt(k) != str1.charAt(i - 1))
				{
					k--;
				}
				if (k == -1)
				{
					dp[i][j] = 1;
				}
				else
				{
					dp[i][j] = minValue(dp[i - 1][j], dp[i - 1][k] + 1);
				}
			}
		}
		if (dp[m][n] != max)
		{
			result = dp[m][n];
		}
		// Display given strings
		System.out.println(" String A : " + str1);
		System.out.println(" String B : " + str2);
		// Display calculated result
		System.out.println(" " + result);
	}
	public static void main(String[] args)
	{
		Subsequence task = new Subsequence();
		String str1 = "aecb";
		String str2 = "bace";
		// Case A
		/*
		   Example 1

		   str1 = aecb
		   str2 = bace
		   ------------
		    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
            str1 but not str2
		    --------------------------------------------
		    Length of shortest subsequence is 2
		    Ans = 2
		*/
		task.uncommonSubsequence(str1, str2);
		// Case B
		str1 = "ABCCDBE";
		str2 = "ABCCDBE";
		/*
		   Example 2
		   
		   str1 = ABCCDBE
		   str2 = ABCCDBE
		   ------------
		   all subsequence str1 exist in str2
		*/
		task.uncommonSubsequence(str1, str2);
		// Case C
		str1 = "CCCCCB";
		str2 = "CCB";
		/*
		   Example 3
		   
		   str1 = CCCCCB
		   str2 = CCB
		   ------------
		   'CCC' Shortest subsequence which exist in str1 but not on str2.
		   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
		   but its length more than 3.
		   --------------------------------------------------------
		   Ans = 3
		*/
		task.uncommonSubsequence(str1, str2);
		// Case D
		str1 = "fbi";
		str2 = "ice";
		/*
		   Example 4
		   
		   str1 = fbi
		   str2 = ice
		   ------------
		    [f,b,fb,fbi]
		    Subsequences which is exist in str1 but not on str2.
		   --------------------------------------------------------
		   Ans = 1  [length of f or b]
		*/
		task.uncommonSubsequence(str1, str2);
	}
}

Output

 String A : aecb
 String B : bace
 2
 String A : ABCCDBE
 String B : ABCCDBE
 0
 String A : CCCCCB
 String B : CCB
 3
 String A : fbi
 String B : ice
 1
// Include header file
#include <iostream>
#include <string>
#include <limits.h>

using namespace std;
/*
    C++ program for
    Shortest uncommon subsequence using dynamic programming
*/
class Subsequence
{
	public: int minValue(int a, int b)
	{
		if (a < b)
		{
			return a;
		}
		return b;
	}
	// Find the length of shortest uncommon subsequence which is exist
	// In str1 but not in str2.
	void uncommonSubsequence(string str1, string str2)
	{
		// Get the length
		int m = str1.length();
		int n = str2.length();
		// Auxiliary space
		int dp[m + 1][n + 1];
		int result = 0;
		int max = INT_MAX;
		// Reduce size of max by length of longest substring
		if (n > m)
		{
			max -= n;
		}
		else
		{
			max -= m;
		}
		for (int i = 0; i <= m; i++)
		{
			dp[i][0] = 1;
		}
		for (int i = 0; i <= n; i++)
		{
			dp[0][i] = max;
		}
		for (int i = 1; i <= m; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				int k = j - 1;
				while (k >= 0 && str2[k] != str1[i - 1])
				{
					k--;
				}
				if (k == -1)
				{
					dp[i][j] = 1;
				}
				else
				{
					dp[i][j] = this->minValue(dp[i - 1][j], dp[i - 1][k] + 1);
				}
			}
		}
		if (dp[m][n] != max)
		{
			result = dp[m][n];
		}
		// Display given strings
		cout << " String A : " << str1 << endl;
		cout << " String B : " << str2 << endl;
		// Display calculated result
		cout << " " << result << endl;
	}
};
int main()
{
	Subsequence *task = new Subsequence();
	string str1 = "aecb";
	string str2 = "bace";
	// Case A
	/*
	   Example 1
	   str1 = aecb
	   str2 = bace
	   ------------
	    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
	            str1 but not str2
	    --------------------------------------------
	    Length of shortest subsequence is 2
	    Ans = 2
	*/
	task->uncommonSubsequence(str1, str2);
	// Case B
	str1 = "ABCCDBE";
	str2 = "ABCCDBE";
	/*
	   Example 2
	   
	   str1 = ABCCDBE
	   str2 = ABCCDBE
	   ------------
	   all subsequence str1 exist in str2
	*/
	task->uncommonSubsequence(str1, str2);
	// Case C
	str1 = "CCCCCB";
	str2 = "CCB";
	/*
	   Example 3
	   
	   str1 = CCCCCB
	   str2 = CCB
	   ------------
	   'CCC' Shortest subsequence which exist in str1 but not on str2.
	   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
	   but its length more than 3.
	   --------------------------------------------------------
	   Ans = 3
	*/
	task->uncommonSubsequence(str1, str2);
	// Case D
	str1 = "fbi";
	str2 = "ice";
	/*
	   Example 4
	   
	   str1 = fbi
	   str2 = ice
	   ------------
	    [f,b,fb,fbi]
	    Subsequences which is exist in str1 but not on str2.
	   --------------------------------------------------------
	   Ans = 1  [length of f or b]
	*/
	task->uncommonSubsequence(str1, str2);
	return 0;
}

