Find next Smaller of next Greater in an array
Last Updated :
05 Apr, 2023
Given array of integer, find the next smaller of next greater element of every element in array.
Note : Elements for which no greater element exists or no smaller of greater element exist, print -1.
Examples:
Input : arr[] = {5, 1, 9, 2, 5, 1, 7}
Output: 2 2 -1 1 -1 -1 -1
Explanation :
Next Greater -> Right Smaller
5 -> 9 9 -> 2
1 -> 9 9 -> 2
9 -> -1 -1 -> -1
2 -> 5 5 -> 1
5 -> 7 7 -> -1
1 -> 7 7 -> -1
7 -> -1 -1 -> -1
Input : arr[] = {4, 8, 2, 1, 9, 5, 6, 3}
Output : 2 5 5 5 -1 3 -1 -1
A simple solution is to iterate through all elements. For every element, find the next greater element of current element and then find right smaller element for current next greater element.
Code-
C++
// C++ Program to find Right smaller element of next
// greater element
#include<bits/stdc++.h>
using namespace std;
// Function to find Right smaller element of next greater
// element
void nextSmallerOfNextGreater(int arr[], int n)
{
vector<int> vec;
//For 1st n-1 elements of vector
for(int i=0;i<n-1;i++){
int temp=arr[i];
int next=-1;
int ans=-1;
for(int j=i+1;j<n;j++){
if(arr[j]>temp){
next=j;
break;
}
}
if(next==-1){vec.push_back(-1);}
else{
for(int j=next+1;j<n;j++){
if(arr[j]<arr[next]){
ans=j;
break;
}
}
if(ans==-1){vec.push_back(-1);}
else{vec.push_back(arr[ans]);}
}
}
vec.push_back(-1);//For last element of vector
for(auto x: vec){
cout<<x<<" ";
}
cout<<endl;
}
// Driver program
int main()
{
int arr[] = {5, 1, 9, 2, 5, 1, 7};
int n = sizeof(arr)/sizeof(arr[0]);
nextSmallerOfNextGreater(arr, n);
return 0;
}
// Java Program to find Right smaller element of next
// greater element
import java.util.*;
public class Main {
// Function to find Right smaller element of next greater element
static void nextSmallerOfNextGreater(int arr[], int n) {
ArrayList<Integer> vec = new ArrayList<Integer>();
// For 1st n-1 elements of vector
for(int i = 0; i < n - 1; i++) {
int temp = arr[i];
int next = -1;
int ans = -1;
for(int j = i + 1; j < n; j++) {
if(arr[j] > temp) {
next = j;
break;
}
}
if(next == -1) {
vec.add(-1);
}
else {
for(int j = next + 1; j < n; j++) {
if(arr[j] < arr[next]) {
ans = j;
break;
}
}
if(ans == -1) {
vec.add(-1);
}
else {
vec.add(arr[ans]);
}
}
}
vec.add(-1); // For last element of vector
for(int x : vec) {
System.out.print(x + " ");
}
System.out.println();
}
// Driver program
public static void main(String[] args) {
int arr[] = {5, 1, 9, 2, 5, 1, 7};
int n = arr.length;
nextSmallerOfNextGreater(arr, n);
}
}
# Function to find Right smaller element of next greater element
def nextSmallerOfNextGreater(arr, n):
vec = []
# For 1st n-1 elements of vector
for i in range(n-1):
temp = arr[i]
next = -1
ans = -1
for j in range(i+1, n):
if arr[j] > temp:
next = j
break
if next == -1:
vec.append(-1)
else:
for j in range(next+1, n):
if arr[j] < arr[next]:
ans = j
break
if ans == -1:
vec.append(-1)
else:
vec.append(arr[ans])
vec.append(-1) # For last element of vector
for x in vec:
print(x, end=" ")
print()
# Driver program
arr = [5, 1, 9, 2, 5, 1, 7]
n = len(arr)
nextSmallerOfNextGreater(arr, n)
using System;
using System.Collections.Generic;
public class MainClass
{
// Function to find Right smaller element of next
// greater element
static void nextSmallerOfNextGreater(int[] arr, int n)
{
List<int> vec = new List<int>();
// For 1st n-1 elements of vector
for (int i = 0; i < n - 1; i++) {
int temp = arr[i];
int next = -1;
int ans = -1;
for (int j = i + 1; j < n; j++) {
if (arr[j] > temp) {
next = j;
break;
}
}
if (next == -1) {
vec.Add(-1);
}
else {
for (int j = next + 1; j < n; j++) {
if (arr[j] < arr[next]) {
ans = j;
break;
}
}
if (ans == -1) {
vec.Add(-1);
}
else {
vec.Add(arr[ans]);
}
}
}
vec.Add(-1); // For last element of vector
foreach(var x in vec) { Console.Write(x + " "); }
Console.WriteLine();
}
// Driver program
