Given a set of integers of N elements. The task is to print the absolute difference of all of the pairwise consecutive elements in a set. Pairwise consecutive pairs of a set of size N are accessed using iterator.
Example:
Input: s = {8, 5, 4, 3, 15, 20}
Output: 1 1 3 7 5
Explanation:
The set is : 3 4 5 8 15 20
The difference between 4 and 3 is 1
The difference between 5 and 4 is 1
The difference between 8 and 5 is 3
The difference between 15 and 8 is 7
The difference between 20 and 15 is 5Input: s = {5, 10, 15, 20}
Output: 5 5 5
Explanation:
The set is : 5 10 15 20
The difference between 10 and 5 is 5
The difference between 15 and 10 is 5
The difference between 20 and 15 is 5
The article Absolute Difference of all pairwise consecutive elements in an array covers the approach to find the absolute difference of all pairwise consecutive elements in an array.
Approach: This problem can be solved using two pointer algorithm. We will be using iterators as the two pointers to iterate the set and check for a given condition. Follow the steps below to understand the solution to the above problem:
- Declare two iterators itr1 and itr2 and both of them point to the beginning element of the set.
- Increment itr2 i.e. itr2++ at the beginning of the loop.
- Subtract values pointed by itr1 and itr2 i.e. *itr2 – *itr1.
- Increment itr1 at the end of the loop, this means *itr1++.
- If itr2 reaches the end of the set, then break the loop and exit.
In C++, the set elements are sorted and duplicates are removed before storing in the memory. Therefore, in the below C++ program the difference between the pairwise consecutive elements is computed on the sorted set as explained in the above examples.
Below is the C++ program implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the// difference of consecutive pairwise// elements in a setvoid display_difference(set<int> s){ // Declaring the set iterator set<int>::iterator itr; // Printing difference between // consecutive elements in a set set<int>::iterator itr1 = s.begin(); set<int>::iterator itr2 = s.begin(); while (1) { itr2++; if (itr2 == s.end()) break; cout << (*itr2 - *itr1) << " "; itr1++; }}// Driver codeint main(){ // Declaring the set set<int> s{ 8, 5, 4, 3, 15, 20 }; // Invoking the display_difference() // function display_difference(s); return 0;} |
Java
// Java program to implement// the above approachimport java.util.ArrayList;import java.util.HashSet;import java.util.List;import java.util.TreeSet;// Function to calculate the// difference of consecutive pairwise// elements in a setclass GFG { static void display_difference(HashSet<Integer> S) { // Printing difference between // consecutive elements in a set TreeSet<Integer> s = new TreeSet<Integer>(S); int itr1 = 0; int itr2 = 0; while (true) { itr2 += 1; ; if (itr2 >= s.size()) { break; } List<Integer> temp = new ArrayList<Integer>(); temp.addAll(s); System.out.print((temp.get(itr2) - temp.get(itr1)) + " "); itr1 += 1; } } // Driver code public static void main(String args[]) { // Declaring the set HashSet<Integer> s = new HashSet<Integer>(); s.add(8); s.add(5); s.add(4); s.add(3); s.add(15); s.add(20); // Invoking the display_difference() // function display_difference(s); }}// This code is contributed by gfgking |
Python3
# Python 3 program to implement# the above approach# Function to calculate the# difference of consecutive pairwise# elements in a setdef display_difference(s): # Printing difference between # consecutive elements in a set itr1 = 0 itr2 = 0 while (1): itr2 += 1 if (itr2 >= len(s)): break print((list(s)[itr2] - list(s)[itr1]), end=" ") itr1 += 1# Driver codeif __name__ == "__main__": # Declaring the set s = set([8, 5, 4, 3, 15, 20]) # Invoking the display_difference() # function display_difference(s) # This code is contributed by ukasp. |
C#
// C# program to implement// the above approach// Function to calculate the// difference of consecutive pairwise// elements in a setusing System;using System.Collections.Generic;public class GFG{ static void display_difference(HashSet<int> S) { // Printing difference between // consecutive elements in a set SortedSet<int> s = new SortedSet<int>(S); int itr1 = 0; int itr2 = 0; while (true) { itr2 += 1; ; if (itr2 >= s.Count) { break; } List<int> temp = new List<int>(); temp.AddRange(s); Console.Write((temp[itr2] - temp[itr1]) + " "); itr1 += 1; } } // Driver code public static void Main(String []args) { // Declaring the set HashSet<int> s = new HashSet<int>(); s.Add(8); s.Add(5); s.Add(4); s.Add(3); s.Add(15); s.Add(20); // Invoking the display_difference() // function display_difference(s); }}// This code is contributed by Rajput-Ji. |
Javascript
<script>// Javascript program to implement// the above approach// Function to calculate the// difference of consecutive pairwise// elements in a setfunction display_difference(s) { // Printing difference between // consecutive elements in a set s = new Set([...s].sort((a, b) => a - b)); let itr1 = 0 let itr2 = 0 while (1) { itr2 += 1 if (itr2 >= s.size) { break } document.write(([...s][itr2] - [...s][itr1]) + " ") itr1 += 1 }}// Driver code// Declaring the setlet s = new Set([8, 5, 4, 3, 15, 20]);// Invoking the display_difference()// functiondisplay_difference(s)// This code is contributed by Saurabh Jaiswal</script> |
Output:
1 1 3 7 5
Time Complexity: O(n)
Auxiliary Space: O(n)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
