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;
vector<string> splitWords(string str)
{
vector<string> words;
stringstream ss(str);
string word;
while (ss >> word)
words.push_back(word);
return words;
}
void 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;
}
int main()
{
string str = "Learn how to code";
int k = 400;
CountWords(str, k);
return 0;
}
|
Java
import java.util.*;
public class Main {
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;
}
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);
}
public static void main(String[] args) {
String str = "Learn how to code";
int k = 400;
countWords(str, k);
}
}
|
Python3
import re
def splitWords(s):
return re.findall(r'\w+', s)
def 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)
str = "Learn how to code"
k = 400
countWords(str, k)
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;
class Program {
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;
}
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);
}
static void Main(string[] args)
{
string str = "Learn how to code";
int k = 400;
CountWords(str, k);
}
}
|
Javascript
function splitWords(str) {
let words = [];
let sc = str.split(" ");
for (let i = 0; i < sc.length; i++) {
words.push(sc[i]);
}
return words;
}
function 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);
}
let str = "Learn how to code";
let k = 400;
countWords(str, k);
|
OutputNumber 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++
#include <bits/stdc++.h>
using namespace std;
void CountWords(string str, int k)
{
int sum = 0;
int NumberOfWords = 0;
int counter = 0;
int len = str.length();
for (int i = 0; i < len; ++i) {
if (str[i] == ' ') {
if (sum < k)
counter++;
sum = 0;
NumberOfWords++;
}
else
sum += str[i];
}
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;
}
int main()
{
string str = "Learn how to code";
int k = 400;
CountWords(str, k);
return 0;
}
|
Java
class GFG
{
static void CountWords(String str, int k)
{
int sum = 0;
int NumberOfWords = 0;
int counter = 0;
int len = str.length();
for (int i = 0; i < len; ++i)
{
if (str.charAt(i) == ' ')
{
if (sum < k)
{
counter++;
}
sum = 0;
NumberOfWords++;
}
else
{
sum += str.charAt(i);
}
}
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));
}
public static void main(String[] args)
{
String str = "Learn how to code";
int k = 400;
CountWords(str, k);
}
}
|
Python 3
def CountWords(str, k):
sum = 0
NumberOfWords = 0
counter = 0
l = len(str)
for i in range(l):
if (str[i] == ' ') :
if (sum < k):
counter += 1
sum = 0
NumberOfWords += 1
else:
sum += ord(str[i])
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)
if __name__ == "__main__":
str = "Learn how to code"
k = 400
CountWords(str, k)
|
C#
using System;
class GFG
{
static void CountWords(String str,
int k)
{
int sum = 0;
int NumberOfWords = 0;
int counter = 0;
int len = str.Length;
for (int i = 0; i < len; ++i)
{
if (str[i]==' ')
{
if (sum < k)
{
counter++;
}
sum = 0;
NumberOfWords++;
}
else
{
sum += str[i];
}
}
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));
}
public static void Main(String[] args)
{
String str = "Learn how to code";
int k = 400;
CountWords(str, k);
}
}
|
PHP
<?php
function CountWords($str, $k)
{
$sum = 0;
$NumberOfWords = 0;
$counter = 0;
$len = strlen($str);
for ($i = 0; $i < $len; ++$i)
{
if ($str[$i] == ' ')
{
if ($sum < $k)
$counter++;
$sum = 0;
$NumberOfWords++;
}
else
$sum += ord($str[$i]);
}
$NumberOfWords++;
if ($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);
}
$str = "Learn how to code";
$k = 400;
CountWords($str, $k);
?>
|
Javascript
<script>
function CountWords(str, k)
{
let sum = 0;
let NumberOfWords = 0;
let counter = 0;
let len = str.length;
for (let i = 0; i < len; ++i) {
if (str[i] == " ") {
if (sum < k) {
counter++;
}
sum = 0;
NumberOfWords++;
}
else {
sum += str.charCodeAt(i);
}
}
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>");
}
let str = "Learn how to code";
let k = 400;
CountWords(str, k);
</script>
|
OutputNumber 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.