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



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.

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// 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<Integer, Boolean> tm = 
                    new TreeMap<Integer, Boolean>();
        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);
    }
}

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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.



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