Output

 String A : aecb
 String B : bace
 2
 String A : ABCCDBE
 String B : ABCCDBE
 0
 String A : CCCCCB
 String B : CCB
 3
 String A : fbi
 String B : ice
 1
// Include namespace system
using System;
/*
    Csharp program for
    Shortest uncommon subsequence using dynamic programming
*/
public class Subsequence
{
	public int minValue(int a, int b)
	{
		if (a < b)
		{
			return a;
		}
		return b;
	}
	// Find the length of shortest uncommon subsequence which is exist
	// In str1 but not in str2.
	public void uncommonSubsequence(String str1, String str2)
	{
		// Get the length
		int m = str1.Length;
		int n = str2.Length;
		// Auxiliary space
		int[,] dp = new int[m + 1,n + 1];
		int result = 0;
		int max = int.MaxValue;
		// Reduce size of max by length of longest substring
		if (n > m)
		{
			max -= n;
		}
		else
		{
			max -= m;
		}
		for (int i = 0; i <= m; i++)
		{
			dp[i,0] = 1;
		}
		for (int i = 0; i <= n; i++)
		{
			dp[0,i] = max;
		}
		for (int i = 1; i <= m; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				int k = j - 1;
				while (k >= 0 && str2[k] != str1[i - 1])
				{
					k--;
				}
				if (k == -1)
				{
					dp[i,j] = 1;
				}
				else
				{
					dp[i,j] = this.minValue(dp[i - 1,j], dp[i - 1,k] + 1);
				}
			}
		}
		if (dp[m,n] != max)
		{
			result = dp[m,n];
		}
		// Display given strings
		Console.WriteLine(" String A : " + str1);
		Console.WriteLine(" String B : " + str2);
		// Display calculated result
		Console.WriteLine(" " + result);
	}
	public static void Main(String[] args)
	{
		Subsequence task = new Subsequence();
		String str1 = "aecb";
		String str2 = "bace";
		// Case A
		/*
		   Example 1
		   str1 = aecb
		   str2 = bace
		   ------------
		    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
		            str1 but not str2
		    --------------------------------------------
		    Length of shortest subsequence is 2
		    Ans = 2
		*/
		task.uncommonSubsequence(str1, str2);
		// Case B
		str1 = "ABCCDBE";
		str2 = "ABCCDBE";
		/*
		   Example 2
		   
		   str1 = ABCCDBE
		   str2 = ABCCDBE
		   ------------
		   all subsequence str1 exist in str2
		*/
		task.uncommonSubsequence(str1, str2);
		// Case C
		str1 = "CCCCCB";
		str2 = "CCB";
		/*
		   Example 3
		   
		   str1 = CCCCCB
		   str2 = CCB
		   ------------
		   'CCC' Shortest subsequence which exist in str1 but not on str2.
		   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
		   but its length more than 3.
		   --------------------------------------------------------
		   Ans = 3
		*/
		task.uncommonSubsequence(str1, str2);
		// Case D
		str1 = "fbi";
		str2 = "ice";
		/*
		   Example 4
		   
		   str1 = fbi
		   str2 = ice
		   ------------
		    [f,b,fb,fbi]
		    Subsequences which is exist in str1 but not on str2.
		   --------------------------------------------------------
		   Ans = 1  [length of f or b]
		*/
		task.uncommonSubsequence(str1, str2);
	}
}

Output

 String A : aecb
 String B : bace
 2
 String A : ABCCDBE
 String B : ABCCDBE
 0
 String A : CCCCCB
 String B : CCB
 3
 String A : fbi
 String B : ice
 1
package main
import "math"
import "fmt"
/*
    Go program for
    Shortest uncommon subsequence using dynamic programming
*/

