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Java: How to Check if a String Contains a Substring? Implement Your Own Method

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How to Check if a String Contains a Substring

I have two Strings: str1 and str2. How to check if str2 is contained within str1? You can definitely use Java’s own .contains()  method.

Here by this program we will implement contains() method. Let’s call it as findMe(String subString, String mainString) . Find below implementation.

Result1:

Result2: (notice blank space between find and Me)

Let me know if you know any better way to implement this.

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Comments

  1. package com.maven.practice;

import java.util.Scanner;
public class Test {

	public static void main(String[] args) {
		
		try{
			 Scanner scanner=new Scanner(System.in);
		 	 String mainString=scanner.next();
		 	 String subString=scanner.next();
		 	 scanner.close();
		 	 System.out.println("substring present in main string? : " +containSubString(mainString,subString));
		 }
		 catch(Exception e)
		 {
			 System.out.println("enter valid input");
		 }
		 
	}
	
	public static boolean containSubString(String mainString, String subString)
	{
		int lengthMain=mainString.length();     //length of main string
		int lengthSub=subString.length();        //length of sub string
		char main[]=mainString.toCharArray();     //convert string to char array
		char sub[]=subString.toCharArray();
		int k=0, j=0;
		while(k<lengthMain)         //loop to traverse the main string
		{
			int i=k;
			while(j<lengthSub)       //loop for substring
			{
				if(main[i]!=sub[j])
				{
					
					j=0;
					break;
				}
				else
				{
					i++;  j++;
				}
				
			}
			k++;
			if(j==lengthSub)
			{
				System.out.println("substring present at index" +(k-1)); //cuz starting index is zero
				return true;
			}
		}
		return false;
	}
}

    – Screenshot1.
    – Screenshot2.

    Reply

  2. if you watch the code carefully it does not work, you need to update your code for condition where you need to back track the matching process as in this case, consider “aabbc” as the substring and the main string is “aaabbcc”, the substring is present in this main string but your code will make it false as soon as it encountered with the third ‘a’ in the main string.

    Reply

    • Hi Shreya – thanks for your feedback. Nice corner case. Would you be so kind to provide a fix for this scenario? Looking forward for your update.

      Also, If you have nice idea about new tutorials then let us know as we are encouraging Guest post authors for Java language.

      Reply

      • yes, ofcourse! there are 3 ways to solve this problem, I am sharing the simplest solution. You can also use a stack.

        Reply

      • package com.maven.practice;

import java.util.Scanner;
public class Test {

	public static void main(String[] args) {
		
		try{
			 Scanner scanner=new Scanner(System.in);
		 	 String mainString=scanner.next();
		 	 String subString=scanner.next();
		 	 scanner.close();
		 	 System.out.println("substring present in main string? : " +containSubString(mainString,subString));
		 }
		 catch(Exception e)
		 {
			 System.out.println("enter valid input");
		 }
		 
	}
	
	public static boolean containSubString(String mainString, String subString)
	{
		int lengthMain=mainString.length();
		int lengthSub=subString.length();
		System.out.println(lengthSub);
		char main[]=mainString.toCharArray();
		char sub[]=subString.toCharArray();
		
		int k=0, j=0;
		while(k<lengthMain)
		{
			int i=k;
			while(j<lengthSub)
			{
				if(main[i]!=sub[j])
				{
					
					j=0;
					break;
				}
				else
				{
					i++;
					j++;
				}
				
			}
			k++;
			
			if(j==lengthSub)
			{
				System.out.println("substring present at index" +(k-1)); //cuz starting index is zero
				return true;
			}
			
		}
		return false;
		
	}
	
}
        Reply

      • package com.maven.practice;

import java.util.Scanner;
public class Test {

	public static void main(String[] args) {
		
		try{
			 Scanner scanner=new Scanner(System.in);
		 	 String mainString=scanner.next();
		 	 String subString=scanner.next();
		 	 scanner.close();
		 	 System.out.println("substring present in main string? : " +containSubString(mainString,subString));
		 }
		 catch(Exception e)
		 {
			 System.out.println("enter valid input");
		 }
	}
	
	public static boolean containSubString(String mainString, String subString)
	{
		int lengthMain=mainString.length();
		int lengthSub=subString.length();
		System.out.println(lengthSub);
		char main[]=mainString.toCharArray();
		char sub[]=subString.toCharArray();
		
		int k=0, j=0;
		while(k<lengthMain)
		{
			int i=k;
			while(j<lengthSub)
			{
				if(main[i]!=sub[j])
				{
					
					j=0;
					break;
				}
				else
				{
					i++;
					j++;
				}				
			}
			k++;			
			if(j==lengthSub)
			{
				System.out.println("substring present at index" +(k-1)); //cuz starting index is zero
				return true;
			}			
		}
		return false;		
	}	
}
        Reply

  5. Why did you use “max” instead of just using mainString.length() in the for loop. By using “max” we will be reducing the total number of characters to be checked, yes? What if the subString is present at the end of the mainString. I don’t get it. Explain? –
    ” checkrecusion: for (int i = 0; i <= max; i++) "

    Reply

    • Hi Aniruddha – here max is a difference between mainString and subString and that’s the max we need to iterate through loop. Here max value will be always less than mainString.lenght().

       max = mainString.length() - subString.length();
      Reply

  6. This is wrong. Your program doesn’t work for duplicates.. it does not work if a string contains duplicate characters. try it with substring = “aab” and main string = “aaab”

    it gives false while it should be true.

    The contains function should be implemented using suffix trees

    Reply

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