Find closest greater value for every element in array
Last Updated :
09 Sep, 2022
Given an array of integers, find the closest greater element for every element. If there is no greater element then print -1
Examples:
Input : arr[] = {10, 5, 11, 6, 20, 12}
Output : 11 6 12 10 -1 20
Input : arr[] = {10, 5, 11, 10, 20, 12}
Output :z 11 10 12 11 -1 20
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 greater element. The time complexity of this solution is O(n*n)
A better solution is to use sorting. We sort all elements, then for every element, traverse toward right until we find a greater element (Note that there can be multiple occurrences of an element).
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
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void closetGreater( int arr[], int n)
{
set< int > ts;
for ( int i = 0; i < n; i++)
ts.insert(arr[i]);
for ( int i = 0; i < n; i++)
{
auto greater = ts.upper_bound(arr[i]);
if (greater == ts.end())
cout << -1 << " " ;
else
cout << *greater << " " ;
}
}
int main( int argc, char const *argv[])
{
int arr[] = {10, 5, 11, 10, 20, 12};
int n = sizeof (arr) / sizeof (arr[0]);
closetGreater(arr, n);
return 0;
}
|
Java
import java.util.*;
class TreeSetDemo {
public static void closestGreater( int [] arr)
{
TreeSet<Integer> ts = new TreeSet<Integer>();
for ( int i = 0 ; i < arr.length; i++)
ts.add(arr[i]);
for ( int i = 0 ; i < arr.length; i++) {
Integer greater = ts.higher(arr[i]);
if (greater == null )
System.out.print(- 1 + " " );
else
System.out.print(greater + " " );
}
}
public static void main(String[] args)
{
int [] arr = { 10 , 5 , 11 , 10 , 20 , 12 };
closestGreater(arr);
}
}
|
Python3
def upper_bound(s, val):
temp = list (s).copy()
temp.sort()
try :
index_value = temp.index(val)
if (index_value = = len (temp) - 1 ):
return 0
return temp[index_value + 1 ]
except ValueError:
return arr[ 0 ]
def closetGreater(arr, n):
ts = set ()
for i in range (n):
ts.add(arr[i])
for i in range (n):
greater = upper_bound(ts, arr[i])
if (greater = = 0 ):
print ( - 1 ,end = " " )
else :
print (greater,end = " " )
arr = [ 10 , 5 , 11 , 10 , 20 , 12 ]
n = len (arr)
closetGreater(arr, n)
|
C#
using System;
using System.Collections.Generic;
public class TreeSetDemo {
public static void closestGreater( int [] arr)
{
SortedSet< int > ts = new SortedSet< int >();
for ( int i = 0; i < arr.Length; i++)
ts.Add(arr[i]);
for ( int i = 0; i < arr.Length; i++) {
int greater = upper_bound(ts, arr[i]);
if (greater == -1)
Console.Write(-1 + " " );
else
Console.Write(greater + " " );
}
}
public static int upper_bound(SortedSet< int > s, int val)
{
List< int > temp = new List< int >();
temp.AddRange(s);
temp.Sort();
if (temp.IndexOf(val) + 1 == temp.Count)
return -1;
else
return temp[temp.IndexOf(val) + 1];
}
public static void Main(String[] args)
{
int [] arr = { 10, 5, 11, 10, 20, 12 };
closestGreater(arr);
}
}
|
Javascript
<script>
function closetGreater(arr, n) {
let ts = new Set();
for (let i = 0; i < n; i++)
ts.add(arr[i]);
for (let i = 0; i < n; i++) {
let greater = upper_bound(ts, arr[i]);
if (!greater)
document.write(-1 + " " );
else
document.write(greater + " " );
}
}
function upper_bound(s, val) {
let temp = [...s];
temp.sort((a, b) => a - b);
return temp[temp.indexOf(val) + 1];
}
let arr = [10, 5, 11, 10, 20, 12];
let n = arr.length;
closetGreater(arr, n);
</script>
|
Time Complexity: O(n Log n)
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