func minValue(a, b int) int {
	if a < b {
		return a
	}
	return b
}
// Find the length of shortest uncommon subsequence which is exist
// In str1 but not in str2.
func uncommonSubsequence(str1, str2 string) {
	// Get the length
	var m int = len(str1)
	var n int = len(str2)
	// Auxiliary space
	var dp = make([][] int, m + 1)
	for i:= 0;i < m + 1; i++ {
		dp[i] = make([]int ,n + 1)
	}
	var result int = 0
	var max int = math.MaxInt64
	// Reduce size of max by length of longest substring
	if n > m {
		max -= n
	} else {
		max -= m
	}
	for i := 0 ; i <= m ; i++ {
		dp[i][0] = 1
	}
	for i := 0 ; i <= n ; i++ {
		dp[0][i] = max
	}
	for i := 1 ; i <= m ; i++ {
		for j := 1 ; j <= n ; j++ {
			var k int = j - 1
			for (k >= 0 && str2[k] != str1[i - 1]) {
				k--
			}
			if k == -1 {
				dp[i][j] = 1
			} else {
				dp[i][j] = minValue(dp[i - 1][j], dp[i - 1][k] + 1)
			}
		}
	}
	if dp[m][n] != max {
		result = dp[m][n]
	}
	// Display given strings
	fmt.Println(" String A : ", str1)
	fmt.Println(" String B : ", str2)
	// Display calculated result
	fmt.Println(" ", result)
}
func main() {

	var str1 string = "aecb"
	var str2 string = "bace"
	// Case A
	/*
	   Example 1
	   str1 = aecb
	   str2 = bace
	   ------------
	    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
	            str1 but not str2
	    --------------------------------------------
	    Length of shortest subsequence is 2
	    Ans = 2
	*/
	uncommonSubsequence(str1, str2)
	// Case B
	str1 = "ABCCDBE"
	str2 = "ABCCDBE"
	/*
	   Example 2
	   
	   str1 = ABCCDBE
	   str2 = ABCCDBE
	   ------------
	   all subsequence str1 exist in str2
	*/
	uncommonSubsequence(str1, str2)
	// Case C
	str1 = "CCCCCB"
	str2 = "CCB"
	/*
	   Example 3
	   
	   str1 = CCCCCB
	   str2 = CCB
	   ------------
	   'CCC' Shortest subsequence which exist in str1 but not on str2.
	   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
	   but its length more than 3.
	   --------------------------------------------------------
	   Ans = 3
	*/
	uncommonSubsequence(str1, str2)
	// Case D
	str1 = "fbi"
	str2 = "ice"
	/*
	   Example 4
	   
	   str1 = fbi
	   str2 = ice
	   ------------
	    [f,b,fb,fbi]
	    Subsequences which is exist in str1 but not on str2.
	   --------------------------------------------------------
	   Ans = 1  [length of f or b]
	*/
	uncommonSubsequence(str1, str2)
}

Output

 String A :  aecb
 String B :  bace
  2
 String A :  ABCCDBE
 String B :  ABCCDBE
  0
 String A :  CCCCCB
 String B :  CCB
  3
 String A :  fbi
 String B :  ice
  1

<?php
/*
    Php program for
    Shortest uncommon subsequence using dynamic programming
*/
class Subsequence
{
	public	function minValue($a, $b)
	{
		if ($a < $b)
		{
			return $a;
		}
		return $b;
	}
	// Find the length of shortest uncommon subsequence which is exist
	// In str1 but not in str2.
	public	function uncommonSubsequence($str1, $str2)
	{
		// Get the length
		$m = strlen($str1);
		$n = strlen($str2);
		// Auxiliary space
		$dp = array_fill(0, $m + 1, array_fill(0, $n + 1, 0));
		$result = 0;
		$max = PHP_INT_MAX;
		// Reduce size of max by length of longest substring
		if ($n > $m)
		{
			$max -= $n;
		}
		else
		{
			$max -= $m;
		}
		for ($i = 0; $i <= $m; $i++)
		{
			$dp[$i][0] = 1;
		}
		for ($i = 0; $i <= $n; $i++)
		{
			$dp[0][$i] = $max;
		}
		for ($i = 1; $i <= $m; $i++)
		{
			for ($j = 1; $j <= $n; $j++)
			{
				$k = $j - 1;
				while ($k >= 0 && $str2[$k] != $str1[$i - 1])
				{
					$k--;
				}
				if ($k == -1)
				{
					$dp[$i][$j] = 1;
				}
				else
				{
					$dp[$i][$j] = $this->minValue($dp[$i - 1][$j], $dp[$i - 1][$k] + 1);
				}
			}
		}
		if ($dp[$m][$n] != $max)
		{
			$result = $dp[$m][$n];
		}
		// Display given strings
		echo(" String A : ".$str1.
			"\n");
		echo(" String B : ".$str2.
			"\n");
		// Display calculated result
		echo(" ".$result.
			"\n");
	}
}

function main()
{
	$task = new Subsequence();
	$str1 = "aecb";
	$str2 = "bace";
	// Case A
	/*
	   Example 1
	   str1 = aecb
	   str2 = bace
	   ------------
	    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
	            str1 but not str2
	    --------------------------------------------
	    Length of shortest subsequence is 2
	    Ans = 2
	*/
	$task->uncommonSubsequence($str1, $str2);
	// Case B
	$str1 = "ABCCDBE";
	$str2 = "ABCCDBE";
	/*
	   Example 2
	   
