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 = “neveropen for neveropen is best place to learn”, S2 = “bad place”
Output: neveropen
“neveropen” is the most frequent word in S1 and is also not present in S2.
The frequency of “neveropen” is 2Input: 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:
- Iterate through string S2 and create a map and insert all of the words in it to the map.
- Iterate through string S1 and check whether the word is not present in the map created in the previous step or not.
- If the word satisfies the condition then update the answer if the frequency of the same word is maximum till now.
- 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 S2string 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 programint main(){ string S1 = "neveropen for neveropen 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 = "neveropen for neveropen 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 Codeif __name__ == "__main__": S1 = "neveropen for neveropen is best place to learn" S2 = "bad place" print(smallestFreq(S1, S2)) # This code is contributed # by Rituraj Jain |
C#
// C# implementation of above approachusing 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 = "neveropen for neveropen 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 S2function 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 Codelet S1 = "neveropen for neveropen is best place to learn";let S2 = "bad place";document.write(smallestFreq(S1, S2));// This code is contributed by avanitrachhadiya2155</script> |
Output
neveropen
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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