Minimum Insertion to Form a Palindrome in Java
Java dynamic programming technique is a relevant computational method for finding out the number of insertions required to create a palindrome from a given string.
Palindrome: A group of characters that reads the same from both forward and backward is called as Palindrome. Words like "Madam," "Level," "Mom," "Dad," and "Civic" are examples of palindromes sense they do not change while read backward.
Recursive Approach:
The problem can be solved using the recursive approach to finding a minimum insertion to make a palindrome, which divides the problem into smaller subproblems and solves them recursively. Eventually, using the smaller recursive calls and iterative decision-making, the solution is found by the Recursive Approach.
The recursive functions occur when the left and right index intersect or meet. At this point, the method returns 0, and no more insertions are needed. There is no need to insert any characters if the characters at the left and right index are the same. The function calls itself recursively the usage updated indexes (left 1 and right - 1) to verify the inner substring. A character is inserted at the beginning (left + 1 and right). Adding a character (left and right - 1) at the end. The minimum insertion required at the moment is represented with the help of adding 1 to the minimum of each of these choices.
Filename: MinimumInsertionsPalindrome.java
public class MinimumInsertionsPalindrome
{
static int insertions(String str) // Minimum number of insertions by recursive function
{
return x(str, 0, str.length() - 1);
}
static int x(String str, int intialize, int end)
{
if (intialize >= end) // base case
{
return 0;
}
if (str.charAt(intialize) == str.charAt(end))
{
return x(str, intialize + 1, end - 1);
}
else
{
return 1 + Math.min(x(str, intialize + 1, end), x(str, intialize, end - 1)); //check wheather the first and last characters are same
}
}
public static void main(String[] args)
{
String str = "divider";
String str1 = "rotator";
int MinimumInsertions = insertions(str);
int MinimumInsertions1 = insertions(str1);
System.out.println("Minimum Insertion to Form a Palindrome is: " + MinimumInsertions);
System.out.println("Minimum Insertion to Form a Palindrome is: " + MinimumInsertions1);
}
}
Output:
Minimum Insertion to Form a Palindrome is: 2
Minimum Insertion to Form a Palindrome is: 0
Time Complexity:
Without optimization, the worst-case time complexity is exponential, or O(2^n), where 'n' is the input string length. The algorithm's recursive structure and possible overlapping subproblems lead to this exponential time complexity.
Space Complexity:
The recursive stack has an O(n) space complexity.
Note: Recursion provides an easy and simple approach to handling the problems. Due to the possibility of repeated computations, it may not be the most effective approach for longer sequences. By decreasing unnecessary computations, using memoization or dynamic programming methods can improve efficiency.
Dynamic Programming:
Using an effective algorithmic approach called Dynamic Programming, problems can be solved by dividing them into simpler, less overlapping subproblems and storing the solutions to the subproblems to prevent repeating calculations. When used in finding the minimum number of insertions needed to create a palindrome in Java from a given string, dynamic programming effectively computes and stores solutions for smaller issues in order to solve the major problem. The approach frequently creates a table of minimum insertions needed to make a palindrome for different substrings within the provided string using a bottom-up dynamic programming technique.
Filename: DynamicProgramming.java
import java.util.*;
public class DynamicProgramming
{
public static void main(String[] args)
{
String str = "divider";
int x = str.length();
int[][] dp = new int[x][x]; // for finding the minimum number of insertions, it is used dp function
for(int i = 0; i < dp.length; i++) // finding the table
{
for(int j = 0; j < dp[0].length; j++)
{
dp[i][j] = -1;
}
}
int y = getMinInsertion(str, 0, x-1, dp);
System.out.println("Minimum Insertion to Form a Palindrome using DynamicProgramming is: " + dp[0][x-1]);
}
public static int getMinInsertion(String str, int a, int b, int[][] dp)
{
if(a == b)
return dp[a][b] = 0;
if(dp[a][b]!=-1)
return dp[a][b];
if(str.charAt(a) == str.charAt(b))
return dp[a][b] = getMinInsertion(str, a+1, b-1, dp);
else
{
int result1 = getMinInsertion(str, a+1, b, dp);
int result2 = getMinInsertion(str, a, b-1, dp);
return dp[a][b] = Math.min(result1, result2) + 1;
}
}
}
Output:
Minimum Insertion to Form a Palindrome using DynamicProgramming is: 2
Time Complexity: Even with large input sizes, the dynamic programming method to finding the minimum number of insertions needed to generate a palindrome from a
2D Array:
Using dynamic programming to identify the minimum number of insertions needed to create a palindrome in Java, solutions for subproblems are stored in a 2D array, commonly represented as dp[][]. The array eventually contributes to the larger problems by effectively computing and storing the minimum number of insertions
For quickly computing and storing solutions for substrings, the 2D array is required to find the minimum number of insertions needed to create a palindrome. By efficiently creating solutions for smaller subproblems and using those solutions to resolve large subproblems, it allows for the use of the dynamic programming methodology. In conclusion, it finds the minimum number of insertions required in Java to generate a palindrome from the given string.