	   str1 = ABCCDBE
	   str2 = ABCCDBE
	   ------------
	   all subsequence str1 exist in str2
	*/
	$task->uncommonSubsequence($str1, $str2);
	// Case C
	$str1 = "CCCCCB";
	$str2 = "CCB";
	/*
	   Example 3
	   
	   str1 = CCCCCB
	   str2 = CCB
	   ------------
	   'CCC' Shortest subsequence which exist in str1 but not on str2.
	   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
	   but its length more than 3.
	   --------------------------------------------------------
	   Ans = 3
	*/
	$task->uncommonSubsequence($str1, $str2);
	// Case D
	$str1 = "fbi";
	$str2 = "ice";
	/*
	   Example 4
	   
	   str1 = fbi
	   str2 = ice
	   ------------
	    [f,b,fb,fbi]
	    Subsequences which is exist in str1 but not on str2.
	   --------------------------------------------------------
	   Ans = 1  [length of f or b]
	*/
	$task->uncommonSubsequence($str1, $str2);
}
main();

Output

 String A : aecb
 String B : bace
 2
 String A : ABCCDBE
 String B : ABCCDBE
 0
 String A : CCCCCB
 String B : CCB
 3
 String A : fbi
 String B : ice
 1
/*
    Node JS program for
    Shortest uncommon subsequence using dynamic programming
*/
class Subsequence
{
	minValue(a, b)
	{
		if (a < b)
		{
			return a;
		}
		return b;
	}
	// Find the length of shortest uncommon subsequence which is exist
	// In str1 but not in str2.
	uncommonSubsequence(str1, str2)
	{
		// Get the length
		var m = str1.length;
		var n = str2.length;
		// Auxiliary space
		var dp = Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
		var result = 0;
		var max = Number.MAX_VALUE;
		// Reduce size of max by length of longest substring
		if (n > m)
		{
			max -= n;
		}
		else
		{
			max -= m;
		}
		for (var i = 0; i <= m; i++)
		{
			dp[i][0] = 1;
		}
		for (var i = 0; i <= n; i++)
		{
			dp[0][i] = max;
		}
		for (var i = 1; i <= m; i++)
		{
			for (var j = 1; j <= n; j++)
			{
				var k = j - 1;
				while (k >= 0 && str2.charAt(k) != str1.charAt(i - 1))
				{
					k--;
				}
				if (k == -1)
				{
					dp[i][j] = 1;
				}
				else
				{
					dp[i][j] = this.minValue(dp[i - 1][j], dp[i - 1][k] + 1);
				}
			}
		}
		if (dp[m][n] != max)
		{
			result = dp[m][n];
		}
		// Display given strings
		console.log(" String A : " + str1);
		console.log(" String B : " + str2);
		// Display calculated result
		console.log(" " + result);
	}
}

function main()
{
	var task = new Subsequence();
	var str1 = "aecb";
	var str2 = "bace";
	// Case A
	/*
	   Example 1
	   str1 = aecb
	   str2 = bace
	   ------------
	    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
	            str1 but not str2
	    --------------------------------------------
	    Length of shortest subsequence is 2
	    Ans = 2
	*/
	task.uncommonSubsequence(str1, str2);
	// Case B
	str1 = "ABCCDBE";
	str2 = "ABCCDBE";
	/*
	   Example 2
	   
	   str1 = ABCCDBE
	   str2 = ABCCDBE
	   ------------
	   all subsequence str1 exist in str2
	*/
	task.uncommonSubsequence(str1, str2);
	// Case C
	str1 = "CCCCCB";
	str2 = "CCB";
	/*
	   Example 3
	   
	   str1 = CCCCCB
	   str2 = CCB
	   ------------
	   'CCC' Shortest subsequence which exist in str1 but not on str2.
	   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
	   but its length more than 3.
	   --------------------------------------------------------
	   Ans = 3
	*/
	task.uncommonSubsequence(str1, str2);
	// Case D
	str1 = "fbi";
	str2 = "ice";
	/*
	   Example 4
	   
	   str1 = fbi
	   str2 = ice
	   ------------
	    [f,b,fb,fbi]
	    Subsequences which is exist in str1 but not on str2.
	   --------------------------------------------------------
	   Ans = 1  [length of f or b]
	*/
	task.uncommonSubsequence(str1, str2);
}
main();

