//Time Complexity O(bns) | Space Complexity O(n)

// predefined method indexOf(String str)

// b -> length of big String

//n -> size of the list of small String

//s -> max length of small string among the m=small syting in the list

import java.util.*;

class A

{

public static void main(String[] args)

{

//check given list of strings is contained in given String

String bigString = "this is a big string";

String[] smallStrings = new String[]{"this","yo","is","a","bigger","string","kapper"};

//accordingly return boolean array

boolean[] result = getSearchResult(bigString,smallStrings);

System.out.println("Search Result is");

System.out.println(Arrays.toString(result));

}

public static boolean[] getSearchResult(String big,String[] smalls)

{

//initialize result array

boolean[] result = new boolean[smalls.length];

for(int i = 0 ; i < result.length ; i++)

{

//populate the result array accordingly if the tsing at inex i in the smalls array

//is contained in the big String

result[i] = checkIfContained(big,smalls[i]);

}

return result;

}

public static boolean checkIfContained(String big,String small)

{

//it returns the index of first occurance of small string in big String

//if small string is not present in big it returns -1

int idx = big.indexOf(small);

//according to idx value return true or false;

return idx == -1 ? false : true;

}

}