Given n sentences. The task is to count the number of words that appear in all of these sentences. Note that every word consists of only lowercase English alphabets.
Examples:
Input: arr[] = {
“there is a cow”,
“cow is our mother”,
“cow gives us milk and milk is sweet”,
“there is a boy who loves cow”}
Output: 2
Only the words “is” and “cow” appear in all of the sentences.
Input: arr[] = {
“abc aac abcd ccc”,
“ac aa abc cca”,
“abca aaac abcd ccc”}
Output: 0
Naive Approach: The naive approach is to take every word of the first sentence and compare it with the rest of the lines if it is also present in all lines then increment the count.
Efficient Approach: For all the words of the first sentence, we can check if it is also present in all the other sentences in constant time by using a map. Store all the words of the first sentence in a map and check how many of these stored words are present in all the other sentences.
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count of common words // in all the sentences int commonWords(vector<string> S) { int m, n, i, j; // To store separate words string str; // It will be used to check if a word is present // in a particuler string unordered_map<string, bool > has; // To store all the words of first string vector<pair<string, bool > > ans; pair<string, bool > tmp; // m will store number of strings in given vector m = S.size(); i = 0; // Extract all words of first string and store it in ans while (i < S[0].size()) { str = "" ; while (i < S[0].size() && S[0][i] != ' ' ) { str += S[0][i]; i++; } // Increase i to get at starting index // of the next word i++; // If str is not empty store it in map if (str != "" ) { tmp = make_pair(str, true ); ans.push_back(tmp); } } // Start from 2nd line check if any word of // the first string did not match with // some word in the current line for (j = 1; j < m; j++) { has.clear(); i = 0; while (i < S[j].size()) { str = "" ; while (i < S[j].size() && S[j][i] != ' ' ) { str += S[j][i]; i++; } i++; if (str != "" ) { has[str] = true ; } } // Check all words of this vector // if it is not present in current line // make it false for ( int k = 0; k < ans.size(); k++) { if (ans[k].second != false && has[ans[k].first] == false ) { ans[k].second = false ; } // This line is used to consider only distinct words else if (ans[k].second != false && has[ans[k].first] == true ) { has[ans[k].first] = false ; } } } // This function will print // the count of common words int cnt = 0; // If current word appears in all the sentences for ( int k = 0; k < ans.size(); k++) { if (ans[k].second == true ) cnt++; } return cnt; } // Driver code int main() { vector<string> S; S.push_back( "there is a cow" ); S.push_back( "cow is our mother" ); S.push_back( "cow gives us milk and milk is sweet" ); S.push_back( "there is a boy who loves cow" ); cout << commonWords(S); return 0; } |
Java
// Java implementation of the approach import java.util.HashMap; class GFG { // Function to return the count of // common words in all the sentences static int commonWords(String[] s) { int m, i, j; // To store separate words String str; // It will be used to check if a word // is present in a particuler string HashMap<String, Boolean> has = new HashMap<>(); // To store all the words of first string String[] ans1 = new String[ 100 ]; boolean [] ans2 = new boolean [ 100 ]; int track = 0 ; // m will store number of strings // in given vector m = s.length; i = 0 ; // Extract all words of first string // and store it in ans while (i < s[ 0 ].length()) { str = "" ; while (i < s[ 0 ].length() && s[ 0 ].charAt(i) != ' ' ) { str += s[ 0 ].charAt(i); i++; } // Increase i to get at starting index // of the next word i++; // If str is not empty store it in map if (str.compareTo( "" ) != 0 ) { ans1[track] = str; ans2[track] = true ; track++; } } // Start from 2nd line check if any word of // the first string did not match with // some word in the current line for (j = 1 ; j < m; j++) { has.clear(); i = 0 ; while (i < s[j].length()) { str = "" ; while (i < s[j].length() && s[j].charAt(i) != ' ' ) { str += s[j].charAt(i); i++; } i++; if (str.compareTo( "" ) != 0 ) has.put(str, true ); } // Check all words of this vector // if it is not present in current line // make it false for ( int k = 0 ; k < track; k++) { // System.out.println(has.get(ans1[k])); if (ans2[k] != false && !has.containsKey(ans1[k])) ans2[k] = false ; // This line is used to consider // only distinct words else if (ans2[k] != false && has.containsKey(ans1[k]) && has.get(ans1[k]) == true ) has.put(ans1[k], false ); } } // This function will print // the count of common words int cnt = 0 ; // If current word appears // in all the sentences for ( int k = 0 ; k < track; k++) if (ans2[k] == true ) cnt++; return cnt; } // Driver Code public static void main(String[] args) { String[] s = { "there is a cow" , "cow is our mother" , "cow gives us milk and milk is sweet" , "there is a boy who loves cow" }; System.out.println(commonWords(s)); } } // This code is contributed by // sanjeev2552 |
