HackerRank Closest Numbers Problem Solution Yashwant Parihar, April 26, 2023May 6, 2023 In this post, we will solve HackerRank Closest Numbers Problem Solution.Sorting is useful as the first step in many different tasks. The most common task is to make finding things easier, but there are other uses as well. In this case, it will make it easier to determine which pair or pairs of elements have the smallest absolute difference between them.Examplearr = [5, 2, 3, 4, 1]Sorted, arr’ = [1, 2, 3, 4, 5]. Several pairs have the minimum difference of 1:[(1, 2), (2, 3), (3, 4), (4, 5)]. Return the array [1, 2, 2, 3, 3, 4, 4, 5].NoteAs shown in the example, pairs may overlap.Given a list of unsorted integers, arr, find the pair of elements that have the smallest absolute difference between them. If there are multiple pairs, find them all.Function DescriptionComplete the closestNumbers function in the editor below.closestNumbers has the following parameter(s):int arr[n]: an array of integersReturns– int[]: an array of integers as describedInput FormatThe first line contains a single integer n. the length of arr.The second line contains n space-separated integers, arr[i].Output FormatSample Input 010 -20 -3916237 -357920 -3620601 7374819 -7330761 30 6246457 -6461594 266854 Sample Output 0-20 30 Explanation 0(30) – (-20) = 50, which is the smallest difference.Sample Input 112 -20 -3916237 -357920 -3620601 7374819 -7330761 30 6246457 -6461594 266854 -520 -470 Sample Output 1-520 -470 -20 30 Explanation 1(-470) – (-520) = 30 – (-20) = 50, which is the smallest difference.Sample Input 24 5 4 3 2 Sample Output 22 3 3 4 4 5 Explanation 2Here, the minimum difference is 1. Valid pairs are (2, 3), (3, 4), and (4, 5).HackerRank Closest Numbers Problem SolutionClosest Numbers C Solution#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> long int cmpfunc (const void * a, const void * b) { return ( *(long int*)a - *(long int*)b ); } int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ int n,i=0; long int a[200000],diff,small=100000000; scanf("%d",&n); for(i=0;i<n;i++) scanf("%ld",&a[i]); qsort(a,n,sizeof(long int),cmpfunc); for(i=1;i<n;i++) { diff = a[i]-a[i-1]; if(diff<0) diff*=-1; if(diff<small) { small = diff; } } for(i=1;i<n;i++) { diff = a[i]-a[i-1]; if(diff<0) diff*=-1; if(diff==small) printf("%ld %ld ",a[i-1],a[i]); } return 0; }Closest Numbers C++ Solution#include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> int n, a[200005]; int main() { scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&a[i]); std::sort(a,a+n); int diff = 2e9; for(int i=0;i+1<n;i++) diff = std::min(diff, a[i+1]-a[i]); for(int i=0;i+1<n;i++) if(a[i+1]-a[i]==diff) printf("%d %d ",a[i],a[i+1]); return 0; }Closest Numbers C Sharp Solutionusing System; using System.Collections.Generic; using System.Linq; using System.IO; class Solution { const int MAX = 20000002; static int[] bit = new int[MAX]; static bool[] flag = new bool[MAX]; static int min = MAX; static List<KeyValuePair<int, int>> L; static void add_bit(int n) { int max = n; flag[n] = true; n += n & -n; while (n < MAX && bit[n] < max) { bit[n] = max; n += n & -n; } } static void find_pair(int n) { int diff = 0, N = n, k = 0; while (n > 0) { if (n != N && flag[n]) { diff = N - n; k = n; break; } if (bit[n] > 0) { diff = N - bit[n]; k = bit[n]; break; } n -= n & -n; } if (diff > 0) { if (diff < min) { min = diff; L = new List<KeyValuePair<int, int>>(); L.Add(new KeyValuePair<int, int>(k - 10000001, N - 10000001)); } else if (diff == min) { L.Add(new KeyValuePair<int, int>(k - 10000001, N - 10000001)); } } } static void Main(String[] args) { int N = int.Parse(Console.ReadLine()); string[] s = Console.ReadLine().Split(); int[] a = new int[N]; for (int i = 0; i < N; i++) { a[i] = int.Parse(s[i]) + 10000001; } for (int i = 0; i < N; i++) { add_bit(a[i]); } for (int i = 0; i < N; i++) { find_pair(a[i]); } int length = L.Count; L = L.OrderBy(x => x.Key).ToList(); for (int i = 0; i < length; i++) { Console.Write(L[i].Key + " "); Console.Write(L[i].Value + " "); } } }Closest Numbers Java Solutionimport java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.function.*; import java.util.regex.*; import java.util.stream.*; import static java.util.stream.Collectors.joining; import static java.util.stream.Collectors.toList; class Result { /* * Complete the 'closestNumbers' function below. * * The function is expected to return an INTEGER_ARRAY. * The function accepts INTEGER_ARRAY arr as parameter. */ public static List<Integer> closestNumbers(List<Integer> arr) { // Write your code here // Write your code here Collections.sort(arr); Map<Integer, List<Integer>> map = new TreeMap<>(); for (int i = 1; i < arr.size(); i++) { map.computeIfAbsent(arr.get(i) - arr.get(i - 1), k -> new ArrayList<>()) .add(i); } List<Integer> ans = new ArrayList<>(); if (!map.isEmpty()) { List<Integer> ps = map.values() .iterator() .next(); for (Integer p : ps) { ans.add(arr.get(p - 1)); ans.add(arr.get(p)); } } return ans; } } public class Solution { public static void main(String[] args) throws IOException { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in)); BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); int n = Integer.parseInt(bufferedReader.readLine().trim()); List<Integer> arr = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" ")) .map(Integer::parseInt) .collect(toList()); List<Integer> result = Result.closestNumbers(arr); bufferedWriter.write( result.stream() .map(Object::toString) .collect(joining(" ")) + "\n" ); bufferedReader.close(); bufferedWriter.close(); } }Closest Numbers JavaScript Solutionfunction processData(input) { //Enter your code here var lines = input.split("\n"); var N = lines[0]; var nums = lines[1].split(" "); nums.sort(function(a,b){return a-b}); var result = ""; var maxdiff = Infinity; for (var i=1;i<nums.length;i++) { if ((nums[i]-nums[i-1]) < maxdiff) { maxdiff = nums[i]-nums[i-1]; result = nums[i-1] + " " + nums[i]; } else if ((nums[i]-nums[i-1]) == maxdiff) { result += " " + nums[i-1] + " " + nums[i]; } } console.log(result); } process.stdin.resume(); process.stdin.setEncoding("ascii"); _input = ""; process.stdin.on("data", function (input) { _input += input; }); process.stdin.on("end", function () { processData(_input); });Closest Numbers Python Solutionn = int(input()) array = list(map(int,input().split())) array.sort() min_diff = min(array[i+1]-array[i] for i in range(len(array)-1)) print(' '.join(str(array[i])+' '+str(array[i+1]) for i in range(len(array)-1) if array[i+1]-array[i] == min_diff))Other SolutionsHackerRank The Love-Letter Mystery SolutionHackerRank Find the Median Problem Solution c C# C++ HackerRank Solutions java javascript python CcppCSharpHackerrank Solutionsjavajavascriptpython