Output

 String A : aecb
 String B : bace
 2
 String A : ABCCDBE
 String B : ABCCDBE
 0
 String A : CCCCCB
 String B : CCB
 3
 String A : fbi
 String B : ice
 1
import sys
#    Python 3 program for
#    Shortest uncommon subsequence using dynamic programming
class Subsequence :
	def minValue(self, a, b) :
		if (a < b) :
			return a
		
		return b
	
	#  Find the length of shortest uncommon subsequence which is exist
	#  In str1 but not in str2.
	def uncommonSubsequence(self, str1, str2) :
		#  Get the length
		m = len(str1)
		n = len(str2)
		#  Auxiliary space
		dp = [[0] * (n + 1) for _ in range(m + 1) ]
		result = 0
		max = sys.maxsize
		#  Reduce size of max by length of longest substring
		if (n > m) :
			max -= n
		else :
			max -= m
		
		i = 0
		while (i <= m) :
			dp[i][0] = 1
			i += 1
		
		i = 0
		while (i <= n) :
			dp[0][i] = max
			i += 1
		
		i = 1
		while (i <= m) :
			j = 1
			while (j <= n) :
				k = j - 1
				while (k >= 0 and str2[k] != str1[i - 1]) :
					k -= 1
				
				if (k == -1) :
					dp[i][j] = 1
				else :
					dp[i][j] = self.minValue(dp[i - 1][j], dp[i - 1][k] + 1)
				
				j += 1
			
			i += 1
		
		if (dp[m][n] != max) :
			result = dp[m][n]
		
		#  Display given strings
		print(" String A : ", str1)
		print(" String B : ", str2)
		#  Display calculated result
		print(" ", result)
	

def main() :
	task = Subsequence()
	str1 = "aecb"
	str2 = "bace"
	#  Case A
	#   Example 1
	#   str1 = aecb
	#   str2 = bace
	#   ------------
	#    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
	#            str1 but not str2
	#    --------------------------------------------
	#    Length of shortest subsequence is 2
	#    Ans = 2
	task.uncommonSubsequence(str1, str2)
	#  Case B
	str1 = "ABCCDBE"
	str2 = "ABCCDBE"
	#   Example 2
	#   str1 = ABCCDBE
	#   str2 = ABCCDBE
	#   ------------
	#   all subsequence str1 exist in str2
	task.uncommonSubsequence(str1, str2)
	#  Case C
	str1 = "CCCCCB"
	str2 = "CCB"
	#   Example 3
	#   str1 = CCCCCB
	#   str2 = CCB
	#   ------------
	#   'CCC' Shortest subsequence which exist in str1 but not on str2.
	#   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
	#   but its length more than 3.
	#   --------------------------------------------------------
	#   Ans = 3
	task.uncommonSubsequence(str1, str2)
	#  Case D
	str1 = "fbi"
	str2 = "ice"
	#   Example 4
	#   str1 = fbi
	#   str2 = ice
	#   ------------
	#    [f,b,fb,fbi]
	#    Subsequences which is exist in str1 but not on str2.
	#   --------------------------------------------------------
	#   Ans = 1  [length of f or b]
	task.uncommonSubsequence(str1, str2)

if __name__ == "__main__": main()

Output

 String A :  aecb
 String B :  bace
  2
 String A :  ABCCDBE
 String B :  ABCCDBE
  0
 String A :  CCCCCB
 String B :  CCB
  3
 String A :  fbi
 String B :  ice
  1
#    Ruby program for
#    Shortest uncommon subsequence using dynamic programming
class Subsequence 
	def minValue(a, b) 
		if (a < b) 
			return a
		end

		return b
	end

	#  Find the length of shortest uncommon subsequence which is exist
	#  In str1 but not in str2.
	def uncommonSubsequence(str1, str2) 
		#  Get the length
		m = str1.length
		n = str2.length
		#  Auxiliary space
		dp = Array.new(m + 1) {Array.new(n + 1) {0}}
		result = 0
		max = (2 ** (0. size * 8 - 2))
		#  Reduce size of max by length of longest substring
		if (n > m) 
			max -= n
		else
 
			max -= m
		end

		i = 0
		while (i <= m) 
			dp[i][0] = 1
			i += 1
		end

		i = 0
		while (i <= n) 
			dp[0][i] = max
			i += 1
		end

		i = 1
		while (i <= m) 
			j = 1
			while (j <= n) 
				k = j - 1
				while (k >= 0 && str2[k] != str1[i - 1]) 
					k -= 1
				end

				if (k == -1) 
					dp[i][j] = 1
				else
 
					dp[i][j] = self.minValue(
                      dp[i - 1][j], dp[i - 1][k] + 1
                    )
				end

				j += 1
			end

			i += 1
		end

		if (dp[m][n] != max) 
			result = dp[m][n]
		end

		#  Display given strings
		print(" String A : ", str1, "\n")
		print(" String B : ", str2, "\n")
		#  Display calculated result
		print(" ", result, "\n")
	end

