Print all strings in the given array that occur as the substring in the given string
Last Updated :
29 Mar, 2022
Improve
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++ 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.
// C++ program of the above approachusing namespace std;// Returns true if s1 is substring of s2int 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 stringvoid 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 Codeint 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 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
# 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# 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
<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)
