Print distinct absolute differences of all possible pairs from a given array

Given an array, arr[] of size N, the task is to find the distinct absolute differences of all possible pairs of the given array.

Examples:

Input: arr[] = { 1, 3, 6 } 
Output: 2 3 5 
Explanation:
abs(arr[0] – arr[1]) = 2 
abs(arr[1] – arr[2]) = 3 
abs(arr[0] – arr[2]) = 5 

Input: arr[] = { 5, 6, 7, 8, 14, 19, 21, 22 } 
Output: 1 2 3 5 6 7 8 9 11 12 13 14 15 16 17

Naive Approach: The simplest approach to solve this problem is to generate all possible pairs of the given array and insert the absolute difference of each pair in a Set. Finally, print all the elements of the set. 
Time Complexity: O(N² * log(N))
Auxiliary Space: O(N²)

Approach: The above approach can be optimized using Bitset. Follow the steps below to solve the problem:

• Initialize a Bitset, say bset, where bset[i] check if i is present in the array or not.
• Traverse the array arr[] and store all the array elements in the bset.
• Initialize a Bitset, say diff, where diff[i] stores if the absolute difference of there exists any pair in the array whose value equal to i or not.
• Find the largest element of the array, say Max
• Iterate over the range [0, Max]. In every i^th iteration check if bset[i] is true or not. If found to be true, then insert the absolute difference of i with all other array elements using diff = diff | (bset >> i).
• Finally, iterate over the range [0, Max] and check if diff[i] is true or not. If found to be true, then print i.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;
#define Max 100005

// Function to find all distinct
// absolute difference of all
// possible pairs of the array
void printUniqDif(int n, int a[])
{

    // bset[i]: Check if i is present
    // in the array or not
    bitset<Max> bset;

    // diff[i]: Check if there exists a
    // pair whose absolute difference is i
    bitset<Max> diff;

    // Traverse the array, arr[]
    for (int i = 0; i < n; i++) {

        // Add in bitset
        bset.set(a[i]);
    }

    // Iterate over the range[0, Max]
    for (int i = 0; i <= Max; i++) {

        // If i-th bit is set
        if (bset[i]) {

            // Insert the absolute difference
            // of all possible pairs whose
            // first element is arr[i]
            diff = diff | (bset >> i);
        }
    }

    // Stores count of set bits
    int X = bset.count();

    // If there is at least one
    // duplicate element in arr[]
    if (X != n) {

        cout << 0 << " ";
    }

    // Printing the distinct absolute
    // differences of all possible pairs
    for (int i = 1; i <= Max; i++) {

        // If i-th bit is set
        if (diff[i]) {
            cout << i << " ";
        }
    }
}

// Driver Code
int main()
{

    // Given array
    int a[] = { 1, 4, 6 };

    // Given size
    int n = sizeof(a) / sizeof(a[0]);

    // Function Call
    printUniqDif(n, a);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG
{
  static int Max = 100005;

  // Function to find all distinct
  // absolute difference of all
  // possible pairs of the array
  static void printUniqDif(int n, int[] a)
  {
    // bset[i] Check if i is present
    // in the array or not
    int[] bset = new int[33];

    // diff[i] Check if there exists a
    // pair whose absolute difference is i
    int diff = 0;

    // Traverse the array, arr[]
    for (var i = 0; i < n; i++)
      bset[a[i]] = 1;

    // Iterate over the range[0, Max]
    int d = 0;

    for (var i = 0; i < 33; i++)
      d = d | (bset[i] << i);

    for (var i = 0; i < 33; i++)
    {
      // If i-th bit is set
      if (bset[i] == 1)

        // Insert the absolute difference
        // of all possible pairs whose
        // first element is arr[i]
        diff |= (d >> i);
    }

    // Stores count of set bits
    int X =  0;
    for (int i = 0; i < 33; i++)
        if (bset[i] == 1)
            X++;

    String Y =Integer.toBinaryString(diff);

    // If there is at least one
    // duplicate element in arr[]
    if (X != n)
      System.out.print("0 ");

    // Printing the distinct absolute
    // differences of all possible pairs
    for (var i = 1; i < Y.length(); i++)
    {

      // If i-th bit is set
      if (Y.charAt(i) == '1')
        System.out.print(i + " ");
    }
  }

  // Driver Code
  public static void main(String[] args)
  {

    // Given array
    int[] a = {1, 4, 6};

    // Given size
    int n = a.length;

    // Function Call
    printUniqDif(n, a);
  }
}
// This code is contributed by phasing17

