Given a string, the task is to count the number of words whose sum of Ascii values is less than and greater than or equal to given k.
Examples:
Input: str = "Learn how to code", k = 400 Output: Number of words having sum of ascii less than k = 2 Number of words having sum of ascii greater than or equal to k = 2 Input: str = "Geeks for Geeks", k = 400 Output: Number of words having sum of ascii less than k = 1 Number of words having sum of ascii greater than or equal to k = 2
Brute Force Approach:
The approach is to solve this problem is to first split the given string into individual words and calculate the sum of ASCII values of each word. Then, we can iterate through each word and check whether its sum of ASCII values is less than or greater than or equal to the given value k. Finally, we can count the number of words that satisfy each condition and print the results.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;// Function to split the given string into individual wordsvector<string> splitWords(string str){ vector<string> words; stringstream ss(str); string word; while (ss >> word) words.push_back(word); return words;}// Function to count the wordsvoid CountWords(string str, int k){ vector<string> words = splitWords(str); int lessThanK = 0; int greaterThanOrEqualK = 0; for (string word : words) { int sum = 0; for (char c : word) sum += c; if (sum < k) lessThanK++; else greaterThanOrEqualK++; } cout << "Number of words having sum of ascii less than k = " << lessThanK << endl; cout << "Number of words having sum of ascii greater than or equal to k = " << greaterThanOrEqualK << endl;}// Driver codeint main(){ string str = "Learn how to code"; int k = 400; CountWords(str, k); return 0;} |
Java
import java.util.*;public class Main { // Function to split the given string into individual words static List<String> splitWords(String str) { List<String> words = new ArrayList<>(); Scanner sc = new Scanner(str); while (sc.hasNext()) { words.add(sc.next()); } return words; } // Function to count the words static void countWords(String str, int k) { List<String> words = splitWords(str); int lessThanK = 0; int greaterThanOrEqualK = 0; for (String word : words) { int sum = 0; for (char c : word.toCharArray()) { sum += (int) c; } if (sum < k) { lessThanK++; } else { greaterThanOrEqualK++; } } System.out.println("Number of words having sum of ascii less than k = " + lessThanK); System.out.println("Number of words having sum of ascii greater than or equal to k = " + greaterThanOrEqualK); } // Driver code public static void main(String[] args) { String str = "Learn how to code"; int k = 400; countWords(str, k); }} |
Python3
# Importing the regular expression moduleimport re# Function to split the given string into individual wordsdef splitWords(s): return re.findall(r'\w+', s)# Function to count the wordsdef countWords(s, k): words = splitWords(s) lessThanK = 0 greaterThanOrEqualK = 0 for word in words: sum = 0 for c in word: sum += ord(c) if sum < k: lessThanK += 1 else: greaterThanOrEqualK += 1 print("Number of words having sum of ascii less than k =", lessThanK) print("Number of words having sum of ascii greater than or equal to k =", greaterThanOrEqualK)# Driver codestr = "Learn how to code"k = 400countWords(str, k) |
C#
using System;using System.Collections.Generic;using System.Linq;using System.Text.RegularExpressions;class Program { // Function to split the given string into individual // words static List<string> SplitWords(string str) { List<string> words = new List<string>(); Regex regex = new Regex(@"\w+"); MatchCollection matches = regex.Matches(str); foreach(Match match in matches) { words.Add(match.Value); } return words; } // Function to count the words static void CountWords(string str, int k) { List<string> words = SplitWords(str); int lessThanK = 0; int greaterThanOrEqualK = 0; foreach(string word in words) { int sum = 0; foreach(char c in word) { sum += c; } if (sum < k) { lessThanK++; } else { greaterThanOrEqualK++; } } Console.WriteLine( "Number of words having sum of ascii less than k = " + lessThanK); Console.WriteLine( "Number of words having sum of ascii greater than or equal to k = " + greaterThanOrEqualK); } // Driver code static void Main(string[] args) { string str = "Learn how to code"; int k = 400; CountWords(str, k); }} |
Javascript
// Function to split the given string into individual wordsfunction splitWords(str) { let words = []; let sc = str.split(" "); for (let i = 0; i < sc.length; i++) { words.push(sc[i]); } return words;}// Function to count the wordsfunction countWords(str, k) { let words = splitWords(str); let lessThanK = 0; let greaterThanOrEqualK = 0; for (let word of words) { let sum = 0; for (let c of word.split('')) { sum += c.charCodeAt(0); } if (sum < k) { lessThanK++; } else { greaterThanOrEqualK++; } } console.log("Number of words having sum of ascii less than k = " + lessThanK); console.log("Number of words having sum of ascii greater than or equal to k = " + greaterThanOrEqualK);}// Driver codelet str = "Learn how to code";let k = 400;countWords(str, k); |
Output
Number of words having sum of ascii less than k = 2 Number of words having sum of ascii greater than or equal to k = 2
Time Complexity: O(n*m), where n is the length of the string and m is the average length of each word in the string.
Space Complexity: O(m), as we are using a vector to store each word in the string.
