Closest greater element for every array element from another array

Given two arrays a[] and b[], we need to build an array c[] such that every element c[i] of c[] contains a value from a[] which is greater than b[i] and is closest to b[i]. If a[] has no greater element than b[i], then value of c[i] is -1. All arrays are of same size. 

Examples:

Input : a[] = [ 2, 6, 5, 7, 0]  
        b[] = [1, 3, 2, 5, 8]
Output : c[] = [2, 5, 5, 7, -1]

Input : a[] = [ 2, 6, 5, 7, 0]   
        b[] = [0, 2, 3, 5, 1]
Output : c[] = [2, 5, 5, 6, 2]

Naive Approach : For each element in b[], we traverse the whole of a[] and try to find the closest greater element, and save the result for each search. This will cost time complexity of O(n^2). 

Efficient Approach : Sort the array a[], and for each b[i], apply binary search in sorted array a[]. For this method our time complexity will be O(nlogn).

Note: For closest greater element we can use upper_bound(). 

Below is the implementation. 

C++

// CPP to find result from target array
// for closest element
#include <bits/stdc++.h>
using namespace std;

// Function for printing resultant array
void closestResult(int a[], int b[], int n)
{
    // change arr[] to vector
    vector<int> vect(a, a + n);

    // sort vector for ease
    sort(vect.begin(), vect.end());

    // iterator for upper_bound
    vector<int>::iterator up;

    // vector for result
    vector<int> c;

    // calculate resultant array
    for (int i = 0; i < n; i++) {

        // check upper bound element
        up = upper_bound(vect.begin(), vect.end(), b[i]);

        // if no element found push -1
        if (up == vect.end())
            c.push_back(-1);

        // Else push the element
        else
            c.push_back(*up); // add to resultant
    }

    cout << "Result = ";
    for (auto it = c.begin(); it != c.end(); it++)
        cout << *it << " ";
}

// driver program
int main()
{
    int a[] = { 2, 5, 6, 1, 8, 9 };
    int b[] = { 2, 1, 0, 5, 4, 9 };
    int n = sizeof(a) / sizeof(a[0]);
    closestResult(a, b, n);
    return 0;
}

Java

// Java to find result from target array
// for closest element
import java.util.*;

class GFG
{

    // Function for printing resultant array
    static void closestResult(Integer[] a,
                              int[] b, int n)
    {

        // change arr[] to Set
        TreeSet<Integer> vect = new TreeSet<>(Arrays.asList(a));

        // vector for result
        Vector<Integer> c = new Vector<>();

        // calculate resultant array
        for (int i = 0; i < n; i++)
        {

            // check upper bound element
            Integer up = vect.higher(b[i]);

            // if no element found push -1
            if (up == null)
                c.add(-1);

            // Else push the element
            else
                c.add(up); // add to resultant
        }

        System.out.print("Result = ");
        for (int i : c)
            System.out.print(i + " ");
    }

    // Driver Code
    public static void main(String[] args)
    {
        Integer[] a = { 2, 5, 6, 1, 8, 9 };
        int[] b = { 2, 1, 0, 5, 4, 9 };
        int n = a.length;

        closestResult(a, b, n);
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Python implementation to find result
# from target array for closest element

import bisect

# Function for printing resultant array
def closestResult(a, b, n):

    # sort list for ease
    a.sort()

    # list for result
    c = []

    # calculate resultant array
    for i in range(n):

        # check location of upper bound element
        up = bisect.bisect_right(a, b[i])

        # if no element found push -1
        if up == n:
            c.append(-1)

        # else push the element
        else:
            c.append(a[up]) # add to resultant

    print("Result = ", end = "")
    for i in c:
        print(i, end = " ")


# Driver code
if __name__ == "__main__":
    a = [2,5,6,1,8,9]
    b = [2,1,0,5,4,9]
    n = len(a)
    closestResult(a, b, n)

# This code is contributed by
# sanjeev2552

C#

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;

// C# to find result from target array
// for closest element
class HelloWorld {

  // Function for printing resultant array
  public static void closestResult(int[] a, int[] b, int n)
  {
    HashSet<int> vect = new HashSet<int>();
    for(int i = 0; i < n; i++){
      vect.Add(a[i]);
    }

    // array for result
    List<int> c = new List<int>();

    // calculate resultant array
    for (int i = 0; i < n; i++) {
      // check upper bound element
      int up = -1;
      foreach(int elem in vect){
        if(elem > b[i] && (up == -1 || elem < up)){
          up = elem;
        }
      }

      // if no element found push -1
      if (up == -1) c.Add(-1);
      // Else push the element
      else c.Add(up); // add to resultant
    }

    Console.Write("Result = ");
    foreach(var i in c){
      Console.Write(i + " ");
    }
  }

  static void Main() {
    int[] a = { 2, 5, 6, 1, 8, 9 };
    int[] b = { 2, 1, 0, 5, 4, 9 };
    int n = a.Length;
    closestResult(a, b, n);
  }
}

// The code is contributed by Nidhi goel.

Javascript

// JavaScript to find result from target array
// for closest element
function closestResult(a, b, n) {
// change a[] to Set
let vect = new Set(a);
// array for result
let c = [];

// calculate resultant array
for (let i = 0; i < n; i++) {
// check upper bound element
let up = null;
vect.forEach((elem) => {
if (elem > b[i] && (up == null || elem < up)) {
up = elem;
}
});
// if no element found push -1
if (up == null) c.push(-1);
// Else push the element
else c.push(up); // add to resultant
}

console.log("Result =", c.join(" "));
}

// Driver Code
let a = [2, 5, 6, 1, 8, 9];
let b = [2, 1, 0, 5, 4, 9];
let n = a.length;

closestResult(a, b, n);

Output:

Result = 5 2 1 6 5 -1

Time Complexity: O(n log₂(n))
Auxiliary Space: O(n)