cantonios Test case 5 seems to be incorrect.
6 -5 15 25 71 63
There are only 5 numbers provided, not 6. Anyone else getting this? I seem to get full points without it though.
apazala Yup. Sample Test #2 when you "Run", Test #5 when you submit. It should be corrected but here is an easy Java workaround:
int n = scanner.nextInt(); int[] arr = new int[n]; int i = 0; try{ for (; i < n; i++) { arr[i] = scanner.nextInt(); } } catch(NoSuchElementException ex){ arr = Arrays.copyOf(arr, i); }MohammedShoaib Yup Same.
atulkala20 yes ,testcase 5 seems to ignored by the author.
dimonb It causes segmentation fault in c++ stub
alexanderfanz @HackerRank should review this!!!
Coder_AMiT sort and then check the diff of two adjecent numbers.. keep tracking the min value, update the both pairs in a list, if the new pair is having less than last min value, then delete the list, update with new pairs,, if the new pair is having exact value, add them in list too.
n = input() nums = list(map(int,input().strip().split())) nums.sort() lowestDiff = nums[1]-nums[0] res = [nums[0],nums[1]] for i in range(1,len(nums)-1): if nums[i+1]-nums[i] < lowestDiff: res = [] res.append(nums[i]) res.append(nums[i+1]) lowestDiff = abs(nums[i+1]-nums[i]) elif nums[i+1]-nums[i] == lowestDiff: res.append(nums[i]) res.append(nums[i+1]) print(*res)
lincolndu PHP solution
sort($arr); $minDiff=$arr[1]-$arr[0]; $res=[]; for ($i=0; $i < $n-1; $i++) { if($arr[$i+1]-$arr[$i] < $minDiff ){ $res=[]; $res[]=$arr[$i]; $res[]=$arr[$i+1]; $minDiff=abs($arr[$i+1]-$arr[$i]); }elseif($arr[$i+1]-$arr[$i] == $minDiff ){ $res[]=$arr[$i]; $res[]=$arr[$i+1]; } } for ($i=0; $i < count($res); $i++) { echo $res[$i].' '; }
nateshmbhat1 [deleted]
rwan7727 C++:
int n; cin >> n; int a[n]; for (int i=0;i<n;i++) cin >> a[i]; sort(a,a+n); int diff[n]; adjacent_difference(a,a+n,diff); int min=*min_element(diff+1,diff+n); for (int i=1;i<n;i++) { if (diff[i]==min) cout << a[i-1] << " " << a[i] << " "; }mateusz_janda Nice use of STL! I forgot about adjacent_difference
adityachaubey041 is it possible to take a input from user and create a static array of that size ex:- you have taken cin>>n; and creat a static int diff[n];
tanuj_kapoor15 yes it is
sukhbir Cool and simple logic. Thanks.
shanur_cse_nitap when you have already sorted, you dont need to take the absolute value, just saying
lowestDiff = abs(nums[i+1]-nums[i])Coder_AMiT True,, we dont need to do abs()
eric_na I did this in two pass, once to find the min diff, and one more to add the pairs if diff is min diff. It's because I thought it's wasteful to keep adding the pairs and clearing the res list when we see a smaller diff.
hAcKcode i believe it reduces the time complexity, as now we need to iterate just once. Open for discussion..
mittalmrids Yes it reduces the time complexity as we have to perform fewer number of operations .
AghalarovBahruz where is greedy algorithm here ?
mehmetrizaoz There is no need for using abs.
mittalmrids in the if segment lowestDiff = abs(nums[i+1]-nums[i]) we dont need abs as the list is already sorted
GokuSSJ4 For those who are getting Test Case 3 & 4 wrong and their code is correct, you need to output the pairs in ascending order.
narayan_m one simple algorithm:
0. get the inputs from console
1. sort the input array as sortedInput
2. set lowestDiff to maximum integer value
3. set resultArray to empty array
4. for i = 1 to length of sortedInput-1
4a. delta = absolute value (sortedInput[i] - sortedInput[i-1])
4b. if delta < lowestDiff, then
-> lowestDiff = delta
-> clear resultArray (remove all elements from array)
-> add to resultArray sortedInpit[i-1], and then sortedInput[i]
4c. Else, if delta == lowestDiff, then
-> add to resultArray sortedInpit[i-1], and then sortedInput[i]
5. print resultArray as space-separated integer on single line.psdudeman39 Thanks...!!! Even I came up with the same algorithm.
hacker_khan why do we need to find absolute value here?