Approach: Count the number of words having the sum of ASCII values less than k and subtract it from the total number of words to get the number of words having ASCII values to the sum greater than or equal to k. Start traversing the string letter by letter and add the ASCII value to sum. If there is a space then increment the count if the sum is less than k and will also set the sum to 0.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to count the wordsvoid CountWords(string str, int k){ // Sum of ascii values int sum = 0; int NumberOfWords = 0; // Number of words having // sum of ascii less than k int counter = 0; int len = str.length(); for (int i = 0; i < len; ++i) { // If character is a space if (str[i] == ' ') { if (sum < k) counter++; sum = 0; NumberOfWords++; } else // Add the ascii value to sum sum += str[i]; } // Handling the Last word separately NumberOfWords++; if (sum < k) counter++; cout << "Number of words having sum of ASCII" " values less than k = " << counter << endl; cout << "Number of words having sum of ASCII values" " greater than or equal to k = " << NumberOfWords - counter;}// Driver codeint main(){ string str = "Learn how to code"; int k = 400; CountWords(str, k); return 0;} |
Java
// Java implementation of the // above approach class GFG{ // Function to count the words static void CountWords(String str, int k) { // Sum of ascii values int sum = 0; int NumberOfWords = 0; // Number of words having // sum of ascii less than k int counter = 0; int len = str.length(); for (int i = 0; i < len; ++i) { // If character is a space if (str.charAt(i) == ' ') { if (sum < k) { counter++; } sum = 0; NumberOfWords++; } else // Add the ascii value to sum { sum += str.charAt(i); } } // Handling the Last word separately NumberOfWords++; if (sum < k) { counter++; } System.out.println("Number of words having sum " + "of ASCII values less than k = " + counter); System.out.println("Number of words having sum of " + "ASCII values greater than or equal to k = " + (NumberOfWords - counter));}// Driver code public static void main(String[] args) { String str = "Learn how to code"; int k = 400; CountWords(str, k);}}// This code is contributed by RAJPUT-JI |
Python 3
# Python 3 implementation of the # above approach# Function to count the wordsdef CountWords(str, k): # Sum of ascii values sum = 0 NumberOfWords = 0 # Number of words having # sum of ascii less than k counter = 0 l = len(str) for i in range(l): # If character is a space if (str[i] == ' ') : if (sum < k): counter += 1 sum = 0 NumberOfWords += 1 else: # Add the ascii value to sum sum += ord(str[i]) # Handling the Last word separately NumberOfWords += 1 if (sum < k): counter += 1 print("Number of words having sum of ASCII", "values less than k =", counter) print("Number of words having sum of ASCII values", "greater than or equal to k =", NumberOfWords - counter)# Driver codeif __name__ == "__main__": str = "Learn how to code" k = 400 CountWords(str, k)# This code is contributed # by ChitraNayal |
C#
// C# implementation of the // above approach using System;class GFG {// Function to count the words static void CountWords(String str, int k) { // Sum of ascii values int sum = 0; int NumberOfWords = 0; // Number of words having // sum of ascii less than k int counter = 0; int len = str.Length; for (int i = 0; i < len; ++i) { // If character is a space if (str[i]==' ') { if (sum < k) { counter++; } sum = 0; NumberOfWords++; } else // Add the ascii value to sum { sum += str[i]; } } // Handling the Last word // separately NumberOfWords++; if (sum < k) { counter++; } Console.WriteLine("Number of words having sum " + "of ASCII values less than k = " + counter); Console.WriteLine("Number of words having sum of " + "ASCII values greater than or equal to k = " + (NumberOfWords - counter));}// Driver code public static void Main(String[] args){ String str = "Learn how to code"; int k = 400; CountWords(str, k);}}// This code is contributed by RAJPUT-JI |
PHP
<?php// PHP implementation of the above approach// Function to count the wordsfunction CountWords($str, $k){ // Sum of ascii values $sum = 0; $NumberOfWords = 0; // Number of words having // sum of ascii less than k $counter = 0; $len = strlen($str); for ($i = 0; $i < $len; ++$i) { // If character is a space if ($str[$i] == ' ') { if ($sum < $k) $counter++; $sum = 0; $NumberOfWords++; } else // Add the ascii value to sum $sum += ord($str[$i]); } // Handling the Last word separately $NumberOfWords++; if ($sum < $k) $counter++; echo "Number of words having sum of ASCII" . " values less than k = " . $counter . "\n"; echo "Number of words having sum of ASCII " . "values greater than or equal to k = " . ($NumberOfWords - $counter);}// Driver code$str = "Learn how to code";$k = 400;CountWords($str, $k);// This code is contributed by mits?> |
Javascript
<script>// JavaScript implementation of the // above approach// Function to count the wordsfunction CountWords(str, k){ // Sum of ascii values let sum = 0; let NumberOfWords = 0; // Number of words having // sum of ascii less than k let counter = 0; let len = str.length; for (let i = 0; i < len; ++i) { // If character is a space if (str[i] == " ") { if (sum < k) { counter++; } sum = 0; NumberOfWords++; } else { // Add the ascii value to sum sum += str.charCodeAt(i); } } // Handling the Last word separately NumberOfWords++; if (sum < k) { counter++; } document.write("Number of words having sum " + "of ASCII values less than k = " + counter + "<br>"); document.write("Number of words having sum of " + "ASCII values greater than or equal to k = " + (NumberOfWords - counter) + "<br>");}// Driver code let str = "Learn how to code"; let k = 400; CountWords(str, k);// This code is contributed by Surbhi Tyagi.</script> |
Output
Number of words having sum of ASCII values less than k = 2 Number of words having sum of ASCII values greater than or equal to k = 2
Complexity Analysis:
- Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time
- Auxiliary Space: O(1), as we are not using any extra space.
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