end

def main() 
	task = Subsequence.new()
	str1 = "aecb"
	str2 = "bace"
	#  Case A
	#   Example 1
	#   str1 = aecb
	#   str2 = bace
	#   ------------
	#    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
	#            str1 but not str2
	#    --------------------------------------------
	#    Length of shortest subsequence is 2
	#    Ans = 2
	task.uncommonSubsequence(str1, str2)
	#  Case B
	str1 = "ABCCDBE"
	str2 = "ABCCDBE"
	#   Example 2
	#   str1 = ABCCDBE
	#   str2 = ABCCDBE
	#   ------------
	#   all subsequence str1 exist in str2
	task.uncommonSubsequence(str1, str2)
	#  Case C
	str1 = "CCCCCB"
	str2 = "CCB"
	#   Example 3
	#   str1 = CCCCCB
	#   str2 = CCB
	#   ------------
	#   'CCC' Shortest subsequence which exist in str1 but not on str2.
	#   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
	#   but its length more than 3.
	#   --------------------------------------------------------
	#   Ans = 3
	task.uncommonSubsequence(str1, str2)
	#  Case D
	str1 = "fbi"
	str2 = "ice"
	#   Example 4
	#   str1 = fbi
	#   str2 = ice
	#   ------------
	#    [f,b,fb,fbi]
	#    Subsequences which is exist in str1 but not on str2.
	#   --------------------------------------------------------
	#   Ans = 1  [length of f or b]
	task.uncommonSubsequence(str1, str2)
end

main()

Output

 String A : aecb
 String B : bace
 2
 String A : ABCCDBE
 String B : ABCCDBE
 0
 String A : CCCCCB
 String B : CCB
 3
 String A : fbi
 String B : ice
 1

/*
    Scala program for
    Shortest uncommon subsequence using dynamic programming
*/
class Subsequence()
{
	def minValue(a: Int, b: Int): Int = {
		if (a < b)
		{
			return a;
		}
		return b;
	}
	// Find the length of shortest uncommon subsequence which is exist
	// In str1 but not in str2.
	def uncommonSubsequence(str1: String, str2: String): Unit = {
		// Get the length
		var m: Int = str1.length();
		var n: Int = str2.length();
		// Auxiliary space
		var dp: Array[Array[Int]] = Array.fill[Int](m + 1, n + 1)(0);
		var result: Int = 0;
		var max: Int = Int.MaxValue;
		// Reduce size of max by length of longest substring
		if (n > m)
		{
			max -= n;
		}
		else
		{
			max -= m;
		}
		var i: Int = 0;
		while (i <= m)
		{
			dp(i)(0) = 1;
			i += 1;
		}
		i = 0;
		while (i <= n)
		{
			dp(0)(i) = max;
			i += 1;
		}
		i = 1;
		while (i <= m)
		{
			var j: Int = 1;
			while (j <= n)
			{
				var k: Int = j - 1;
				while (k >= 0 && str2.charAt(k) != str1.charAt(i - 1))
				{
					k -= 1;
				}
				if (k == -1)
				{
					dp(i)(j) = 1;
				}
				else
				{
					dp(i)(j) = minValue(dp(i - 1)(j), dp(i - 1)(k) + 1);
				}
				j += 1;
			}
			i += 1;
		}
		if (dp(m)(n) != max)
		{
			result = dp(m)(n);
		}
		// Display given strings
		println(" String A : " + str1);
		println(" String B : " + str2);
		// Display calculated result
		println(" " + result);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Subsequence = new Subsequence();
		var str1: String = "aecb";
		var str2: String = "bace";
		// Case A
		/*
		   Example 1
		   str1 = aecb
		   str2 = bace
		   ------------
		    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
		            str1 but not str2
		    --------------------------------------------
		    Length of shortest subsequence is 2
		    Ans = 2
		*/
		task.uncommonSubsequence(str1, str2);
		// Case B
		str1 = "ABCCDBE";
		str2 = "ABCCDBE";
		/*
		   Example 2
		   
		   str1 = ABCCDBE
		   str2 = ABCCDBE
		   ------------
		   all subsequence str1 exist in str2
		*/
		task.uncommonSubsequence(str1, str2);
		// Case C
		str1 = "CCCCCB";
		str2 = "CCB";
		/*
		   Example 3
		   
		   str1 = CCCCCB
		   str2 = CCB
		   ------------
		   'CCC' Shortest subsequence which exist in str1 but not on str2.
		   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
		   but its length more than 3.
		   --------------------------------------------------------
		   Ans = 3
		*/
		task.uncommonSubsequence(str1, str2);
		// Case D
		str1 = "fbi";
		str2 = "ice";
		/*
		   Example 4
		   
		   str1 = fbi
		   str2 = ice
		   ------------
		    [f,b,fb,fbi]
		    Subsequences which is exist in str1 but not on str2.
		   --------------------------------------------------------
		   Ans = 1  [length of f or b]
		*/
		task.uncommonSubsequence(str1, str2);
	}
}

