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

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  • Difficulty Level : Hard
  • Last Updated : 22 Dec, 2022
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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 ith 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)


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