Hackerrank Closest Numbers problem solution

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.

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 Description

Complete the closestNumbers function in the editor below.

closestNumbers has the following parameter(s):

• int arr[n]: an array of integers

Returns
– int[]: an array of integers as described

Hackerrank Closest Numbers problem solution in Python.

#!/usr/bin/env python3

N = int(input())
arr = [int(x) for x in input().split()]
arr.sort()

result = []
diff = 2**31-1
for i in range(1, len(arr)):
    if arr[i] - arr[i-1] <= diff:
        if arr[i] - arr[i-1] < diff:
            diff = arr[i] - arr[i-1]
            result = []
        result.append(arr[i])
        result.append(arr[i-1])

print(' '.join([str(x) for x in sorted(result)]))

Closest Numbers problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] a = new int[N];
        for(int i = 0; i < N; i++){
            a[i] = sc.nextInt();
        }
        Arrays.sort(a);
        int min = Integer.MAX_VALUE;
        for(int i = 0 ; i< N-1; i++){
            min = Math.min(min, a[i+1] - a[i]);
        }
        StringBuilder sb = new StringBuilder("");
        for(int i = 0; i < N-1; i++){
            if(a[i+1] - a[i] == min){
                sb.append(a[i] + " " + a[i+1] + " ");
            }
        }
        System.out.println(sb.toString());
     }
}

Problem solution in C++ programming.

#include <iostream>
#include <fstream>
#include <sstream>
#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <tuple>
#include <array>
#include <limits>
#include <algorithm>

using namespace std;

int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> values(n, {0, 0});
  for (auto& p: values) {
    cin >> p.first;
    p.second = p.first;
  }
  sort(values.begin(), values.end());
  for (size_t i = 1; i < values.size(); ++i) {
    values[i-1].first = values[i-1].second - values[i].second;
  }
  sort(values.begin(), values.end(),
      [](const pair<int, int>& p1, const pair<int, int>& p2) {
      return abs(p1.first) < abs(p2.first);
  });
  auto it = values.begin();
  int best = abs(it->first);
  vector<int> best_values;
  while (abs(it->first) == best) {
    best_values.push_back(it->second);
    best_values.push_back(it->second - it->first);
    ++it;
  }
  sort(best_values.begin(), best_values.end());
  for (auto it = best_values.begin(), end = best_values.end(); it != end; ++it) {
    if (it != best_values.begin()) {
      cout << " ";
    }
    cout << *it;
  }
  cout << endl;
}

Problem solution in C programming.

#include<stdio.h>
#include<stdlib.h>

void quickSort(int[], int, int);
int partition(int[], int, int);


int main(){
	
	int i,n;
	
	//input
	scanf("%d",&n);
	int *a=(int *)malloc(n*sizeof(int));
	
	for(i=0;i<n;i++)
		scanf("%d",&a[i]);

	//sorting
	quickSort(a,0,n-1);
	
	
	//finding smallest
	int min=a[1]-a[0];
	for(i=2;i<n;i++)
		if(a[i]-a[i-1]<min) min=a[i]-a[i-1];
	
	//printing all pairs
	for(i=1;i<n;i++)
		if(a[i]-a[i-1]==min) printf("%d %d ",a[i-1],a[i]);
	printf("n");

return 0;
}



void quickSort(int a[], int l, int r)
{
   int j;

   if( l < r ) 
   {
   	// divide and conquer
        j = partition( a, l, r);
       quickSort( a, l, j-1);
       quickSort( a, j+1, r);
   }
	
}


int partition(int a[], int l, int r) {
   int pivot, i, j, t;
   pivot = a[l];
   i = l; j = r+1;
		
   while( 1)
   {
   	//for ascending part, erase the line below and use instead those inside the comments: .
   	do ++i; while( a[i] <= pivot && i <= r );
   	do --j; while( a[j] > pivot ); 
   	/*do ++i; while( a[i] >= pivot && i <= r );
   	do --j; while( a[j] < pivot );*/
   	if( i >= j ) break;
   	t = a[i]; a[i] = a[j]; a[j] = t;
   }
   t = a[l]; a[l] = a[j]; a[j] = t;
   return j;
}

Problem solution in JavaScript programming.

function processData(input) {
    input.sort(function (a, b) {
        return a - b;
    });
    
    var max = input[1] - input[0];
    var maxPairs = [input[0], input[1]];
    
    for (var i = 2; i < input.length; i++) {
        var prev = input[i - 1];
        var next = input[i];
        var diff = next - prev;
        
        if (diff < max) {
            max = diff;
            maxPairs = [prev, next];
        } else if (diff === max) {
            maxPairs = maxPairs.concat([prev, next]);
        }
    }
    
    console.log(maxPairs.join(' '));
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
    var array = String(_input.split('n').slice(1)).split(' ').map(function(number) { return Number(number); });
    processData(array);
});