# Find closest value for every element in array

Given an array of integers, find the closest element for every element.

Examples:

Input : arr[] = {10, 5, 11, 6, 20, 12}
Output : 6, -1, 10, 5, 12, 11

Input : arr[] = {10, 5, 11, 10, 20, 12}
Output : 5 -1 10 5 12 11

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to run two nested loops. We pick an outer element one by one. For every picked element, we traverse remaining array and find closest element. Time complexity of this solution is O(n*n)

An efficient solution is to use Self Balancing BST (Implemented as set in C++ and TreeSet in Java). In a Self Balancing BST, we can do both insert and closest greater operations in O(Log n) time.

 // Java program to demonstrate insertions in TreeSet import java.util.*;    class TreeSetDemo {     public static void closestGreater(int[] arr)     {         if (arr.length == -1) {             System.out.print(-1 + " ");             return;         }            // Insert all array elements into a TreeMap.         // A TreeMap value indicates whether an element         // appears once or more.         TreeMap tm =                      new TreeMap();         for (int i = 0; i < arr.length; i++) {                // A value "True" means that the key             // appears more than once.             if (tm.containsKey(arr[i]))                 tm.put(arr[i], true);             else                 tm.put(arr[i], false);         }            // Find smallest greater element for every         // array element         for (int i = 0; i < arr.length; i++) {                // If there are multiple occurrences             if (tm.get(arr[i]) == true)             {                 System.out.print(arr[i] + " ");                 continue;             }                // If element appears only once             Integer greater = tm.higherKey(arr[i]);             Integer lower = tm.lowerKey(arr[i]);             if (greater == null)                 System.out.print(lower + " ");             else if (lower == null)                 System.out.print(greater + " ");             else {                 int d1 = greater - arr[i];                 int d2 = arr[i] - lower;                 if (d1 > d2)                     System.out.print(lower + " ");                 else                     System.out.print(greater + " ");             }         }     }        public static void main(String[] args)     {         int[] arr = { 10, 5, 11, 6, 20, 12, 10 };         closestGreater(arr);     } }

Output:

10 6 12 5 12 11 10

Exercise : Another efficient solution is to use sorting that also works in O(n Log n) time. Write complete algorithm and code for sorting based solution.

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.