Ceiling in right side for every element in an array
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, 10, 20, 12}
Output : 10 10 12 12 -1 -1
Input : arr[] = {50, 20, 200, 100, 30}
Output : 100 30 -1 -1 -1
A simple solution is to run two nested loops. We pick an outer element one by one. For every picked element, we traverse the right side array and find the closest greater or equal element. 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 the 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 ceiling operations in O(Log n) time.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void closestGreater( int arr[], int n)
{
set< int > s;
vector< int > ceilings;
for ( int i = n - 1; i >= 0; i--) {
auto greater = s.lower_bound(arr[i]);
if (greater == s.end())
ceilings.push_back(-1);
else
ceilings.push_back(*greater);
s.insert(arr[i]);
}
for ( int i = n - 1; i >= 0; i--)
cout << ceilings[i] << " " ;
}
int main()
{
int arr[] = { 50, 20, 200, 100, 30 };
closestGreater(arr, 5);
return 0;
}
|
Java
import java.util.*;
class TreeSetDemo {
public static void closestGreater( int [] arr)
{
int n = arr.length;
TreeSet<Integer> ts = new TreeSet<Integer>();
ArrayList<Integer> ceilings = new ArrayList<Integer>(n);
for ( int i = n - 1 ; i >= 0 ; i--) {
Integer greater = ts.ceiling(arr[i]);
if (greater == null )
ceilings.add(- 1 );
else
ceilings.add(greater);
ts.add(arr[i]);
}
for ( int i = n - 1 ; i >= 0 ; i--)
System.out.print(ceilings.get(i) + " " );
}
public static void main(String[] args)
{
int [] arr = { 50 , 20 , 200 , 100 , 30 };
closestGreater(arr);
}
}
|
Python3
import bisect
def closestGreater(arr, n):
s = []
ceilings = []
for i in range (n - 1 , - 1 , - 1 ):
greater = bisect.bisect_left(s, arr[i])
if greater = = len (s):
ceilings.append( - 1 )
else :
ceilings.append(s[greater])
s.insert(greater, arr[i])
for i in range (n - 1 , - 1 , - 1 ):
print (ceilings[i], end = " " )
arr = [ 50 , 20 , 200 , 100 , 30 ]
closestGreater(arr, 5 )
|
C#
using System;
using System.Collections.Generic;
public class TreeSetDemo {
public static void closestGreater( int [] arr)
{
int n = arr.Length;
SortedSet< int > ts = new SortedSet< int >();
List< int > ceilings = new List< int >(n);
for ( int i = n - 1; i >= 0; i--) {
int greater = lower_bound(ts, arr[i]);
if (greater == -1)
ceilings.Add(-1);
else
ceilings.Add(greater);
ts.Add(arr[i]);
}
ceilings.Sort((a,b)=>a-b);
for ( int i = n - 1; i >= 0; i--)
Console.Write(ceilings[i] + " " );
}
public static int lower_bound(SortedSet< int > s, int val)
{
List< int > temp = new List< int >();
temp.AddRange(s);
temp.Sort();
temp.Reverse();
if (temp.IndexOf(val) + 1 == temp.Count)
return -1;
else if (temp[temp.IndexOf(val) +1]>val)
return -1;
else
return temp[temp.IndexOf(val) +1];
}
public static void Main(String[] args)
{
int [] arr = { 50, 20, 200, 100, 30 };
closestGreater(arr);
}
}
|
Javascript
function closestGreater(arr, n) {
let s = [];
let ceilings = [];
for (let i = n - 1; i >= 0; i--) {
let greater = bisect_left(s, arr[i]);
if (greater == s.length) {
ceilings.push(-1);
} else {
ceilings.push(s[greater]);
}
s.splice(greater, 0, arr[i]);
}
let temp = [];
for (let i = n - 1; i >= 0; i--) {
temp.push(ceilings[i]);
}
console.log(temp.join( " " ));
}
function bisect_left(s, x) {
let lo = 0;
let hi = s.length;
while (lo < hi) {
let mid = Math.floor((lo + hi) / 2);
if (s[mid] < x) {
lo = mid + 1;
} else {
hi = mid;
}
}
return lo;
}
let arr = [50, 20, 200, 100, 30];
closestGreater(arr, 5);
|
Time Complexity: O(n Log n)
Auxiliary Space: O(n)
Last Updated :
22 Mar, 2023
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