Question 7 : Find all substrings of String in java?

Solution: Java program to find all substrings of a String.

For example: If the input is “abb”  then the output should be “a”, “b”,”b”, “ab”, “bb”, “abb”

We will use the String class’s subString method to find all substrings.

We will use the String class’s subString method to find all subString

Program:

public class SubstringsOfStringMain
{
 public static void main(String args[]) {
   String str="abbc";
   System.out.println("All substring of abbc are:");
   for (int i = 0; i < str.length(); i++) {
        for (int j = i+1; j <= str.length(); j++) {
              System.out.println(str.substring(i,j));
        }
    }

  Set<String> substrings = new HashSet<>();
  generateSubstrings(str, 0, "", substrings);

  // Print unique substrings
  for (String substring : substrings) {
            System.out.println(substring);
  }
  }

 static void generateSubstrings(String str, int index, String current, Set<String>   substrings) {
        if (index == str.length()) {
            substrings.add(current); // Add the current substring
            return;
        }

        // Include the current character
        generateSubstrings(str, index + 1, current +                       str.charAt(index), substrings);

        // Exclude the current character
        generateSubstrings(str, index + 1, current, substrings);
    }
}

When you run the above program, you will get the following output:

All substring of abbc are:
a
ab
abb
abbc
b
bb
bbc
b
bc
c
 
The above solution is of o(n^3) time complexity. As we have two loops and also String’s substring method has a time complexity of o(n). 

If you want to find all distinct substrings of String, then use HashSet to remove duplicates.

Python code

def find_all_substrings(s: str):
    result = []
    n = len(s)
    for i in range(n):
        for j in range(i + 1, n + 1):
            result.append(s[i:j])
    return result

if __name__ == "__main__":
    for sub in find_all_substrings("abc"):
        print(sub)

📌 Final Comparison

Approach	Time Complexity	Space Complexity	Best Use
Nested loops with substring	O(n²)	O(n²)	⭐ Best general approach
Using recursion (build substrings)	O(n²)	O(n²)	✔ Works, but more complex
Suffix tree (advanced)	O(n²) lower bound*	O(n)	⚠ Complex, not needed here