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sergey [27]
1 year ago
9

given an array of integers a, your task is to count the number of pairs i and j (where 0 ≤ i < j < a.length), such that a[

i] and a[j] are digit anagrams. two integers are considered to be digit anagrams if they contain the same digits. in other words, one can be obtained from the other by rearranging the digits (or trivially, if the numbers are equal). for example, 54275 and 45572 are digit anagrams, but 321 and 782 are not (since they don't contain the same digits). 220 and 22 are also not considered as digit anagrams, since they don't even have the same number of digits.
Computers and Technology
1 answer:
Nimfa-mama [501]1 year ago
4 0

Using the knowledge of computational language in C++ it is possible to write a code that given an array of integers a, your task is to count the number of pairs i and j.

<h3>Writting the code:</h3>

<em>// C++ program for the above approach</em>

<em> </em>

<em>#include <bits/stdc++.h></em>

<em>using namespace std;</em>

<em> </em>

<em>// Function to find the count required pairs</em>

<em>void getPairs(int arr[], int N, int K)</em>

<em>{</em>

<em>    // Stores count of pairs</em>

<em>    int count = 0;</em>

<em> </em>

<em>    // Traverse the array</em>

<em>    for (int i = 0; i < N; i++) {</em>

<em> </em>

<em>        for (int j = i + 1; j < N; j++) {</em>

<em> </em>

<em>            // Check if the condition</em>

<em>            // is satisfied or not</em>

<em>            if (arr[i] > K * arr[j])</em>

<em>                count++;</em>

<em>        }</em>

<em>    }</em>

<em>    cout << count;</em>

<em>}</em>

<em> </em>

<em>// Driver Code</em>

<em>int main()</em>

<em>{</em>

<em>    int arr[] = { 5, 6, 2, 5 };</em>

<em>    int N = sizeof(arr) / sizeof(arr[0]);</em>

<em>    int K = 2;</em>

<em> </em>

<em>    // Function Call</em>

<em>    getPairs(arr, N, K);</em>

<em> </em>

<em>    return 0;</em>

<em>}</em>

See more about C++ code at brainly.com/question/17544466

#SPJ4

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8 0
3 years ago
How to do assignment 5 on edhesive
s2008m [1.1K]

Answer:

//To pass in a string, and pass in a pivot string, and at that moment print in obligatory sequence, the string after pivot string till last as first string, and published beforehand the left slice of string.

import java.util.Scanner;

public class Main {

public static void main(String args[]) {

  Scanner s2 = new Scanner(System.in);

  System.out.println("Enter the parent String:");

  String f2 = s2.nextLine();

  System.out.println("Enter pivot(child string):");

  String piv1 = s2.nextLine();

  int start2 = f2.indexOf(piv1);

  int end2 = piv1.length()+start2;

  if (start2 == -1){

    System.out.println("Error: child string(Pivot) not discovered.");

    return;

}

String first2 = f2.substring(0,start2-1);

  String second2 = f2.substring(end2);

if (f2.charAt(start2-1)==' '){

System.out.println(second2 + " " + piv1 + " " + first2);

return;

}

System.out.println(second2 + piv1 + first2);

}

}

Explanation:

Please check the answer section.

6 0
3 years ago
Why are the keys on the qwerty keyboard arranged the way they are?
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7 0
3 years ago
(JAVA PLS)
Elza [17]

import java.util.Scanner;

public class JavaApplication54 {

   public static int appearance(String word, char letter){

       int count = 0;

       for (int i = 0; i < word.length(); i++){

           if (word.charAt(i) == letter){

               count++;

           }

       }

       return count;

   }

   public static void main(String[] args) {

       Scanner scan = new Scanner(System.in);

       System.out.println("Type the message to be shortened");

       String message = scan.nextLine();

       message = message.toLowerCase();

       int volCount = 0, repCount = 0, num = 0;

       char prevC = ' ';

       String vowels = "aeiou";

       String newMessage = "";

       for (int i = 0; i < message.length(); i++){

           char c = message.charAt(i);

           

           if (vowels.indexOf(c) != -1 && prevC == ' '){

               newMessage += c;

           }

           else if (vowels.indexOf(c) == -1 && prevC != c){

               newMessage += c;

           }

           else if (vowels.indexOf(c) != -1){

               volCount ++;

           }

           else{

               repCount++;

           }

            prevC = c;

       }

       System.out.println("Algorithm 1");

       System.out.println("Vowels removed: "+volCount);

       System.out.println("Repeats removed: "+repCount);

       System.out.println("Algorithm 1 message: "+newMessage);

       System.out.println("Algorithm 1 characters saved: "+(message.length() - newMessage.length()));

       

       String uniqueMessage = "";

       

       for (int i = 0; i<message.length(); i++){

           char w = message.charAt(i);

           if (uniqueMessage.indexOf(w)== -1 && w!=' '){

               uniqueMessage += appearance(message,w)+""+w;

               num++;

           }

       }

       System.out.println("Algorithm 2");

       System.out.println("Unique characters found: "+num);

       System.out.println("Algorithm 2 message: "+uniqueMessage);

       System.out.println("Algorithm 2 characters saved: "+(message.length() - uniqueMessage.length()));

   }

   

}

I hope this helps!

6 0
3 years ago
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