# 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. Ee 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 C++ implementation of above idea

```// 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;
}
```

Output:

```2 2 -1 1 -1 -1 -1
```

Time complexity : O(n)

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