Print all strings in the given array that occur as the substring in the given string

Last Updated : 29 Mar, 2022
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Given an array of string arr[] and a string str, the task is to print all the strings in arr[] that occur as a substring in str.

Example:

Input: str ="geeksforgeeks", arr[] ={ "forg", "geek", "ek", "dog", "sfor"}
Output: 
forg
geek
ek
sfor
Explanation: The strings "forg", "geek", "ek" and "sfor" occur as a substring in str. Therefore, the required count is 4.

Input: str ="abcd", arr[] ={ "aa", "bb", "cc"}
Output: -1

 

Approach: The given problem is an implementation base problem. It can be solved by iterating over the given array of strings and for each string in arr[], check whether it occurs as a substring of str or not using the algorithm discussed in this article. Maintain a variable that stores If no string exists as a substring. In that case, print -1.

Below is the implementation of the above approach:

C++ 
// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Returns true if s1 is substring of s2
int isSubstring(string s1, string s2)
{
  int M = s1.length();
  int N = s2.length();

  /* A loop to slide pat[] one by one */
  for (int i = 0; i <= N - M; i++) {
    int j;

    /* For current index i, check for
        pattern match */
    for (j = 0; j < M; j++)
      if (s2[i + j] != s1[j])
        break;

    if (j == M)
      return i;
  }

  return -1;
}

// Function to print all the strings
// in the given array that occur as
// the substring in the given string
void isSubstr(string Str, string arr[], int len)
{

  // Stores if no string is a
  // substring of str
  int flag = 0;

  // Iterate over the array of strings
  for (int i = 0; i < len; i++)
  {

    // if the current string occur
    // as a substring in Str
    int s = isSubstring(arr[i],Str);
    if (s != -1) {

      // Print string i
      cout << arr[i] <<endl;
      flag = 1;
    }
  }

  // If no substring exist
  if (flag == 0)
    cout<<"-1"<<endl;
}

// Driver Code
int main()
{
  string arr[5]
    = { "forg", "geek", "ek", "dog", "sfo"};
  int len = sizeof(arr)/sizeof(arr[0]);
  string Str = "geeksforgeeks";

  isSubstr(Str, arr, len);

  return 0;
}

// This code is contributed by sanjoy_62.
Java 
// JAVA program of the above approach
import java.util.*;
class GFG
{
  
    // Function to print all the strings
    // in the given array that occur as
    // the substring in the given string
    public static void isSubstr(String Str,
                                ArrayList<String> arr)
    {
      
        // Stores if no string is a
        // substring of str
        int flag = 0;

        // Iterate over the array of strings
        for (int i = 0; i < arr.size(); i++)
        {
          
            // if the current string occur
            // as a substring in Str
            if (Str.indexOf(arr.get(i)) != -1) {

                // Print string i
                System.out.println(arr.get(i));
                flag = 1;
            }
        }
      
        // If no substring exist
        if (flag == 0)
            System.out.print(-1);
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        ArrayList<String> arr
            = new ArrayList<>(Arrays.asList(
                "forg", "geek", "ek", "dog", "sfo"));
        String Str = "geeksforgeeks";

        isSubstr(Str, arr);
    }
}

// This code is contributed by Taranpreet
Python3 
# Python program of the above approach

# Function to print all the strings
# in the given array that occur as
# the substring in the given string


def isSubstr(Str, arr):

    # Stores if no string is a
    # substring of str
    flag = 0

    # Iterate over the array of strings
    for i in arr:

        # if the current string occur
        # as a substring in Str
        if i in Str:

            # Print string i
            print(i)
            flag = 1

    # If no substring exist
    if flag == 0:
        print(-1)


# Driver Code
arr = ["forg", "geek", "ek", "dog", "sfo"]
Str = "geeksforgeeks"

isSubstr(Str, arr)
C# 
// C# program of the above approach
using System;
public class GFG
{

  // Function to print all the strings
  // in the given array that occur as
  // the substring in the given string
  public static void isSubstr(String Str,String[] arr)
  {

    // Stores if no string is a
    // substring of str
    int flag = 0;

    // Iterate over the array of strings
    for (int i = 0; i < arr.Length; i++)
    {

      // if the current string occur
      // as a substring in Str
      if (Str.IndexOf(arr[i]) != -1) {

        // Print string i
        Console.WriteLine(arr[i]);
        flag = 1;
      }
    }

    // If no substring exist
    if (flag == 0)
      Console.Write(-1);
  }

  // Driver Code
  public static void Main(String[] args)
  {
    String[] arr = {"forg", "geek", "ek", "dog", "sfo"};
    String Str = "geeksforgeeks";

    isSubstr(Str, arr);
  }
}

// This code is contributed by 29AjayKumar
JavaScript 
    <script>
        // JavaScript program of the above approach

        // Function to print all the strings
        // in the given array that occur as
        // the substring in the given string
        const isSubstr = (Str, arr) => {

            // Stores if no string is a
            // substring of str
            let flag = 0;

            // Iterate over the array of strings
            for (i in arr) 
            {
            
                // if the current string occur
                // as a substring in Str
                if (Str.indexOf(arr[i]) != -1)
                {
                
                    // Print string i
                    document.write(`${arr[i]}<br/>`);
                    flag = 1;
                }
            }
            
            // If no substring exist
            if (flag == 0)
                document.write(-1);

        }

        // Driver Code
        let arr = ["forg", "geek", "ek", "dog", "sfo"];
        let Str = "geeksforgeeks";

        isSubstr(Str, arr)

        // This code is contributed by rakeshsahni

    </script>

Output
forg
geek
ek
sfo

Time Complexity: O(N2)
Auxiliary Space: O(1)


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