Number of strings in two array satisfy the given conditions

Given two arrays of string arr1[] and arr2[]. For each string in arr2[](say str2), the task is to count numbers string in arr1[](say str1) which satisfy the below conditions:

• The first characters of str1 and str2 must be equal.
• String str2 must contain each character of string str1.

Examples:

Input: arr1[] = {“aaaa”, “asas”, “able”, “ability”, “actt”, “actor”, “access”}, arr2[] = {“aboveyz”, “abrodyz”, “absolute”, “absoryz”, “actresz”, “gaswxyz”} 
Output:
1 
1 
3 
2 
4 
0 
Explanation:
The following is the string in arr1[] which follows the given condition: 
1 valid word for “aboveyz” : “aaaa”. 
1 valid word for “abrodyz” : “aaaa”. 
3 valid words for “absolute” : “aaaa”, “asas”, “able”. 
2 valid words for “absoryz” : “aaaa”, “asas”. 
4 valid words for “actresz” : “aaaa”, “asas”, “actt”, “access”. 
There are no valid words for “gaswxyz” because none of the words in the list contains the letter ‘g’.

Input: arr1[] = {“abbg”, “able”, “absolute”, “abil”, “actresz”, “gaswxyz”}, arr2[] = {“abbgaaa”, “asas”, “able”, “ability”} 
Output:
1 
0 
1 
1

Brute Force Approach: This approach uses a nested loop to iterate through each string in arr1 and arr2, and checks if the first character of each string in arr1 matches the first character of the string in arr2.

Step by step algorithm:

1. Initialize an empty vector of integers called result to store the count of valid strings.
2. For each string str2 in arr2, do the following:
   a. Initialize a counter variable called count to 0.
   b. For each string str1 in arr1, do the following:
   i. Check if the first character of str1 is equal to the first character of str2.
   ii. If they are equal, iterate over each character c in str1, and check if it is present in str2 using the find() function.
   iii. If all characters of str1 are present in str2, increment the count.
   c. Add the final count to the result vector.
3. Return the result vector.

Below is the implementation of above approach:

C++

#include <iostream>
#include <vector>
#include <string>

using namespace std;

vector<int> count_strings(vector<string> arr1, vector<string> arr2) {
    vector<int> result;
    for (string str2 : arr2) {
        int count = 0;
        for (string str1 : arr1) {
            if (str1[0] == str2[0]) {
                bool flag = true;
                for (char c : str1) {
                    if (str2.find(c) == string::npos) {
                        flag = false;
                        break;
                    }
                }
                if (flag) {
                    count++;
                }
            }
        }
        result.push_back(count);
    }
    return result;
}

int main() {
    vector<string> arr1 = {"aaaa", "asas", "able", "ability", "actt", "actor", "access"};
    vector<string> arr2 = {"aboveyz", "abrodyz", "absolute", "absoryz", "actresz", "gaswxyz"};
    vector<int> result = count_strings(arr1, arr2);
    for (int i : result) {
        cout << i << "\n";
    }
    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

public class StringCounter {
    // Function to count strings in arr1 that are compatible
    // with each string in arr2
    public static List<Integer>
    countStrings(List<String> arr1, List<String> arr2)
    {
        List<Integer> result
            = new ArrayList<>(); // List to store the counts
                                 // for each arr2 string
        for (String str2 :
             arr2) { // Loop through each string in arr2
            int count = 0; // Counter for compatible strings
                           // in arr1
            for (String str1 :
                 arr1) { // Loop through each string in arr1
                if (str1.charAt(0)
                    == str2.charAt(
                        0)) { // Check if the first
                              // character matches
                    boolean flag = true; // Flag to track
                                         // compatibility
                    for (char c :
                         str1.toCharArray()) { // Loop
                                               // through
                                               // each
                                               // character
                                               // in str1
                        if (str2.indexOf(c)
                            == -1) { // Check if character
                                     // is not in str2
                            flag = false; // Set flag to
                                          // false
                            break; // Exit the loop as
                                   // compatibility is not
                                   // possible
                        }
                    }
                    if (flag) { // If all characters in str1
                                // are found in str2
                        count++; // Increment the
                                 // compatibility count
                    }
                }
            }
            result.add(
                count); // Store the count for the current
                        // str2 in the result list
        }
        return result; // Return the list containing counts
    }

    public static void main(String[] args)
    {
        // Input lists containing strings
        List<String> arr1
            = List.of("aaaa", "asas", "able", "ability",
                      "actt", "actor", "access");
        List<String> arr2
            = List.of("aboveyz", "abrodyz", "absolute",
                      "absoryz", "actresz", "gaswxyz");

