# 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”

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

 // CPP implementation of above approach #include using namespace std;    // Function to return frequent // word from S1 that isn't // present in S2 string smallestFreq(string S1, string S2) {        map 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 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 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 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 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 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# 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 banned = new Dictionary();            // 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 result = new Dictionary();         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

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.

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