Filename: TwoDArray.java
import java.util.*;
public class TwoDArray
{
public static void main(String[] args)
{
String str = "divider";
int a = str.length();
int[][] dp = new int[a][a]; // for finding the minimum number of insertions dp function is used
for(int x = 0; x < a; x++) // the table is being filled
{
for(int start = 0, end = x; end < a; start++, end++)
{
if(x == 0)
dp[start][end] = 0;
else if(x == 1)
dp[start][end] = str.charAt(start) == str.charAt(end)?0:1;
else
{
if(str.charAt(start) == str.charAt(end))
dp[start][end] = dp[start + 1][end - 1];
else
dp[start][end] = 1 + Math.min(dp[start][end - 1], dp[start + 1][end]);
}
}
}
System.out.println("Minimum Insertion to Form a Palindrome using DynamicProgramming is: " + dp[0][a - 1]);
}
}
Output:
Minimum Insertion to Form a Palindrome using DynamicProgramming is: 2
Time Complexity: The minimum number of insertions needed to create a palindrome in Java using a 2D array and dynamic programming has an O(n^2) time complexity, where 'n' is the length of the input string, where the outer loop executes 'n' times. For each new iteration, the inner loop executes 'n' times in the first, 'n-1' times in the second, 'n-2' times in the third, and so on. In terms of complexity, the total number of iterations for the inner loop is about (n * n) / 2, or O(n^2).
Longest Common Subsequence:
When using dynamic programming in Java to find out the minimum number of insertions needed to create a palindrome from a given string, the term "Longest Common Subsequence (LCS)" is important. This technique achieves an effective solution by combining LCS with the basics of dynamic programming.
The longest character sequence that each string and its reversed form have in common is represented by using the LCS. The minimum number of insertions needed to convert the string into a palindrome can be found by reducing the length of the common subsequence from the string's entire length. With the aim of effectively computing the LCS between strings, dynamic programming is regularly utilized. The subproblem solutions are stored in a 2D array, which makes it possible to calculate the LCS length frequently.
Filename: LongestCommonSubsequence.java
import java.util.*;
public class LongestCommonSubsequence
{
public static void main(String[] args)
{
String str = "divider";
int a = str.length();
int b = getLCS(str,reverse(str));
int x = a - b;
System.out.println("Minimum Insertion to Form a Palindrome using DynamicProgramming is: " + x);
}
public static String reverse(String str)
{
String result = "";
for(int i = str.length()-1; i >= 0; i--)
{
char ch = str.charAt(i);
result+=ch;
}
return result;
}
public static int getLCS(String str1, String str2)
{
int a = str1.length();
int c = str2.length();
int[][] dp = new int[a + 1][c + 1];
for(int i = 1; i <= a; i++)
{
for(int j = 1; j <= c; j++)
{
if(str1.charAt(i-1) == str2.charAt(j-1))
dp[i][j] = dp[i-1][j-1]+1;
else
dp[i][j] = Math.max(dp[i-1][j],dp[i][j-1]);
}
}
return dp[a][c];
}
}
Output:
Minimum Insertion to Form a Palindrome using DynamicProgramming is: 2
Time Complexity: When finding out how few insertions are required to create a palindrome from a given string in Java, the time complexity for obtaining the Longest Common Subsequence (LCS) using dynamic programming is O(n^2), where n is the length of the input string.
Conclusion
The optimal strategy is decided with the help of the specific requirements, limitations, and input sizes. Given possible space issues, dynamic programming, with a wide range of optimization approaches, frequently provides scalable and effective solutions for larger inputs. When selecting a solution for the Java problem of finding the minimum number of insertions required to provide a palindrome, it is important to understand the conflicts among time complexity, space issues, and implementation overhead.