        // Call the countStrings function and store the
        // result
        List<Integer> result = countStrings(arr1, arr2);

        // Print the compatibility counts for each string in
        // arr2
        for (int i : result) {
            System.out.println(i);
        }
    }
}

// This code is contributed by akshitaguprzj3

Python3

def count_strings(arr1, arr2):
    result = []
    for str2 in arr2:
        count = 0
        for str1 in arr1:
            if str1[0] == str2[0]:
                flag = True
                for c in str1:
                    if c not in str2:
                        flag = False
                        break
                if flag:
                    count += 1
        result.append(count)
    return result

if __name__ == '__main__':
    arr1 = ["aaaa", "asas", "able", "ability", "actt", "actor", "access"]
    arr2 = ["aboveyz", "abrodyz", "absolute", "absoryz", "actresz", "gaswxyz"]
    result = count_strings(arr1, arr2)
    for i in result:
        print(i)

#code added by Avinash Wani

C#

using System;
using System.Collections.Generic;

class GFG
{
    static List<int> CountStrings(List<string> arr1, List<string> arr2)
    {
        List<int> result = new List<int>();
        foreach (string str2 in arr2)
        {
            int count = 0;
            foreach (string str1 in arr1)
            {
                if (str1[0] == str2[0])
                {
                    bool flag = true;
                    foreach (char c in str1)
                    {
                        if (str2.IndexOf(c) == -1)
                        {
                            flag = false;
                            break;
                        }
                    }
                    if (flag)
                    {
                        count++;
                    }
                }
            }
            result.Add(count);
        }
        return result;
    }

    static void Main()
    {
        List<string> arr1 = new List<string> { "aaaa", "asas", "able", "ability", "actt", "actor", "access" };
        List<string> arr2 = new List<string> { "aboveyz", "abrodyz", "absolute", "absoryz", "actresz", "gaswxyz" };
        List<int> result = CountStrings(arr1, arr2);
        foreach (int i in result)
        {
            Console.WriteLine(i);
        }
    }
}

Javascript

// Define a function countStrings that takes two arrays of
// strings as input and returns an array of integers.
function countStrings(arr1, arr2) {
    // Create an empty array to store the results.
    const result = [];

    // Loop through each string in arr2.
    for (const str2 of arr2) {
        // Initialize a count variable to keep track of matching strings.
        let count = 0;

        // Loop through each string in arr1.
        for (const str1 of arr1) {
            // Check if the first character of str1 matches the first character of str2.
            if (str1[0] === str2[0]) {
                let flag = true;

                // Loop through each character in str1.
                for (const c of str1) {
                    // Check if the character is not found in str2.
                    if (str2.indexOf(c) === -1) {
                        flag = false;
                        break;
                    }
                }

                // If all characters in str1 are found in str2, increment the count.
                if (flag) {
                    count++;
                }
            }
        }

        // Push the count into the result array.
        result.push(count);
    }

    // Return the final result array.
    return result;
}

// Define the main function.
function main() {
    // Define two arrays of strings.
    const arr1 = ["aaaa", "asas", "able", "ability", "actt", "actor", "access"];
    const arr2 = ["aboveyz", "abrodyz", "absolute", "absoryz", "actresz", "gaswxyz"];

    // Call the countStrings function and store the result.
    const result = countStrings(arr1, arr2);

    // Loop through the result array and print each integer on a new line.
    for (const i of result) {
        console.log(i);
    }
}

// Call the main function to start the program.
main();

Output

1
1
3
2
4
0

Time Complexity: O(N^2) 
Space Complexity: O(1) 

Approach: This problem can be solved using the concept of Bitmasking. Below are the steps:

• Convert each string of the array arr1[] to its corresponding bitmask as shown below:

For string str = "abcd"
the corresponding bitmask conversion is:
characters | value 
    a          0
    b          1
    c          2
    d          3
As per the above characters value, the number is:
value = 20 + 21 + 23 + 24
value = 15.
so the string "abcd" represented as 15.

• Note: While bitmasking each string if the frequency of characters is more than 1, then include the corresponding characters only once.
• Store the frequency of each string in an unordered_map.
• Similarly, convert each string in the arr2[] to the corresponding bitmask and do the following:
  • Instead of calculating all possible words corresponding to arr1[], use the bit operation to find the next valid bitmask using temp = (temp – 1)&val.
  • It produces the next bitmask pattern, reducing one char at a time by producing all possible combinations.
• For each valid permutation, check if it validates the given two conditions and adds the corresponding frequency to the current string stored in an unordered_map to the result.

