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# 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

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 using namespace std;   // Function to find Right smaller element of next greater // element void nextSmallerOfNextGreater(int arr[], int n) {     vector vec;     //For 1st n-1 elements of vector     for(int i=0;itemp){               next=j;               break;           }         }                    if(next==-1){vec.push_back(-1);}         else{           for(int j=next+1;j

## Java

 // 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 vec = new ArrayList();       // 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);   } }

## Python3

 # 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)

## C#

 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 vec = new List();       // 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);   } }

## Javascript

 // 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);

Output

2 2 -1 1 -1 -1 -1

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 using namespace std;   // function find Next greater element void nextGreater(int arr[], int n, int next[], char order) {     // create empty stack     stack 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

 // 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 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

 # 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

## 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 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.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

## Javascript



Output

2 2 -1 1 -1 -1 -1

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.