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. 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)
This article is contributed by Nishant_Singh(Pintu). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