Python3
# Python3 implementation of the approach from collections import defaultdict # Function to return the count of # common words in all the sentences def commonWords(S): # It will be used to check if a word # is present in a particuler string has = defaultdict( lambda : False ) # To store all the words of first string ans = [] # m will store number of strings # in given vector m = len (S) i = 0 # Extract all words of first string # and store it in ans while i < len (S[ 0 ]): string = "" while i < len (S[ 0 ]) and S[ 0 ][i] ! = ' ' : string + = S[ 0 ][i] i + = 1 # Increase i to get at starting # index of the next word i + = 1 # If str is not empty store it in map if string ! = "": ans.append([string, True ]) # Start from 2nd line check if any word # of the first string did not match with # some word in the current line for j in range ( 1 , m): has.clear() i = 0 while i < len (S[j]): string = "" while i < len (S[j]) and S[j][i] ! = ' ' : string + = S[j][i] i + = 1 i + = 1 if string ! = "": has[string] = True # Check all words of this vector # if it is not present in current # line make it false for k in range ( 0 , len (ans)): if (ans[k][ 1 ] ! = False and has[ans[k][ 0 ]] = = False ): ans[k][ 1 ] = False # This line is used to consider # only distinct words elif (ans[k][ 1 ] ! = False and has[ans[k][ 0 ]] = = True ): has[ans[k][ 0 ]] = False # This function will print # the count of common words cnt = 0 # If current word appears in all # the sentences for k in range ( 0 , len (ans)): if ans[k][ 1 ] = = True : cnt + = 1 return cnt # Driver code if __name__ = = "__main__" : S = [] S.append( "there is a cow" ) S.append( "cow is our mother" ) S.append( "cow gives us milk and milk is sweet" ) S.append( "there is a boy who loves cow" ) print (commonWords(S)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the // above approach using System; using System.Collections.Generic; class GFG{ // Function to return the count of // common words in all the sentences static int commonWords(String[] s) { int m, i, j; // To store separate words String str; // It will be used to check // if a word is present in a // particuler string Dictionary<String, Boolean> has = new Dictionary<String, Boolean>(); // To store all the words of // first string String[] ans1 = new String[100]; bool [] ans2 = new bool [100]; int track = 0; // m will store number of // strings in given vector m = s.Length; i = 0; // Extract all words of // first string and store // it in ans while (i < s[0].Length) { str = "" ; while (i < s[0].Length && s[0][i] != ' ' ) { str += s[0][i]; i++; } // Increase i to get at // starting index of the // next word i++; // If str is not empty store // it in map if (str.CompareTo( "" ) != 0) { ans1[track] = str; ans2[track] = true ; track++; } } // Start from 2nd line check if // any word of the first string // did not match with some word // in the current line for (j = 1; j < m; j++) { has.Clear(); i = 0; while (i < s[j].Length) { str = "" ; while (i < s[j].Length && s[j][i] != ' ' ) { str += s[j][i]; i++; } i++; if (str.CompareTo( "" ) != 0) has[str] = true ; } // Check all words of this // vector if it is not present // in current line make it false for ( int k = 0; k < track; k++) { // Console.WriteLine(has[ans1[k])); if (ans2[k] != false && !has.ContainsKey(ans1[k])) ans2[k] = false ; // This line is used to consider // only distinct words else if (ans2[k] != false && has.ContainsKey(ans1[k]) && has[ans1[k]] == true ) has[ans1[k]] = false ; } } // This function will print // the count of common words int cnt = 0; // If current word appears // in all the sentences for ( int k = 0; k < track; k++) if (ans2[k] == true ) cnt++; return cnt; } // Driver Code public static void Main(String[] args) { String[] s = { "there is a cow" , "cow is our mother" , "cow gives us milk" + "and milk is sweet" , "there is a boy who" + "loves cow" }; Console.WriteLine(commonWords(s)); } } // This code is contributed by Rajput-Ji |
2
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