Find next Smaller of next Greater in an array

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. Time taken of this solution is O(n²).

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 nonempty 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 then 
              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 then or equal to arr[i] (if Key is G) 
        // element is greater then 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

// 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 then or equal to arr[i] (if Key is G)  
            // element is greater then 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

# 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 then or equal to arr[i] (if Key is G) 
        # element is greater then 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 

C#

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 then or equal to arr[i] (if Key is G)  
            // element is greater then 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 

Output:

2 2 -1 1 -1 -1 -1

Time complexity : O(n)

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python 3

C#

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