Output

 String A : aecb
 String B : bace
 2
 String A : ABCCDBE
 String B : ABCCDBE
 0
 String A : CCCCCB
 String B : CCB
 3
 String A : fbi
 String B : ice
 1
import Foundation;
/*
    Swift 4 program for
    Shortest uncommon subsequence using dynamic programming
*/
class Subsequence
{
	func minValue(_ a: Int, _ b: Int) -> Int
	{
		if (a < b)
		{
			return a;
		}
		return b;
	}
	// Find the length of shortest uncommon subsequence which is exist
	// In str1 but not in str2.
	func uncommonSubsequence(_ s1: String, _ s2: String)
	{
      	let str1 = Array(s1);
        let str2 = Array(s2);
		// Get the length
		let m: Int = str1.count;
		let n: Int = str2.count;
		// Auxiliary space
		var dp: [
			[Int]
		] = Array(
          repeating: Array(repeating: 0, count: n + 1), count: m + 1
      	);
		var result: Int = 0;
		var max: Int = Int.max;
		// Reduce size of max by length of longest substring
		if (n > m)
		{
			max -= n;
		}
		else
		{
			max -= m;
		}
		var i: Int = 0;
		while (i <= m)
		{
			dp[i][0] = 1;
			i += 1;
		}
		i = 0;
		while (i <= n)
		{
			dp[0][i] = max;
			i += 1;
		}
		i = 1;
		while (i <= m)
		{
			var j: Int = 1;
			while (j <= n)
			{
				var k: Int = j - 1;
				while (k >= 0 && str2[k]  != str1[i - 1])
				{
					k -= 1;
				}
				if (k == -1)
				{
					dp[i][j] = 1;
				}
				else
				{
					dp[i][j] = self.minValue(
                      dp[i - 1][j], dp[i - 1][k] + 1
                    );
				}
				j += 1;
			}
			i += 1;
		}
		if (dp[m][n]  != max)
		{
			result = dp[m][n];
		}
		// Display given strings
		print(" String A : ", s1);
		print(" String B : ", s2);
		// Display calculated result
		print(" ", result);
	}
}
func main()
{
	let task: Subsequence = Subsequence();
	var str1: String = "aecb";
	var str2: String = "bace";
	// Case A
	/*
	   Example 1
	   str1 = aecb
	   str2 = bace
	   ------------
	    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
	            str1 but not str2
	    --------------------------------------------
	    Length of shortest subsequence is 2
	    Ans = 2
	*/
	task.uncommonSubsequence(str1, str2);
	// Case B
	str1 = "ABCCDBE";
	str2 = "ABCCDBE";
	/*
	   Example 2
	   
	   str1 = ABCCDBE
	   str2 = ABCCDBE
	   ------------
	   all subsequence str1 exist in str2
	*/
	task.uncommonSubsequence(str1, str2);
	// Case C
	str1 = "CCCCCB";
	str2 = "CCB";
	/*
	   Example 3
	   
	   str1 = CCCCCB
	   str2 = CCB
	   ------------
	   'CCC' Shortest subsequence which exist in str1 but not on str2.
	   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
	   but its length more than 3.
	   --------------------------------------------------------
	   Ans = 3
	*/
	task.uncommonSubsequence(str1, str2);
	// Case D
	str1 = "fbi";
	str2 = "ice";
	/*
	   Example 4
	   
	   str1 = fbi
	   str2 = ice
	   ------------
	    [f,b,fb,fbi]
	    Subsequences which is exist in str1 but not on str2.
	   --------------------------------------------------------
	   Ans = 1  [length of f or b]
	*/
	task.uncommonSubsequence(str1, str2);
}
main();

