Most frequent word in first String which is not present in second String

Given two string ‘S1’ and ‘S2’, the task is to return the most frequent (which is used the maximum number of times) word from ‘S1’ that is not present in ‘S2’. If more than one word is possible then print lexicographically smallest among them.

Examples:

Input: S1 = “geeks for geeks is best place to learn”, S2 = “bad place” 
Output: geeks 
“geeks” is the most frequent word in S1 and is also not present in S2. 
The frequency of “geeks” is 2

Input: S1 = “the quick brown fox jumps over the lazy dog”, S2 = “the brown fox jumps” 
Output: dog 
All the words have frequency 1. 
The lexicographically smallest word is “dog” 

Approach:

The thought process must begin with the creation of a map to store key-value pair( string, int). Following this begins the extraction of the words from the first string while updating the map and the count. For every word from the second array that is present in the first array, reset the count. Finally, traverse the map and find the word with the highest frequency and get the lexicographically smallest one.

Algorithm:

1. Iterate through string S2 and create a map and insert all of the words in it to the map.
2. Iterate through string S1 and check whether the word is not present in the map created in the previous step or not.
3. If the word satisfies the condition then update the answer if the frequency of the same word is maximum till now.
4. If the frequency of the word is equal to the previously chosen word then update the answer according to lexicographically smallest of the two strings.

Below is the implementation of above approach: 

C++

// CPP implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return frequent
// word from S1 that isn't
// present in S2
string smallestFreq(string S1, string S2)
{

    map<string, int> banned;

    // create map of banned words
    for (int i = 0; i < S2.length(); ++i) {

        string s = "";
        while (i < S2.length() && S2[i] != ' ')
            s += S2[i++];

        banned[s]++;
    }

    map<string, int> result;
    string ans;
    int freq = 0;

    // find smallest and most frequent word
    for (int i = 0; i < S1.length(); ++i) {

        string s = "";
        while (i < S1.length() && S1[i] != ' ')
            s += S1[i++];

        // check if word is not banned
        if (banned[s] == 0) {
            result[s]++;
            if (result[s] > freq
                || (result[s] == freq && s < ans)) {
                ans = s;
                freq = result[s];
            }
        }
    }

    // return answer
    return ans;
}

// Driver program
int main()
{
    string S1 = "geeks for geeks is best place to learn";
    string S2 = "bad place";

    cout << smallestFreq(S1, S2);

    return 0;
}

Java

// Java implementation of above approach
import java.util.HashMap;

class GFG
{

    // Function to return frequent
    // word from S1 that isn't
    // present in S2
    static String smallestFreq(String S1,
                               String S2)
    {
        HashMap<String,
                Integer> banned = new HashMap<>();

        // create map of banned words
        for (int i = 0; i < S2.length(); i++)
        {
            String s = "";
            while (i < S2.length() &&
                       S2.charAt(i) != ' ')
                s += S2.charAt(i++);

            banned.put(s, banned.get(s) == null ?
                      1 : banned.get(s) + 1);
        }

        HashMap<String,
                Integer> result = new HashMap<>();
        String ans = "";
        int freq = 0;

        // find smallest and most frequent word
        for (int i = 0; i < S1.length(); i++)
        {
            String s = "";
            while (i < S1.length() &&
                       S1.charAt(i) != ' ')
                s += S1.charAt(i++);

            // check if word is not banned
            if (banned.get(s) == null)
            {
                result.put(s, result.get(s) == null ? 1 :
                              result.get(s) + 1);
                if (result.get(s) > freq ||
                   (result.get(s) == freq &&
                    s.compareTo(ans) < 0))
                {
                    ans = s;
                    freq = result.get(s);
                }
            }
        }

        // return answer
        return ans;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String S1 = "geeks for geeks is best place to learn";
        String S2 = "bad place";
        System.out.println(smallestFreq(S1, S2));
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Python3 implementation of above approach
from collections import defaultdict