Below is the implementation of the above approach:

C++

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

void findNumOfValidWords(vector<string>& w,
                         vector<string>& p)
{
    // To store the frequency of string
    // after bitmasking
    unordered_map<int, int> m;

    // To store result for each string
    // in arr2[]
    vector<int> res;

    // Traverse the arr1[] and bitmask each
    // string in it
    for (string& s : w) {

        int val = 0;

        // Bitmasking for each string s
        for (char c : s) {
            val = val | (1 << (c - 'a'));
        }

        // Update the frequency of string
        // with it's bitmasking value
        m[val]++;
    }

    // Traverse the arr2[]
    for (string& s : p) {
        int val = 0;

        // Bitmasking for each string s
        for (char c : s) {
            val = val | (1 << (c - 'a'));
        }

        int temp = val;
        int first = s[0] - 'a';
        int count = 0;

        while (temp != 0) {

            // Check if temp is present
            // in an unordered_map or not
            if (((temp >> first) & 1) == 1) {
                if (m.find(temp) != m.end()) {
                    count += m[temp];
                }
            }

            // Check for next set bit
            temp = (temp - 1) & val;
        }

        // Push the count for current
        // string in resultant array
        res.push_back(count);
    }

    // Print the count for each string
    for (auto& it : res) {
        cout << it << '\n';
    }
}

// Driver Code
int main()
{
    vector<string> arr1;
    arr1 = { "aaaa", "asas", "able",
             "ability", "actt",
             "actor", "access" };

    vector<string> arr2;
    arr2 = { "aboveyz", "abrodyz",
             "absolute", "absoryz",
             "actresz", "gaswxyz" };

    // Function call
    findNumOfValidWords(arr1, arr2);
    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{

static void findNumOfValidWords(Vector<String> w,
                                Vector<String> p)
{
  // To store the frequency of String
  // after bitmasking
  HashMap<Integer,
          Integer> m = new HashMap<>();

  // To store result for
  // each string in arr2[]
  Vector<Integer> res = new Vector<>();

  // Traverse the arr1[] and
  // bitmask each string in it
  for (String s : w)
  {
    int val = 0;

    // Bitmasking for each String s
    for (char c : s.toCharArray())
    {
      val = val | (1 << (c - 'a'));
    }

    // Update the frequency of String
    // with it's bitmasking value
    if(m.containsKey(val))
      m.put(val, m.get(val) + 1);
    else
      m.put(val, 1);
  }

  // Traverse the arr2[]
  for (String s : p)
  {
    int val = 0;

    // Bitmasking for each String s
    for (char c : s.toCharArray())
    {
      val = val | (1 << (c - 'a'));
    }

    int temp = val;
    int first = s.charAt(0) - 'a';
    int count = 0;

    while (temp != 0)
    {
      // Check if temp is present
      // in an unordered_map or not
      if (((temp >> first) & 1) == 1)
      {
        if (m.containsKey(temp))
        {
          count += m.get(temp);
        }
      }

      // Check for next set bit
      temp = (temp - 1) & val;
    }

    // Push the count for current
    // String in resultant array
    res.add(count);
  }

  // Print the count for each String
  for (int it : res)
  {
    System.out.println(it);
  }
}

// Driver Code
public static void main(String[] args)
{
  Vector<String> arr1 = new Vector<>();
  arr1.add("aaaa"); arr1.add("asas");
  arr1.add("able"); arr1.add("ability");
  arr1.add("actt"); arr1.add("actor");
  arr1.add("access");

  Vector<String> arr2 = new Vector<>();
  arr2.add("aboveyz"); arr2.add("abrodyz");
  arr2.add("absolute"); arr2.add("absoryz");
  arr2.add("actresz"); arr2.add("gaswxyz");

  // Function call
  findNumOfValidWords(arr1, arr2);
}
}

// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach
from collections import defaultdict

def findNumOfValidWords(w, p):

    # To store the frequency of string
    # after bitmasking
    m = defaultdict(int)

    # To store result for each string
    # in arr2[]
    res = []

    # Traverse the arr1[] and bitmask each
    # string in it
    for s in w:
        val = 0

        # Bitmasking for each string s
        for c in s:
            val = val | (1 << (ord(c) - ord('a')))

        # Update the frequency of string
        # with it's bitmasking value
        m[val] += 1

    # Traverse the arr2[]
    for s in p:
        val = 0

        # Bitmasking for each string s
        for c in s:
            val = val | (1 << (ord(c) - ord('a')))

        temp = val
        first = ord(s[0]) - ord('a')
        count = 0

        while (temp != 0):

            # Check if temp is present
            # in an unordered_map or not
            if (((temp >> first) & 1) == 1):
                if (temp in m):
                    count += m[temp]

