Print all valid words from given dictionary that are possible using Characters of given Array

Given a dictionary of strings dict[] and a character array arr[]. Print all valid words of the dictionary that are possible using characters from the given character array.
Examples:

Input: dict – [“go”, “bat”, “me”, “eat”, “goal”, boy”, “run”] 
arr[] = [‘e’, ‘o’, ‘b’, ‘a’, ‘m’, ‘g’, ‘l’]
Output: “go”, “me”, “goal”.
Explanation: Only all the characters of these three strings are present in the dictionary.

Input: Dict=  [“goa”, “bait”, “mess”, “ate”, “goal”, “girl”, “rain”]
arr[]= [‘s’, ‘o’, ‘t’, ‘a’, ‘e’, ‘g’, ‘l’, ‘i’, ‘r’]
Output: “goa”, “ate”, “goal”, “girl”

Approach: The problem can be solved by checking the characters of each string of the dictionary. If all the characters of a string is present then that string can be formed. Follow the steps mentioned below to solve the problem.

• Concatenate all the characters and make a string.
• Traverse each word in a string and find whether all the characters of each word are present in concatenated string or not.

Follow the illustration shown below for better understanding.

Illustration: Consider the example below.

dict[] = [“go”, “bat”, “eat”, “meat”, “goal”, “boy”, “run”],  
arr[] = [‘e’, ‘o’, ‘b’, ‘a’, ‘m’, ‘g’, ‘l’]

• concatenated string= e o b a m g l
• now if we traverse “go”
• indexOf(“g”) in “eobamgl” is 5
• indexOf(“o”) in “eobamgl” is 1
• This means all the indexes are present. Hence we will print “go”
• If any of the indexes are not present, that will not be included.
• Similarly “me” and “goal” also satisfies the condition.

So the output is “go”, “goal” and “me”.

Below is the implementation of the above approach.

C++

// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the words
void printWord(string str, string s) {
    for (int i = 0; i < str.size(); i++) {
        if (s.find(str[i]) == string::npos) {
            return;
        }
    }
    cout << str << endl;
}

// Function to find the words
void findWord(vector<string> str1, vector<char> str2) {
    string s = "";
    for (int i = 0; i < str2.size(); i++) {
        s += str2[i];
    }
    for (int i = 0; i < str1.size(); i++) {
        printWord(str1[i], s);
    }
}

int main() {
    vector<string> str1 = {"go", "bat", "me", "eat",
                "goal", "boy", "run"};
    vector<char> str2 = {'e', 'o', 'b', 'a', 'm', 'g', 'l'};
    findWord(str1, str2);
    return 0;
}

// This code is contributed by Samim Hossain Mondal.

Java

// Java code to implement the above approach
public class GFG{

  // Function to print the words
  public static void printWord(String str, String s) {
    for (int i = 0; i < str.length(); i++) {
      if (s.indexOf(str.charAt(i)) < 0) {
        return;
      }
    }
    System.out.println(str);
  }

  // Function to find the words
  public static void findWord(String[] str1, char[] str2) {
    String s = "";
    for (int i = 0; i < str2.length; i++) {
      s += str2[i];
    }
    for (int i = 0; i < str1.length; i++) {
      printWord(str1[i], s);
    }
  }

  public static void main(String args[]) {
    String[] str1 = {"go", "bat", "me", "eat", "goal", "boy", "run"};
    char[] str2 = {'e', 'o', 'b', 'a', 'm', 'g', 'l'};
    findWord(str1, str2);
  }
}

// This code is contributed by Saurabh Jaiswal

Python3

# python code to implement the above approach

# Function to print the words
def printWord(str, s):
    for i in range(0, len(str)):
        if (not (str[i] in s)):
            return

    print(str)

# Function to find the words
def findWord(str1, str2):
    s = ""
    for i in str2:
        s += i

    for i in range(0, len(str1)):
        printWord(str1[i], s)

# Driver Code
if __name__ == "__main__":
    str1 = ["go", "bat", "me", "eat", "goal", "boy", "run"]

    str2 = ["e", "o", "b", "a", "m", "g", "l"]

    findWord(str1, str2)

    # This code is contributed by rakeshsahni

C#

// C# code to implement the above approach
using System;

class GFG {

  // Function to print the words
  public static void printWord(string str, string s)
  {
    for (int i = 0; i < str.Length; i++) {
      if (s.IndexOf((str[i])) < 0) {
        return;
      }
    }
    Console.WriteLine(str);
  }

  // Function to find the words
  public static void findWord(string[] str1, char[] str2)
  {
    string s = "";
    for (int i = 0; i < str2.Length; i++) {
      s += str2[i];
    }
    for (int i = 0; i < str1.Length; i++) {
      printWord(str1[i], s);
    }
  }

  public static void Main(string[] args)
  {
    string[] str1 = { "go",   "bat", "me", "eat",
                     "goal", "boy", "run" };
    char[] str2 = { 'e', 'o', 'b', 'a', 'm', 'g', 'l' };
    findWord(str1, str2);
  }
}

// This code is contributed by ukasp.

Javascript

<script>
    // JavaScript code to implement the above approach

    // Function to print the words
    function printWord(str, s) {
        for (var i = 0; i < str.length; i++) {
            if (s.indexOf(str[i]) < 0) {
                return;
            }
        }
        document.write(str);
        document.write("<br>");
    }

    // Function to find the words
    function findWord(str1, str2) {
        var s = "";
        for (var i in str2) {
            s += str2[i];
        }
        for (var i = 0; i < str1.length; i++) {
            printWord(str1[i], s);
        }
    }

    var str1 = ["go", "bat", "me", "eat",
                "goal", "boy", "run"];
    var str2 = ["e", "o", "b", "a", "m", "g", "l"];
    findWord(str1, str2);
</script>

Output

go
me
goal

Time Complexity: O(N * K) where N is the length of the dict[] and k is the length of arr[].
Auxiliary Space: O(1)