public static void Main()
{
int[] arr = { 5, 1, 9, 2, 5, 1, 7 };
int n = arr.Length;
nextSmallerOfNextGreater(arr, n);
}
}
// Function to find Right smaller element of next greater element
function nextSmallerOfNextGreater(arr, n) {
let vec = [];
// For 1st n-1 elements of vector
for (let i = 0; i < n - 1; i++) {
let temp = arr[i];
let next = -1;
let ans = -1;
for (let j = i + 1; j < n; j++) {
if (arr[j] > temp) {
next = j;
break;
}
}
if (next == -1) {
vec.push(-1);
} else {
for (let j = next + 1; j < n; j++) {
if (arr[j] < arr[next]) {
ans = j;
break;
}
}
if (ans == -1) {
vec.push(-1);
} else {
vec.push(arr[ans]);
}
}
}
vec.push(-1); // For last element of vector
for (let x of vec) {
process.stdout.write(x + " ");
}
console.log();
}
// Driver program
let arr = [5, 1, 9, 2, 5, 1, 7];
let n = arr.length;
nextSmallerOfNextGreater(arr, n);
Time Complexity of this solution is O(n2).
Space Complexity: O(1)
An efficient solution takes O(n) time. Notice that it is the combination of Next greater element & next smaller element in array.
Let input array be 'arr[]' and size of array be 'n'
find next greatest element of every element
step 1 : Create an empty stack (S) in which we store the indexes
and NG[] that is user to store the indexes of NGE
of every element.
step 2 : Traverse the array in reverse order
where i goes from (n-1 to 0)
a) While S is non empty and the top element of
S is smaller than or equal to 'arr[i]':
pop S
b) If S is empty
arr[i] has no greater element
NG[i] = -1
c) else we have next greater element
NG[i] = S.top() // here we store the index of NGE
d) push current element index in stack
S.push(i)
Find Right smaller element of every element
step 3 : create an array RS[] used to store the index of
right smallest element
step 4 : we repeat step (1 & 2) with little bit of
modification in step 1 & 2 .
they are :
a). we use RS[] in place of NG[].
b). In step (2.a)
we pop element form stack S while S is not
empty or the top element of S is greater than
or equal to 'arr[i]'
step 5 : compute all RSE of NGE :
where i goes from 0 to n-1
if NG[ i ] != -1 && RS[ NG [ i]] ! =-1
print arr[RS[NG[i]]]
else
print -1
Below is the implementation of above idea
C++
// C++ Program to find Right smaller element of next
// greater element
#include<bits/stdc++.h>
using namespace std;
// function find Next greater element
void nextGreater(int arr[], int n, int next[], char order)
{
// create empty stack
stack<int> S;
// Traverse all array elements in reverse order
// order == 'G' we compute next greater elements of
// every element
// order == 'S' we compute right smaller element of
// every element
for (int i=n-1; i>=0; i--)
{
// Keep removing top element from S while the top
// element is smaller than or equal to arr[i] (if Key is G)
// element is greater than or equal to arr[i] (if order is S)
while (!S.empty() &&
((order=='G')? arr[S.top()] <= arr[i]:
arr[S.top()] >= arr[i]))
S.pop();
// store the next greater element of current element
if (!S.empty())
next[i] = S.top();
// If all elements in S were smaller than arr[i]
else
next[i] = -1;
// Push this element
S.push(i);
}
}
// Function to find Right smaller element of next greater
// element
void nextSmallerOfNextGreater(int arr[], int n)
{
int NG[n]; // stores indexes of next greater elements
int RS[n]; // stores indexes of right smaller elements
// Find next greater element
// Here G indicate next greater element
nextGreater(arr, n, NG, 'G');
// Find right smaller element
// using same function nextGreater()
// Here S indicate right smaller elements
nextGreater(arr, n, RS, 'S');
// If NG[i] == -1 then there is no smaller element
// on right side. We can find Right smaller of next
// greater by arr[RS[NG[i]]]
for (int i=0; i< n; i++)
{
if (NG[i] != -1 && RS[NG[i]] != -1)
cout << arr[RS[NG[i]]] << " ";
else
cout<<"-1"<<" ";
}
}
// Driver program
int main()
{
