Count the number of words having sum of ASCII values less than and greater than k

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

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++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to count the words
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++;
  
    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 code
int main()
{
    string str = "Learn how to code";
    int k = 400;
    CountWords(str, k);
  
    return 0;
}

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Java














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// 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 



















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Python 3














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# Python 3 implementation of the  


# above approach 


  


# Function to count the words 


def 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 code 


if __name__ == "__main__": 


      


    str = "Learn how to code"


    k = 400


    CountWords(str, k) 


  


# This code is contributed  


# by ChitraNayal 



















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C#














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// 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 



















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PHP














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<?php 


// PHP implementation of the above approach 


  


// Function to count the words 


function 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 


?> 



















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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





Time Complexity: O(N)



          
          
          

          

    

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