Find an Array inside another larger Array in Java

Question

I have the following problem that I tried to solve, but I cannot come with a solution:

Write in a class a method that receives as parameters two arrays of numbers a and b and returns true if the elements of b are found in a in the order of their appearance and false otherwise.

If the first string has length 0 the method returns false. If the second string is 0, the method returns true.

The signature of the method is: public static boolean contains(int[] a, int[] b)

Ex:

contains({1, 2, 3, 4, 2, 1, 3, 4}, {2, 3, 4}) -> true

contains({1, 2, 3, 4, 2, 1, 3, 4}, {1, 4, 3}) -> true

contains({1, 2, 3, 4, 2, 1, 3, 4}, {3, 3, 2}) -> false

contains({1, 2, 3, 4, 2, 1, 3, 4}, {1, 1, 2}) -> false

contains({1, 2, 3, 4, 2, 1, 3, 4}, {}) -> true

contains({}, {1, 2, 3}) -> false

My solution until now:

public class TestEA8 {
public static boolean contains(int[] a, int[] b) {
    int i=0;
    int j=0;
    boolean result = false;

    if(a.length == 0){
        System.out.println(false);
        return false;
    }else if(b.length ==0){
        System.out.println(true);
        return true;}

    for (i=0;i<b.length;i++){
        for (j=0;j<a.length;j++){
            if (b[i] == a[j]){
                result = true;
                break;
                }
            }
            if(j==a.length){
                result =false;
            }
        }
        System.out.println(result);
        return  result;
    }
}

Thank you,

Research hint: you're trying to find whether b is a subsequence of a. — Kevin Anderson
– Kevin Anderson, Commented Apr 24, 2020 at 14:33
Seems to me like a case for Collection.containsAll(): docs.oracle.com/javase/8/docs/api/java/util/… — Usagi Miyamoto
– Usagi Miyamoto, Commented Apr 27, 2020 at 12:37

a.eugene · Accepted Answer · 2020-04-24 14:48:58Z

2

You can use Collections.indexOfSubList() method.

Example:

public static boolean contains(int[] a, int[] b) {
    return findArray(a, b) > -1;
}

public static int findArray(int[] a, int[] b) {
    return Collections.indexOfSubList(arrayToList(a), arrayToList(b));
}

public static List<Integer> arrayToList(int[] array) {
    return Arrays.stream(array)
        .boxed()
        .collect(Collectors.toList());
}

answered Apr 24, 2020 at 14:48

a.eugene

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Comments

Sarangan · Accepted Answer · 2020-04-24 14:43:54Z

This Code will solve your problem:

use a variable temp where it will be the tracking index of first array and moves forward in every check.

 public static boolean contains(int[] a, int[] b) {
    int i=0;
    int j=0;
    boolean result = false;

    if(a.length == 0){
        System.out.println(false);
        return false;
    }else if(b.length ==0){
        System.out.println(true);
        return true;}

    int count = 0;
    int temp = 0;
    for (i=0;i<b.length;i++){
        for (j= temp;j<a.length;j++){
            if (b[i] == a[j]){
                count = count+1;
                temp++;
                break;
            }
            temp++;
        }
    }

    if (count == b.length){
        result = true;
    } else {
        result =false;
    }
    System.out.println(result);
    return  result;
}

tevemadar · Accepted Answer · 2020-04-27 12:32:23Z

0

You can walk through the first array, "ticking out" the expected element from the second one when you encounter it. If you reach the end of the second array, return true, otherwise false:

public static boolean contains(int[] a, int[] b) {
  int i=0,j=0;
  while(i<a.length && j<b.length){
    if(a[i]==b[j])
      j++;
    i++;
  }
  return j==b.length;
}

Test with the provided example data: https://ideone.com/JRBvkm

answered Apr 27, 2020 at 12:32

tevemadar

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