int arr[] = {5, 1, 9, 2, 5, 1, 7};
int n = sizeof(arr)/sizeof(arr[0]);
nextSmallerOfNextGreater(arr, n);
return 0;
}
// Java Program to find Right smaller element of next
// greater element
import java.util.Stack;
public class Main {
// function find Next greater element
public static void nextGreater(int arr[], int next[], char order)
{
// create empty stack
Stack<Integer> stack=new Stack<>();
// Traverse all array elements in reverse order
// order == 'G' we compute next greater elements of
// every element
// order == 'S' we compute right smaller element of
// every element
for (int i=arr.length-1; i>=0; i--)
{
// Keep removing top element from S while the top
// element is smaller than or equal to arr[i] (if Key is G)
// element is greater than or equal to arr[i] (if order is S)
while (!stack.isEmpty() && ((order=='G')? arr[stack.peek()] <= arr[i]:arr[stack.peek()] >= arr[i]))
stack.pop();
// store the next greater element of current element
if (!stack.isEmpty())
next[i] = stack.peek();
// If all elements in S were smaller than arr[i]
else
next[i] = -1;
// Push this element
stack.push(i);
}
}
// Function to find Right smaller element of next greater
// element
public static void nextSmallerOfNextGreater(int arr[])
{
int NG[]=new int[arr.length]; // stores indexes of next greater elements
int RS[]=new int[arr.length]; // stores indexes of right smaller elements
// Find next greater element
// Here G indicate next greater element
nextGreater(arr, NG, 'G');
// Find right smaller element
// using same function nextGreater()
// Here S indicate right smaller elements
nextGreater(arr, RS, 'S');
// If NG[i] == -1 then there is no smaller element
// on right side. We can find Right smaller of next
// greater by arr[RS[NG[i]]]
for (int i=0; i< arr.length; i++)
{
if (NG[i] != -1 && RS[NG[i]] != -1)
System.out.print(arr[RS[NG[i]]]+" ");
else
System.out.print("-1 ");
}
}
public static void main(String args[]) {
int arr[] = {5, 1, 9, 2, 5, 1, 7};
nextSmallerOfNextGreater(arr);
}
}
//This code is contributed by Gaurav Tiwari
# Python 3 Program to find Right smaller element of next
# greater element
# function find Next greater element
def nextGreater(arr, n, next, order):
S = []
# Traverse all array elements in reverse order
# order == 'G' we compute next greater elements of
# every element
# order == 'S' we compute right smaller element of
# every element
for i in range(n-1,-1,-1):
# Keep removing top element from S while the top
# element is smaller than or equal to arr[i] (if Key is G)
# element is greater than or equal to arr[i] (if order is S)
while (S!=[] and (arr[S[len(S)-1]] <= arr[i]
if (order=='G') else arr[S[len(S)-1]] >= arr[i] )):
S.pop()
# store the next greater element of current element
if (S!=[]):
next[i] = S[len(S)-1]
# If all elements in S were smaller than arr[i]
else:
next[i] = -1
# Push this element
S.append(i)
# Function to find Right smaller element of next greater
# element
def nextSmallerOfNextGreater(arr, n):
NG = [None]*n # stores indexes of next greater elements
RS = [None]*n # stores indexes of right smaller elements
# Find next greater element
# Here G indicate next greater element
nextGreater(arr, n, NG, 'G')
# Find right smaller element
# using same function nextGreater()
# Here S indicate right smaller elements
nextGreater(arr, n, RS, 'S')
# If NG[i] == -1 then there is no smaller element
# on right side. We can find Right smaller of next
# greater by arr[RS[NG[i]]]
for i in range(n):
if (NG[i] != -1 and RS[NG[i]] != -1):
print(arr[RS[NG[i]]],end=" ")
else:
print("-1",end=" ")
# Driver program
if __name__=="__main__":
arr = [5, 1, 9, 2, 5, 1, 7]
n = len(arr)
nextSmallerOfNextGreater(arr, n)
# this code is contributed by ChitraNayal
using System;
using System.Collections.Generic;
// C# Program to find Right smaller element of next
// greater element
public class GFG {
