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Java Program for Check if given string can be formed by two other strings or their permutations

  • Last Updated : 24 Jan, 2022

Given a string str and an array of strings arr[], the task is to check if the given string can be formed by any of the string pair from the array or their permutations.

Examples:

Input: str = “amazon”, arr[] = {“loa”, “azo”, “ft”, “amn”, “lka”}
Output: Yes
The chosen strings are “amn” and “azo”
which can be rearranged as “amazon”.

Input: str = “geeksforgeeks”, arr[] = {“geeks”, “geek”, “for”}
Output: No

Java




// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
// Function that returns true if str can be
// generated from any permutation of the
// two strings selected from the given vector
static boolean isPossible(Vector<String> v, String str)
{
  
    // Sort the given string
    str = sortString(str);
  
    // Select two strings at a time from given vector
    for (int i = 0; i < v.size() - 1; i++) 
    {
        for (int j = i + 1; j < v.size(); j++)
        {
  
            // Get the concatenated string
            String temp = v.get(i) + v.get(j);
  
            // Sort the resultant string
            temp = sortString(temp);
  
            // If the resultant string is equal
            // to the given string str
            if (temp.compareTo(str) == 0
            {
                return true;
            }
        }
    }
  
    // No valid pair found
    return false;
}
  
// Method to sort a string alphabetically 
public static String sortString(String inputString) 
    // convert input string to char array 
    char tempArray[] = inputString.toCharArray(); 
      
    // sort tempArray 
    Arrays.sort(tempArray); 
      
    // return new sorted string 
    return new String(tempArray); 
  
// Driver code
public static void main(String[] args) 
{
    String str = "amazon";
    String []arr = { "fds", "oxq", "zoa", "epw", "amn" };
    Vector<String> v = new Vector<String>(Arrays.asList(arr));
  
    if (isPossible(v, str))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by Rajput-Ji


Method 2: Counting sort can be used to reduce the running time of the above approach. Counting sort uses a table to store the count of each character. We have 26 alphabets hence we make an array of size 26 to store counts of each character in the string. Then take the characters in increasing order to get the sorted string.

Below is the implementation of the above approach:

Java




// Java implementation of the approach 
import java.util.*;
  
class GFG 
{
static int MAX = 26
  
// Function to sort the given string 
// using counting sort 
static String countingsort(char[] s) 
      
    // Array to store the count of each character 
    int []count = new int[MAX]; 
    for (int i = 0; i < s.length; i++) 
    
        count[s[i] - 'a']++; 
    
    int index = 0
  
    // Insert characters in the string 
    // in increasing order 
    for (int i = 0; i < MAX; i++)
    
        int j = 0
        while (j < count[i]) 
        
            s[index++] = (char)(i + 'a'); 
            j++; 
        
    
        return String.valueOf(s);
  
// Function that returns true if str can be 
// generated from any permutation of the 
// two strings selected from the given vector 
static boolean isPossible(Vector<String> v, 
                                 String str) 
  
    // Sort the given string 
    str=countingsort(str.toCharArray()); 
  
    // Select two strings at a time from given vector 
    for (int i = 0; i < v.size() - 1; i++) 
    
        for (int j = i + 1; j < v.size(); j++) 
        
  
            // Get the concatenated string 
            String temp = v.get(i) + v.get(j); 
  
            // Sort the resultant string 
            temp = countingsort(temp.toCharArray()); 
  
            // If the resultant string is equal 
            // to the given string str 
            if (temp.equals(str)) 
            
                return true
            
        
    
  
    // No valid pair found 
    return false
  
// Driver code 
public static void main(String[] args) 
{
    String str = "amazon"
    String []arr = { "fds", "oxq", "zoa", "epw", "amn" };
    Vector<String> v = new Vector<String>(Arrays.asList(arr));
  
    if (isPossible(v, str)) 
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by 29AjayKumar


Please refer complete article on Check if given string can be formed by two other strings or their permutations for more details!

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