# Function to return frequent
# word from S1 that isn't
# present in S2
def smallestFreq(S1, S2):

    banned = defaultdict(lambda:0)

    i = 0

    # create map of banned words
    while i < len(S2):

        s = ""
        while i < len(S2) and S2[i] != ' ':
            s += S2[i]
            i += 1

        i += 1
        banned[s] += 1

    result = defaultdict(lambda:0)
    ans = ""
    freq = 0
    i = 0

    # find smallest and most frequent word
    while i < len(S1):

        s = ""
        while i < len(S1) and S1[i] != ' ':
            s += S1[i]
            i += 1

        i += 1

        # check if word is not banned
        if banned[s] == 0:
            result[s] += 1

            if (result[s] > freq or
               (result[s] == freq and s < ans)):
                ans = s
                freq = result[s]

    # return answer
    return ans

# Driver Code
if __name__ == "__main__":

    S1 = "geeks for geeks is best place to learn"
    S2 = "bad place"

    print(smallestFreq(S1, S2))

# This code is contributed
# by Rituraj Jain

C#

// C# implementation of above approach
using System;
using System.Collections.Generic;

class GFG
{

    // Function to return frequent
    // word from S1 that isn't
    // present in S2
    static String smallestFreq(String S1,
                               String S2)
    {
        Dictionary<String,
                   int> banned = new Dictionary<String,
                                                int>();

        // create map of banned words
        for (int i = 0; i < S2.Length; i++)
        {
            String s = "";
            while (i < S2.Length &&
                    S2[i] != ' ')
                s += S2[i++];

            if(banned.ContainsKey(s))
            {
                var val = banned[s];
                banned.Remove(s);
                banned.Add(s, val + 1);
            }
            else
            {
                banned.Add(s, 1);
            }
        }

        Dictionary<String,
                   int> result = new Dictionary<String,
                                                int>();
        String ans = "";
        int freq = 0;

        // find smallest and most frequent word
        for (int i = 0; i < S1.Length; i++)
        {
            String s = "";
            while (i < S1.Length &&
                    S1[i] != ' ')
                s += S1[i++];

            // check if word is not banned
            if (!banned.ContainsKey(s))
            {
                if(result.ContainsKey(s))
                {
                    var val = result[s];
                    result.Remove(s);
                    result.Add(s, val + 1);
                }
                else
                {
                    result.Add(s, 1);
                }
                if (result[s] > freq ||
                   (result[s] == freq &&
                    s.CompareTo(ans) < 0))
                {
                    ans = s;
                    freq = result[s];
                }
            }
        }

        // return answer
        return ans;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        String S1 = "geeks for geeks is best place to learn";
        String S2 = "bad place";
        Console.WriteLine(smallestFreq(S1, S2));
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript implementation of above approach

 // Function to return frequent
    // word from S1 that isn't
    // present in S2
function smallestFreq(S1,S2)
{
    let banned = new Map();

        // create map of banned words
        for (let i = 0; i < S2.length; i++)
        {
            let s = "";
            while (i < S2.length &&
                       S2[i] != ' ')
                s += S2[i++];

            banned.set(s, banned[s] == null ?
                      1 : banned.get(s) + 1);
        }

        let result = new Map();
        let ans = "";
        let freq = 0;

        // find smallest and most frequent word
        for (let i = 0; i < S1.length; i++)
        {
            let s = "";
            while (i < S1.length &&
                       S1[i] != ' ')
                s += S1[i++];

            // check if word is not banned
            if (banned.get(s) == null)
            {
                result.set(s, result.get(s) == null ? 1 :
                              result.get(s) + 1);
                if (result.get(s) > freq ||
                   (result.get(s) == freq &&
                    s < (ans) ))
                {
                    ans = s;
                    freq = result.get(s);
                }
            }
        }

        // return answer
        return ans;
}

// Driver Code
let S1 = "geeks for geeks is best place to learn";
let S2 = "bad place";
document.write(smallestFreq(S1, S2));


// This code is contributed by avanitrachhadiya2155

</script>

Output

geeks

Complexity Analysis:

• Time Complexity: O(n), where n is the length of the string. 
  • A single traversal of the string is needed.
• Space Complexity: O(n).
  • There can be at most n words in a string. The map requires O(n) space to store the strings.