// function find Next greater element
public static void nextGreater(int []arr, int []next, char order)
{
// create empty stack
Stack<int> stack=new Stack<int>();
// Traverse all array elements in reverse order
// order == 'G' we compute next greater elements of
// every element
// order == 'S' we compute right smaller element of
// every element
for (int i=arr.Length-1; i>=0; i--)
{
// Keep removing top element from S while the top
// element is smaller than or equal to arr[i] (if Key is G)
// element is greater than or equal to arr[i] (if order is S)
while (stack.Count!=0 && ((order=='G')? arr[stack.Peek()] <= arr[i]:arr[stack.Peek()] >= arr[i]))
stack.Pop();
// store the next greater element of current element
if (stack.Count!=0)
next[i] = stack.Peek();
// If all elements in S were smaller than arr[i]
else
next[i] = -1;
// Push this element
stack.Push(i);
}
}
// Function to find Right smaller element of next greater
// element
public static void nextSmallerOfNextGreater(int []arr)
{
int []NG=new int[arr.Length]; // stores indexes of next greater elements
int []RS=new int[arr.Length]; // stores indexes of right smaller elements
// Find next greater element
// Here G indicate next greater element
nextGreater(arr, NG, 'G');
// Find right smaller element
// using same function nextGreater()
// Here S indicate right smaller elements
nextGreater(arr, RS, 'S');
// If NG[i] == -1 then there is no smaller element
// on right side. We can find Right smaller of next
// greater by arr[RS[NG[i]]]
for (int i=0; i< arr.Length; i++)
{
if (NG[i] != -1 && RS[NG[i]] != -1)
Console.Write(arr[RS[NG[i]]]+" ");
else
Console.Write("-1 ");
}
}
public static void Main() {
int []arr = {5, 1, 9, 2, 5, 1, 7};
nextSmallerOfNextGreater(arr);
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript Program to find Right smaller element of next
// greater element
// function find Next greater element
function nextGreater(arr,next,order)
{
// create empty stack
let stack = [];
// Traverse all array elements in reverse order
// order == 'G' we compute next greater elements of
// every element
// order == 'S' we compute right smaller element of
// every element
for (let i = arr.length - 1; i >= 0; i--)
{
// Keep removing top element from S while the top
// element is smaller than or equal to arr[i] (if Key is G)
// element is greater than or equal to arr[i] (if order is S)
while (stack.length!=0 && ((order=='G')? arr[stack[stack.length-1]] <= arr[i] : arr[stack[stack.length-1]] >= arr[i]))
stack.pop();
// store the next greater element of current element
if (stack.length != 0)
next[i] = stack[stack.length - 1];
// If all elements in S were smaller than arr[i]
else
next[i] = -1;
// Push this element
stack.push(i);
}
}
// Function to find Right smaller element of next greater
// element
function nextSmallerOfNextGreater(arr)
{
let NG = new Array(arr.length); // stores indexes of next greater elements
let RS = new Array(arr.length); // stores indexes of right smaller elements
for(let i = 0; i < arr.length; i++)
{
NG[i] = 0;
RS[i] = 0;
}
// Find next greater element
// Here G indicate next greater element
nextGreater(arr, NG, 'G');
// Find right smaller element
// using same function nextGreater()
// Here S indicate right smaller elements
nextGreater(arr, RS, 'S');
// If NG[i] == -1 then there is no smaller element
// on right side. We can find Right smaller of next
// greater by arr[RS[NG[i]]]
for (let i = 0; i < arr.length; i++)
{
if (NG[i] != -1 && RS[NG[i]] != -1)
document.write(arr[RS[NG[i]]] + " ");
else
document.write("-1 ");
}
}
// Driver code
let arr = [5, 1, 9, 2, 5, 1, 7];
nextSmallerOfNextGreater(arr);
// This code is contributed by rag2127
</script>
Time complexity : O(n), where n is the size of the given array.
Auxiliary Space: O(n), where n is the size of the given array.
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