检查是否可以使用递归将一个字符串转换为另一个字符串

Question

I am trying to solve the below method by using recursion

https://www.geeksforgeeks.org/check-possible-transform-one-string-another/

    static boolean abbreviation(String a, String b,int m, int n) {
    
    char[] x = a.toCharArray();
   
  
    if(m ==0|| n==0)
    return true;

    if(m==0 && n!=0)
    return false;

    if(n==0){
        for(int i=0; i < x.length; i++){
            
            //if any character is not in lower case, return false
            if( Character.isLowerCase( x[i] )){
                System.out.println(x[i]);
            return true;  
        
            }
        }
                
         return false;
    }
    if((a.charAt(m-1)==b.charAt(n-1))||
    (Character.toUpperCase(a.charAt(m-1))==b.charAt(n-1))){
       return abbreviation(a,b,m-1,n-1);
    }else return abbreviation(a,b,m-1,n) &&  abbreviation(a,b,m,n-1);
    

  
}

I am getting true instead of false for below inputs

Input 1

AbCdE

AFE

Input 2

beFgH

EFG

Answer 1

You have 3 cases when traversing characters of two string

• Character matched (if lowercase then doing uppercase)
• Charcter not matched and first string character is lowercase
• Charcter not matched and first string character is uppercase

And base case if all character checked in first string then check all matched or not.

  boolean abbreviation(String a, String b, int m, int n) {
    if (m == 0)
      return n == 0;

    if (n > 0 && Character.toUpperCase(a.charAt(m - 1)) == b.charAt(n - 1)) {
      return abbreviation(a, b, m - 1, n - 1);
    } else if(!Character.isUpperCase(a.charAt(m - 1))) {
      return abbreviation(a, b, m - 1, n);
    } else {
      return false;
    }
  }

Demo:

System.out.println(abbreviation("AbCdE", "AFE", 5, 3));
System.out.println(abbreviation("beFgH", "EFG", 5, 3));
System.out.println(abbreviation("beFgh", "EFG", 5, 3));
System.out.println(abbreviation("daBcd", "ABC", 5, 3));

Output:

false
false
true
true