Given an array of words and a string, we need to count all words that are present in given string.
Examples:
Input : words[] = { "welcome", "to", "geeks", "portal"} str = "geeksforgeeks is a computer science portal for geeks." Output : 2 Two words "portal" and "geeks" is present in str. Input : words[] = {"Save", "Water", "Save", "Yourself"} str = "Save" Output :1
Steps:
- Extract each word from string.
- For each word, check if it is in word array(by creating set/map). If present, increment result.
Implementation:
C++
#include<bits/stdc++.h> using namespace std;
// function to count occurrence int countOccurrence(string word[], int n, string str)
{ // counter
int counter = 0;
// for extracting words
regex reg( "[a-zA-Z]+" );
sregex_iterator it(str.begin(), str.end(), reg);
// HashSet for quick check whether
// a word in str present in word[] or not
unordered_set<string> hs;
for ( int i=0; i<n; i++)
hs.insert(word[i]);
while (it != sregex_iterator())
{
if (hs.find(it->str()) != hs.end())
++counter;
it++;
}
return counter;
} int main()
{ string word[] = { "welcome" , "to" , "geeks" , "portal" };
int n = sizeof (word)/ sizeof (word[0]);
string str = "geeksforgeeks is a computer science portal for geeks." ;
cout << countOccurrence(word, n, str);
return 0;
} // This code contributed by Ajax |
Java
// Java program to count number // of words present in a string import java.util.HashSet;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test {
static int countOccurrence(String[] word, String str)
{
// counter
int counter = 0 ;
// for extracting words
Pattern p = Pattern.compile( "[a-zA-Z]+" );
Matcher m = p.matcher(str);
// HashSet for quick check whether
// a word in str present in word[] or not
HashSet<String> hs = new HashSet<String>();
for (String string : word) {
hs.add(string);
}
while (m.find()) {
if (hs.contains(m.group()))
counter++;
}
return counter;
}
public static void main(String[] args)
{
String word[]
= { "welcome" , "to" , "geeks" , "portal" };
String str
= "geeksforgeeks is a computer science portal for geeks." ;
System.out.println(countOccurrence(word, str));
}
} |
Python3
import re
# function to count occurrence def countOccurrence(word, n, str ):
# counter
counter = 0
# for extracting words
reg = re. compile ( "[a-zA-Z]+" )
it = re.finditer(reg, str )
# HashSet for quick check whether
# a word in str present in word[] or not
hs = set (word)
for match in it:
if match.group() in hs:
counter + = 1
return counter
word = [ "welcome" , "to" , "geeks" , "portal" ]
n = len (word)
str = "geeksforgeeks is a computer science portal for geeks."
print (countOccurrence(word, n, str ))
|
Javascript
// function to count occurrence of words in a string function countOccurrence(word, n, str) {
// counter
let counter = 0;
// for extracting words
let reg = new RegExp( "[a-zA-Z]+" , "g" );
let match;
// HashMap for quick check whether
// a word in str present in word[] or not
let hm = new Map();
for (let i = 0; i < n; i++) {
hm.set(word[i], true );
}
while ((match = reg.exec(str)) !== null ) {
if (hm.has(match[0])) {
counter++;
}
}
return counter;
} let word = [ "welcome" , "to" , "geeks" , "portal" ];
let n = word.length; let str = "geeksforgeeks is a computer science portal for geeks." ;
console.log(countOccurrence(word, n, str)); // This code is contributed by Shivam Tiwari |
C#
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;
public class GFG{
static void Main( string [] args)
{
string [] word = { "welcome" , "to" , "geeks" , "portal" };
int n = word.Length;
string str = "geeksforgeeks is a computer science portal for geeks." ;
Console.WriteLine(CountOccurrence(word, n, str));
}
static int CountOccurrence( string [] word, int n, string str)
{
int counter = 0;
Regex reg = new Regex( @"[a-zA-Z]+" );
MatchCollection matches = reg.Matches(str);
HashSet< string > hs = new HashSet< string >(word);
foreach (Match match in matches)
{
if (hs.Contains(match.Value))
{
counter++;
}
}
return counter;
}
}
} // This code is contributed by Shivam Tiwari |
Output
2
Time Complexity: O(N), N is the length of the word.
Auxiliary Space: O(1).
Method: First iterate the words using for loop and Check if that word is present in the given string or not if present then increments the count value. Finally, print the count value.
Implementation:
C++
// C++ program to count number // of words present in a string #include<bits/stdc++.h> using namespace std;
int main()
{ vector<string> w({ "welcome" , "to" , "geeks" , "portal" });
string s = "geeksforgeeks is a computer science portal for geeks." ;
int c = 0;
// iterating words
for ( int i=0; i<w.size(); i++)
{
// checking if word is present in the string or
// not
if (s.find(w[i]) != string::npos)
{
// if present then it increments the count
// value
c = c+1;
}
}
// printing the result
cout<< c <<endl;
return 0;
} // This code is contributed by Srj_27 |
Java
// A Java program to implement above approach import java.util.*;
public class GFG {
public static void main(String[] args)
{
String[] w = { "welcome" , "to" , "geeks" , "portal" };
String s
= "geeksforgeeks is a computer science portal for geeks." ;
int c = 0 ;
// iterating words
for ( int i = 0 ; i < w.length; i++) {
// checking if word is present in the string or
// not
if (s.contains(w[i])) {
// if present then it increments the count
// value
c = c + 1 ;
}
}
// printing the result
System.out.println(c);
}
} // This code is contributed by Karandeep1234 |
Python3
# python code to count words present in a string w = [ "welcome" , "to" , "geeks" , "portal" ]
s = "geeksforgeeks is a computer science portal for geeks."
c = 0
#iterating words for i in w:
#checking if word is present in the string or not
if i in s:
# if present then it increments the count value
c + = 1
# printing the result print (c)
# this code is contributed by gangarajula laxmi |
C#
using System;
public class GFG {
static public void Main()
{
string [] w = { "welcome" , "to" , "geeks" , "portal" };
string s
= "geeksforgeeks is a computer science portal for geeks." ;
int c = 0;
// iterating words
for ( int i = 0; i < w.Length; i++) {
// checking if word is present in the string or
// not
if (s.Contains(w[i])) {
// if present then it increments the count
// value
c = c + 1;
}
}
// printing the result
Console.WriteLine(c);
}
} // This code is contributed by Potta Lokesh |
Javascript
// Javascript program to count number // of words present in a string let w= [ "welcome" , "to" , "geeks" , "portal" ];
let s = "geeksforgeeks is a computer science portal for geeks." ;
let c = 0;
// iterating words
for (let i = 0; i < w.length; i++)
{
// checking if word is present in the string or
// not
if (s.indexOf(w[i]) != -1)
{
// if present then it increments the count
// value
c = c+1;
}
}
// printing the result
console.log(c)
// This code is contributed by poojaagarwal2. |
Output
2
Time Complexity: O(N), N is the size of w.
Auxiliary Space:: O(1).
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