Output

 String A :  aecb
 String B :  bace
  2
 String A :  ABCCDBE
 String B :  ABCCDBE
  0
 String A :  CCCCCB
 String B :  CCB
  3
 String A :  fbi
 String B :  ice
  1
/*
    Kotlin program for
    Shortest uncommon subsequence using dynamic programming
*/
class Subsequence
{
	fun minValue(a: Int, b: Int): Int
	{
		if (a < b)
		{
			return a;
		}
		return b;
	}
	// Find the length of shortest uncommon subsequence which is exist
	// In str1 but not in str2.
	fun uncommonSubsequence(str1: String, str2: String): Unit
	{
		// Get the length
		val m: Int = str1.length;
		val n: Int = str2.length;
		// Auxiliary space
		var dp: Array < Array < Int >> = Array(m + 1)
		{
			Array(n + 1)
			{
				0
			}
		};
		var result: Int = 0;
		var max: Int = Int.MAX_VALUE;
		// Reduce size of max by length of longest substring
		if (n > m)
		{
			max -= n;
		}
		else
		{
			max -= m;
		}
		var i: Int = 0;
		while (i <= m)
		{
			dp[i][0] = 1;
			i += 1;
		}
		i = 0;
		while (i <= n)
		{
			dp[0][i] = max;
			i += 1;
		}
		i = 1;
		while (i <= m)
		{
			var j: Int = 1;
			while (j <= n)
			{
				var k: Int = j - 1;
				while (k >= 0 && str2.get(k) != str1.get(i - 1))
				{
					k -= 1;
				}
				if (k == -1)
				{
					dp[i][j] = 1;
				}
				else
				{
					dp[i][j] = this.minValue(dp[i - 1][j], dp[i - 1][k] + 1);
				}
				j += 1;
			}
			i += 1;
		}
		if (dp[m][n] != max)
		{
			result = dp[m][n];
		}
		// Display given strings
		println(" String A : " + str1);
		println(" String B : " + str2);
		// Display calculated result
		println(" " + result);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Subsequence = Subsequence();
	var str1: String = "aecb";
	var str2: String = "bace";
	// Case A
	/*
	   Example 1
	   str1 = aecb
	   str2 = bace
	   ------------
	    [ab,cb,eb,aecb,acb,ecb] Subsequences exists in 
	            str1 but not str2
	    --------------------------------------------
	    Length of shortest subsequence is 2
	    Ans = 2
	*/
	task.uncommonSubsequence(str1, str2);
	// Case B
	str1 = "ABCCDBE";
	str2 = "ABCCDBE";
	/*
	   Example 2
	   
	   str1 = ABCCDBE
	   str2 = ABCCDBE
	   ------------
	   all subsequence str1 exist in str2
	*/
	task.uncommonSubsequence(str1, str2);
	// Case C
	str1 = "CCCCCB";
	str2 = "CCB";
	/*
	   Example 3
	   
	   str1 = CCCCCB
	   str2 = CCB
	   ------------
	   'CCC' Shortest subsequence which exist in str1 but not on str2.
	   Note ['CCCC',CCCCCB,CCCCB] is also uncommon subsequence 
	   but its length more than 3.
	   --------------------------------------------------------
	   Ans = 3
	*/
	task.uncommonSubsequence(str1, str2);
	// Case D
	str1 = "fbi";
	str2 = "ice";
	/*
	   Example 4
	   
	   str1 = fbi
	   str2 = ice
	   ------------
	    [f,b,fb,fbi]
	    Subsequences which is exist in str1 but not on str2.
	   --------------------------------------------------------
	   Ans = 1  [length of f or b]
	*/
	task.uncommonSubsequence(str1, str2);
}

Output

 String A : aecb
 String B : bace
 2
 String A : ABCCDBE
 String B : ABCCDBE
 0
 String A : CCCCCB
 String B : CCB
 3
 String A : fbi
 String B : ice
 1

Explanation

Let's go through the code and understand how it solves the problem.

In the uncommonSubsequence method, we start by initializing the variables and the 2D array dp. The variable max is set to the maximum integer value, and then it is reduced by the length of the longest substring between str1 and str2. This is done to ensure that the length of the uncommon subsequence is not greater than the length of str1 or str2.

We then initialize the first row of dp to 1 because an empty string is always a subsequence of any string. The first column is initialized to max because an empty str2 would make any non-empty subsequence of str1 uncommon.

Next, we iterate over the remaining cells of dp and fill them based on the rules mentioned in the algorithm. We use the minValue method to get the minimum value between two integers.

If the characters at the current positions in str1 and str2 are the same, we set the current cell value to the minimum of the previous row value (dp[i-1][j]) and the previous diagonal cell value plus one (dp[i-1][k] + 1), where k is the last occurrence of the character in str2.

If the characters are different, we search for the last occurrence of the character at the current position in str1 in str2. If it exists, we set the current cell value to the minimum of the previous row value and the value at the last occurrence position plus one. Otherwise, we set it to 1.

After filling all the cells, we check if the bottom-right cell of dp is equal to max. If it is not, it means there is an uncommon subsequence, and we assign its value to the result variable.

Finally, we display the given strings str1 and str2, as well as the calculated result.

The output of the code matches the expected results for the given test cases.

Time Complexity

The time complexity of this algorithm is O(m * n), where m and n are the lengths of str1 and str2 respectively. This is because we have a nested loop that iterates over the cells of the dp array.

As for the space complexity, it is O(m * n) as well since we use a 2D array of the same size as the input strings.

Conclusion

In this article, we discussed the problem of finding the length of the shortest uncommon subsequence using dynamic programming. We explained the problem statement, presented an algorithm, and provided a step-by-step solution. We also analyzed the time complexity of the code. Dynamic programming is a powerful technique for solving optimization problems, and understanding how it can be applied to different scenarios helps us become better problem solvers.

Last updated on by Kalkicode




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