Question

Please update this java code so 1. A scanner can be used to import the files. Then compute and test with a randomly generated array instead of a file. Compute all possible distinctive pair of values from an an array. [A,B,C, A]

A,B

A,C

B,C

- i.e. (A,B)==(B,A)

(A, A) is not valid

hint: sort the array first

Example: input [1, 2, 3, 2, 3, 4, 3] result 4 (with distinct pairs) and 8 (with symmetric pairs
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

public class Exercise8 {

public static void main(String[] args) {
In in = new In(args[0]);
int[] values = in.readAllInts();
StdOut.println(countNumberOfPairs(values));

// Tests
int[] values1 = {1, 2, 4, 1, 2, 1, 2, 4, 5, 1, 2, 4, 5, 1, 2 ,5, 6, 7, 7, 8, 2, 1, 2, 4, 5};
StdOut.println("Equal pairs 1: " + countNumberOfPairs(values1) + " Expected: 49");

int[] values2 = {1, 1, 1};
StdOut.println("Equal pairs 2: " + countNumberOfPairs(values2) + " Expected: 3");
}

// O(n lg n) solution
private static int countNumberOfPairs(int[] values) {
Arrays.sort(values);

int count = 0;
int currentFrequency = 1;

for (int i = 1; i < values.length; i++) {
if (values[i] == values[i - 1]) {
currentFrequency++;
} else {
if (currentFrequency > 1) {
count += (currentFrequency - 1) * currentFrequency / 2;
currentFrequency = 1;
}
}
}

if (currentFrequency > 1) {
count += (currentFrequency - 1) * currentFrequency / 2;
}

return count;
}

// O(n) solution
private static int countNumberOfPairs2(int[] values) {

Map<Integer, Integer> valuesMap = new HashMap<>();
int equalNumbersCount = 0;

for(int i = 0; i < values.length; i++) {
int count = 0;
if (valuesMap.containsKey(values[i])) {
count = valuesMap.get(values[i]);
}
count++;
valuesMap.put(values[i], count);
}

for(int numberKey : valuesMap.keySet()) {
if (valuesMap.get(numberKey) > 1) {
int n = valuesMap.get(numberKey);
equalNumbersCount += (n - 1) * n / 2;
}
}

return equalNumbersCount;
}

}

Answer 1

ANS.,)

import java.io.File;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Exercise8 {

   public static void main(String args[]){
       Scanner sc = new Scanner(System.in);
       System.out.println("Enter file with full path for reading data :");
       String filePath=sc.next();
       File file = new File(filePath);
       try{
       Scanner scRead = new Scanner(file);
       // Read line by line
       while (scRead.hasNextLine()) {
           String line=scRead.next();
           //convert the line into array
           String arrTemp[]=line.trim().split(",");
           int values[]=new int[arrTemp.length];
           for(int i=0;i<arrTemp.length;i++){
   int num = Integer.parseInt(arrTemp[i]);
   values[i]=num;
           }
           System.out.println("Equal distinct pairs :"+countNumberOfPairs(values));
   }
   sc.close();
       }catch(Exception e){
           e.printStackTrace();
       }

      
   }
  
   private static int countNumberOfPairs(int[] values) {
       int count = 0;
       Arrays.sort(values);
       int currentFrequency = 1;
       for (int i = 1; i < values.length; i++) {
       if (values[i] == values[i - 1]) {
       currentFrequency++;
       } else {
       if (currentFrequency > 1) {
       count += (currentFrequency - 1) * currentFrequency / 2;
       currentFrequency = 1;
       }
       }
       }

       if (currentFrequency > 1) {
       count += (currentFrequency - 1) * currentFrequency / 2;
       }
      
       return count;
   }
  
  

      
}

Input :

Enter file with full path for reading data :
D:\\tempfile.txt

in that file data is :

1,2,3,2,3,4,3
1,1,1

Output :

Equal distinct pairs :4
Equal distinct pairs :3

Explanation : The above code is reading file line by line and splitted by comma(,) and assign to array. And should be numbers only , otherwise you can get exceptions.