            # Check for next set bit
            temp = (temp - 1) & val

        # Push the count for current
        # string in resultant array
        res.append(count)

    # Print the count for each string
    for it in res:
        print(it)

# Driver Code
if __name__ == "__main__":

    arr1 = [ "aaaa", "asas", "able",
             "ability", "actt",
             "actor", "access" ]

    arr2 = [ "aboveyz", "abrodyz",
             "absolute", "absoryz",
             "actresz", "gaswxyz" ]

    # Function call
    findNumOfValidWords(arr1, arr2)

# This code is contributed by chitranayal

C#

// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{

static void findNumOfValidWords(List<String> w,
                                List<String> p)
{
  // To store the frequency of String
  // after bitmasking
  Dictionary<int,
             int> m = new Dictionary<int,
                                     int>();

  // To store result for
  // each string in arr2[]
  List<int> res = new List<int>();

  // Traverse the arr1[] and
  // bitmask each string in it
  foreach (String s in w)
  {
    int val = 0;

    // Bitmasking for each String s
    foreach (char c in s.ToCharArray())
    {
      val = val | (1 << (c - 'a'));
    }

    // Update the frequency of String
    // with it's bitmasking value
    if(m.ContainsKey(val))
      m[val] = m[val] + 1;
    else
      m.Add(val, 1);
  }

  // Traverse the arr2[]
  foreach (String s in p)
  {
    int val = 0;

    // Bitmasking for each String s
    foreach (char c in s.ToCharArray())
    {
      val = val | (1 << (c - 'a'));
    }

    int temp = val;
    int first = s[0] - 'a';
    int count = 0;

    while (temp != 0)
    {
      // Check if temp is present
      // in an unordered_map or not
      if (((temp >> first) & 1) == 1)
      {
        if (m.ContainsKey(temp))
        {
          count += m[temp];
        }
      }

      // Check for next set bit
      temp = (temp - 1) & val;
    }

    // Push the count for current
    // String in resultant array
    res.Add(count);
  }

  // Print the count
  // for each String
  foreach (int it in res)
  {
    Console.WriteLine(it);
  }
}

// Driver Code
public static void Main(String[] args)
{
  List<String> arr1 = new List<String>();
  arr1.Add("aaaa"); arr1.Add("asas");
  arr1.Add("able"); arr1.Add("ability");
  arr1.Add("actt"); arr1.Add("actor");
  arr1.Add("access");

  List<String> arr2 = new List<String>();
  arr2.Add("aboveyz"); arr2.Add("abrodyz");
  arr2.Add("absolute"); arr2.Add("absoryz");
  arr2.Add("actresz"); arr2.Add("gaswxyz");

  // Function call
  findNumOfValidWords(arr1, arr2);
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// Javascript program for the above approach

function findNumOfValidWords(w, p)
{
    // To store the frequency of string
    // after bitmasking
    var m = new Map();

    // To store result for each string
    // in arr2[]
    var res = [];

    // Traverse the arr1[] and bitmask each
    // string in it
    w.forEach(s => {
        var val = 0;

        // Bitmasking for each string s
        s.split('').forEach(c => {
            val = val | (1 << (c.charCodeAt(0) - 'a'.charCodeAt(0)));
        });

        // Update the frequency of string
        // with it's bitmasking value
        if(m.has(val))
            m.set(val, m.get(val)+1)
        else
            m.set(val, 1)
    });

    // Traverse the arr2[]
    p.forEach(s => {
        var val = 0;
        s.split('').forEach(c => {
            val = val | (1 << (c.charCodeAt(0) - 'a'.charCodeAt(0)));
        });

        var temp = val;
        var first = s[0].charCodeAt(0) - 'a'.charCodeAt(0);
        var count = 0;

        while (temp != 0) {

            // Check if temp is present
            // in an unordered_map or not
            if (((temp >> first) & 1) == 1) {
                if (m.has(temp)) {
                    count += m.get(temp);
                }
            }

            // Check for next set bit
            temp = (temp - 1) & val;
        }

        // Push the count for current
        // string in resultant array
        res.push(count);
    });

    // Print the count for each string
    res.forEach(it => {

        document.write( it + "<br>");
    });
}

// Driver Code
var arr1 = ["aaaa", "asas", "able",
         "ability", "actt",
         "actor", "access" ];

var arr2 = [ "aboveyz", "abrodyz",
         "absolute", "absoryz",
         "actresz", "gaswxyz" ];
// Function call
findNumOfValidWords(arr1, arr2);

</script>

Output

1
1
3
2
4
0

Time Complexity: O(N) 
Space Complexity: O(N)