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

Last updated on  by  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.

Would love to know what you think. Chime in and share it as a comment. Don’t forget to share this guide!
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  • Shreya Sharma

    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;
    }
    https://uploads.disquscdn.com/images/05dbc172a8b0120ae472d978add6b963b8e53747b02e2591d4c2f63715836a48.jpg https://uploads.disquscdn.com/images/58a47d77c9635e3c764458572473d1eb705c662f78b5c0a0e2b9de344eef414b.jpg
    }

  • Shreya Sharma

    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.

    • 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.

      • Shreya Sharma

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

        • Thank you so much Shreya. I’ll take a look at all details and will update post with your recommendation tomorrow.

      • Shreya Sharma

        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;

        }

        }

      • Shreya Sharma

        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;

        }

        }

  • sharan

    How to do without inbuilt function

    • Hi Sharan – we already have it in above example.

      Just use method which says: // Implement your own Contains Method with Recursion

  • sagar

    thank you very usefull

  • Aniruddha Ramgir

    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++) "

    • 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();
  • Mohammad Ahmad

    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

  • T

    What is the time complexity of this algorithm?

  • Jack

    what is the theme used here?

  • Dave Henderson

    Curious why not use the already created ‘contains’ method? Is there a benefit to using your own?

    • Hi Dave,
      Above example shows implementation of “contains” method only. Just incase in an interview, somebody ask us to implement the same..

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