Crunchify

Largest free Technical and Blogging resource site for Beginner. We help clients transform their great ideas into reality!

Advertisement
Blog   Java and J2EE Tutorials   Java: How to Check if a String Contains a Substring? Implement Your Ow ...

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

Last Updated on by   App Shah   18 comments

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.

Join the Discussion

Share & leave us some comments on what you think about this topic or if you like to add something.

Share:

Other Popular Articles...

  1. How to Reverse a String in Java?
  2. Java Program to Replace/Remove Characters from String
  3. 4 Ways to Check if Array Contains a Specific Value in Java8? IntStream and Arrays.asList Tutorial
  4. How to Find All Permutations of a String in Java – Example
  5. Best way to Find Duplicate Character from a String in Java
  6. How to Implement your own InetAddress.isReachable(String address, int port, int timeout) method in Java?
Avatar for App Shah

About App Shah

I'm an Engineer by profession, Blogger by passion & Founder of Crunchify, LLC, the largest free blogging & technical resource site for beginners. Love SEO, SaaS, #webperf, WordPress, Java. With over 16 millions+ pageviews/month, Crunchify has changed the life of over thousands of individual around the globe teaching Java & Web Tech for FREE. Get latest update on and .

Subscribe To Newsletter…

Get Early Access To New Articles, Plugins, Discount Codes And Brief Updates About What's New With Crunchify! Join Over 16 Million Monthly Readers...

Comments

  1. Avatar for Shreya SharmaShreya Sharma says 

    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. Avatar for Shreya SharmaShreya Sharma says

    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

    • Avatar for App ShahApp Shah says

      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

      • Avatar for Shreya SharmaShreya Sharma says

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

        Reply

      • Avatar for Shreya SharmaShreya Sharma says 

        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

      • Avatar for Shreya SharmaShreya Sharma says 

        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. Avatar for Aniruddha RamgirAniruddha Ramgir says

    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

    • Avatar for App ShahApp Shah says

      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. Avatar for Mohammad AhmadMohammad Ahmad says

    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

Leave a Reply

Your email address will not be published. Required fields are marked *

Recently Crunched…