Python3

# Python3 program for the above approach
Max = 100005

# Function to find all distinct
# absolute difference of all
# possible pairs of the array
def printUniqDif(n, a):

    # bset[i]: Check if i is present
    # in the array or not
    bset = [0 for i in range(33)]

    # diff[i]: Check if there exists a
    # pair whose absolute difference is i
    diff = 0

    # Traverse the array, arr[]
    for i in range(n):
        bset[a[i]] = 1

    # Iterate over the range[0, Max]
    d = 0

    for i in range(1,33):
        d = d | (bset[i]<<i)

    for i in range(33):

        # If i-th bit is set
        if (bset[i]):

            # Insert the absolute difference
            # of all possible pairs whose
            # first element is arr[i]
            diff = diff | d >> i
            # print(bin(diff))

    # Stores count of set bits
    X, Y = bset.count(1), str(bin(diff)[2:])

    # If there is at least one
    # duplicate element in arr[]
    if (X != n):

        print(0, end=" ")

    # Printing the distinct absolute
    # differences of all possible pairs
    for i in range(1, len(Y)):

        # If i-th bit is set
        if (Y[i] == '1'):
            print(i, end = " ")
# Driver Code
if __name__ == '__main__':

    # Given array
    a = [1, 4, 6]

    # Given size
    n = len(a)

    # Function Call
    printUniqDif(n, a)

    # This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Linq;
using System.Collections.Generic;

class GFG
{
  static int Max = 100005;

  // Function to find all distinct
  // absolute difference of all
  // possible pairs of the array
  static void printUniqDif(int n, int[] a)
  {
    // bset[i] Check if i is present
    // in the array or not
    int[] bset = new int[33];

    // diff[i] Check if there exists a
    // pair whose absolute difference is i
    int diff = 0;

    // Traverse the array, arr[]
    for (var i = 0; i < n; i++)
      bset[a[i]] = 1;

    // Iterate over the range[0, Max]
    int d = 0;

    for (var i = 0; i < 33; i++)
      d = d | (bset[i] << i);

    for (var i = 0; i < 33; i++)
    {
      // If i-th bit is set
      if (bset[i] == 1)

        // Insert the absolute difference
        // of all possible pairs whose
        // first element is arr[i]
        diff |= (d >> i);
    }

    // Stores count of set bits
    int X =  bset.Count(f => f == 1);

    string Y = Convert.ToString(diff, 2);

    // If there is at least one
    // duplicate element in arr[]
    if (X != n)
      Console.Write("0 ");

    // Printing the distinct absolute
    // differences of all possible pairs
    for (var i = 1; i < Y.Length; i++)
    {
      // If i-th bit is set
      if (Y[i] == '1')
        Console.Write(i + " ");
    }
  }

  // Driver Code

  public static void Main(string[] args)
  {
    // Given array
    int[] a = {1, 4, 6};

    // Given size
    int n = a.Length;

    // Function Call
    printUniqDif(n, a);
  }
}

// This code is contributed by phasing17

Javascript

// JS program for the above approach
let Max = 100005

// Function to find all distinct
// absolute difference of all
// possible pairs of the array
function printUniqDif(n, a)
{
    // bset[i] Check if i is present
    // in the array or not
    let bset = new Array(33).fill(0)

    // diff[i] Check if there exists a
    // pair whose absolute difference is i
    let diff = 0

    // Traverse the array, arr[]
    for (var i = 0; i < n; i++)
        bset[a[i]] = 1

    // Iterate over the range[0, Max]
    let d = 0

    for (var i = 0; i < 33; i++)
        d = d | (bset[i] << i)

    for (var i = 0; i < 33; i++)
    {
        // If i-th bit is set
        if (bset[i] == 1)

            // Insert the absolute difference
            // of all possible pairs whose
            // first element is arr[i]
            diff |= (d >> i)
    }

    // Stores count of set bits
    let X = bset.filter(x => x == 1).length

    let Y = diff.toString(2)



    // If there is at least one
    // duplicate element in arr[]
    if (X != n)
        process.stdout.write("0 ")

    // Printing the distinct absolute
    // differences of all possible pairs
    for (var i = 1; i < Y.length; i++)

        // If i-th bit is set
        if (Y.charAt(i) == '1')
            process.stdout.write(i + " ")
}

// Driver Code

// Given array
let a = [1, 4, 6]

// Given size
let n = a.length

// Function Call
printUniqDif(n, a)


// This code is contributed by phasing17

Output:

2 3 5

Time Complexity:O(N + Max), where Max is the largest element of the array.
Auxiliary Space: O(Max)