# 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
*/
// Case B
str1 = "ABCCDBE";
str2 = "ABCCDBE";
/*
Example 2

str1 = ABCCDBE
str2 = ABCCDBE
------------
all subsequence str1 exist in 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
*/
// 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]
*/
}
}``````

#### 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
*/
// Case B
str1 = "ABCCDBE";
str2 = "ABCCDBE";
/*
Example 2

str1 = ABCCDBE
str2 = ABCCDBE
------------
all subsequence str1 exist in 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
*/
// 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]
*/
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
*/
// Case B
str1 = "ABCCDBE";
str2 = "ABCCDBE";
/*
Example 2

str1 = ABCCDBE
str2 = ABCCDBE
------------
all subsequence str1 exist in 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
*/
// 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]
*/
}
}``````

#### 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
*/
// Case B
\$str1 = "ABCCDBE";
\$str2 = "ABCCDBE";
/*
Example 2

str1 = ABCCDBE
str2 = ABCCDBE
------------
all subsequence str1 exist in 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
*/
// 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]
*/
}
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
*/
// Case B
str1 = "ABCCDBE";
str2 = "ABCCDBE";
/*
Example 2

str1 = ABCCDBE
str2 = ABCCDBE
------------
all subsequence str1 exist in 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
*/
// 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]
*/
}
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() :
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
#  Case B
str1 = "ABCCDBE"
str2 = "ABCCDBE"
#   Example 2
#   str1 = ABCCDBE
#   str2 = ABCCDBE
#   ------------
#   all subsequence str1 exist in 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
#  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]

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()
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
#  Case B
str1 = "ABCCDBE"
str2 = "ABCCDBE"
#   Example 2
#   str1 = ABCCDBE
#   str2 = ABCCDBE
#   ------------
#   all subsequence str1 exist in 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
#  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]
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
*/
// Case B
str1 = "ABCCDBE";
str2 = "ABCCDBE";
/*
Example 2

str1 = ABCCDBE
str2 = ABCCDBE
------------
all subsequence str1 exist in 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
*/
// 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]
*/
}
}``````

#### 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
*/
// Case B
str1 = "ABCCDBE";
str2 = "ABCCDBE";
/*
Example 2

str1 = ABCCDBE
str2 = ABCCDBE
------------
all subsequence str1 exist in 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
*/
// 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]
*/
}
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
*/
// Case B
str1 = "ABCCDBE";
str2 = "ABCCDBE";
/*
Example 2

str1 = ABCCDBE
str2 = ABCCDBE
------------
all subsequence str1 exist in 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
*/
// 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]
*/
}``````

#### 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.

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