Closest greater element for every array element from another array
Last Updated :
09 Mar, 2023
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 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
# 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
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 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)
Similar Reads
Map elements of an array to elements of another array Given two arrays A and B of positive integers, elements of array B can be mapped to elements of array A only if both the elements have same value. The task is to compute the positions in array A to which elements of array B will be mapped. Print NA if mapping for a particular element cannot be done.
6 min read
Print X array elements closest to the Kth smallest element in the array Given two integers K, X, and an array arr[] consisting of N distinct elements, the task is to find X elements closest to the Kth smallest element from the given array. Examples: Input: arr[] = {1, 2, 3, 4, 10}, K = 3, X = 2Output: 2 3Explanation: Kth smallest element present in the given array is 3
15+ min read
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 : 11 6 12 5 12 11 Input : arr[] = {10, 5, 11, 10, 20, 12} Output : 10 10 12 10 12 11 A simple solution is to run two nested loops. We pick an outer element one by one. For
11 min read
Find closest smaller value for every element in array Given an array of integers, find the closest smaller element for every element. If there is no smaller element then print -1 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
4 min read
Maximize product of two closest numbers of other array for every element in given array Given arrays arr1[] of size M and arr2[] of size N having length at least 2, the task is for every element in arr1[], maximize the product of two elements in arr2[] which are closest to the element in arr1[]. The closest elements must be present on distinct indices. Example: Input: arr1 = [5, 10, 17
15 min read