Mohit_MR int n; cin>>n; vector <int > v(n); for(int i=0;i<n;i++) cin>>v[i]; sort(v.begin(),v.end()); int max=100000000; int id=0; for(int i=1;i<n;i++) { if(abs(v[i]-v[i-1])<max){max=abs(v[i]-v[i-1]); id=i-1; } } for(int i=id;i<n-1;i++) { if(abs(v[i+1]-v[i])==max){ cout<<v[i]<<" "<<v[i+1]<<" ";} } return 0; handle the index of smallest element to creak it
SHRI_RANG I have written exact same code in java but test cases 3 and 4 are not passing. Please help me.
import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; 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 d =a[1]-a[0],id=0; for(int i=1;i<n;i++) { if(d>Math.abs(a[i]-a[i-1])) { d = Math.abs(a[i]-a[i-1]); id = i-1; } } for(int i=id;i<n-1;i++) { if(d==Math.abs(a[i]-a[i+1])) System.out.print(a[i]+" "+a[i+1]); } } }
us241077 int d =a[1]-a[0] This line will cause troubles in higher cases
jishangarg How? can u explain plz
the_monologue import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;
public class Solution {
public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); int a[]=new int[n]; int b[]=new int[2*n]; int min = Integer.MAX_VALUE,k=0; for(int i=0;i<n;i++) a[i]=sc.nextInt(); Arrays.sort(a); for(int i=0;i<n-1;i++) { if(Math.abs(a[i]-a[i+1])==min) { b[k++]=a[i]; b[k++]=a[i+1]; } else if(Math.abs(a[i]-a[i+1])<min) { k=0; b[k++]=a[i]; b[k++]=a[i+1]; min=Math.abs(a[i]-a[i+1]); } } for(int i=0;i<k;i++) System.out.print(b[i]+" "); }}
xiguang_liu If the array is already sorted, there is no need to use abs for a[i] - a[i-1].
Mohit_MR true
iamuday better code
svvaghela import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { static String findMinDiff(int[] arr, int n) { Arrays.sort(arr); StringBuilder s = new StringBuilder(""); int diff = Integer.MAX_VALUE; for (int i=0; i<n-1; i++){ if (arr[i+1] - arr[i] == diff){ s.append(arr[i] + " "+ arr[i+1] + " "); } if (arr[i+1] - arr[i] < diff){ diff = arr[i+1] - arr[i]; s.setLength(0); s.append(arr[i] + " "+ arr[i+1] + " "); } } return s.toString(); } public static void main(String[] args) { Scanner s = new Scanner(System.in); int N = s.nextInt(); int arr[] = new int[N]; for (int i = 0; i < N; i++) { arr[i] = s.nextInt(); } System.out.println(findMinDiff(arr, arr.length)); } }
thesharpshooter Dont we need to take care of the case when the difference is 0?
Jay_Rockss Yes ,if there is such case we have to consider 0 also. But I think there is no such test case here:)
Balwinder1012 in approx O(nlogn + 2n) time
First sort the array with quick sort which takes nlogn time, then apply
Long diff = Long.MAX_VALUE; String output = "": for(int i=0;i<n-1;i++) if(a[i+1]-a[i] <= diff) diff = a[i+1]-a[i]; for(int i=0;i<n-1;i++) if(a[i+1]-a[i]==diff) output += a[i] + " " +a[i+1]+" "; System.out.println(output);
stewartcj Running brute force algorithm in C# for a lab. Test case 4 has timeout error. My algorithm is O(n^2) [as far as I know]
static int[] closestNumbers(int[] arr) { int smallestDistance = Int32.MaxValue; int distance; List<int> result = new List<int>(); sw = Stopwatch.StartNew(); Array.Sort(arr); for (int i = 0; i < arr.Length; i++) { for (int j = i + 1; j < arr.Length; j++) { distance = Math.Abs(arr[i] - arr[j]); if (distance < smallestDistance) { smallestDistance = distance; result.Clear(); result.Add(arr[i]); result.Add(arr[j]); } else if (distance == smallestDistance) { result.Add(arr[i]); result.Add(arr[j]); } } } result.Sort(); sw.Stop(); WriteLine("Time taken in brute force: " + (sw.ElapsedMilliseconds / 1000.0)); return result.ToArray(); }
shansa09 Here test case input is wrong 6 -5 15 25 71 63 There are 5 elements and count is given as 6 and hence not able to run my code as crashing while loading data
atulkala20 yes, you are right I have encountered the same problem.
Sort 237 Discussions, By:
Please Login in order to post a comment