Replace every element with the least greater element on its right

Given an array of integers, replace every element with the least greater element on its right side in the array. If there are no greater element on the right side, replace it with -1.
Examples: 

Input: [8, 58, 71, 18, 31, 32, 63, 92, 
         43, 3, 91, 93, 25, 80, 28]
Output: [18, 63, 80, 25, 32, 43, 80, 93, 
         80, 25, 93, -1, 28, -1, -1]

A naive method is to run two loops. The outer loop will one by one pick array elements from left to right. The inner loop will find the smallest element greater than the picked element on its right side. Finally, the outer loop will replace the picked element with the element found by inner loop. The time complexity of this method will be O(n2).
A tricky solution would be to use Binary Search Trees. We start scanning the array from right to left and insert each element into the BST. For each inserted element, we replace it in the array by its inorder successor in BST. If the element inserted is the maximum so far (i.e. its inorder successor doesn’t exist), we replace it by -1.

Below is the implementation of the above idea –

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// C++ program to replace every element with the
// least greater element on its right
#include <bits/stdc++.h>
using namespace std;
  
// A binary Tree node
struct Node
{
    int data;
    Node *left, *right;
};
  
// A utility function to create a new BST node
Node* newNode(int item)
{
    Node* temp = new Node;
    temp->data = item;
    temp->left = temp->right = NULL;
  
    return temp;
}
  
/* A utility function to insert a new node with
   given data in BST and find its successor */
void insert(Node*& node, int data, Node*& succ)
{
    /* If the tree is empty, return a new node */
    if (node == NULL)
        node = newNode(data);
  
    // If key is smaller than root's key, go to left
    // subtree and set successor as current node
    if (data < node->data)
    {
        succ = node;
        insert(node->left, data, succ);
    }
  
    // go to right subtree
    else if (data > node->data)
        insert(node->right, data, succ);
}
  
// Function to replace every element with the
// least greater element on its right
void replace(int arr[], int n)
{
    Node* root = NULL;
  
    // start from right to left
    for (int i = n - 1; i >= 0; i--)
    {
        Node* succ = NULL;
  
        // insert current element into BST and
        // find its inorder successor
        insert(root, arr[i], succ);
  
        // replace element by its inorder
        // successor in BST
        if (succ)
            arr[i] = succ->data;
        else    // No inorder successor
            arr[i] = -1;
    }
}
  
// Driver Program to test above functions
int main()
{
    int arr[] = { 8, 58, 71, 18, 31, 32, 63, 92,
                  43, 3, 91, 93, 25, 80, 28 };
    int n = sizeof(arr)/ sizeof(arr[0]);
  
    replace(arr, n);
  
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
  
    return 0;
}
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// Java program to replace every element with 
// the least greater element on its right
import java.io.*;
  
class BinarySearchTree{
      
// A binary Tree node
class Node
{
    int data;
    Node left, right;
  
    Node(int d)
    {
        data = d;
        left = right = null;
    }
}
  
// Root of BST
static Node root;
static Node succ;
  
// Constructor
BinarySearchTree()
{
    root = null;
    succ = null;
}
  
// A utility function to insert a new node with
// given data in BST and find its successor 
Node insert(Node node, int data)
{
      
    // If the tree is empty, return a new node 
    if (node == null)
    {
        node = new Node(data);
    }
  
    // If key is smaller than root's key,
    // go to left subtree and set successor
    // as current node
    if (data < node.data) 
    {
        succ = node;
        node.left = insert(node.left, data);
    }
  
    // Go to right subtree
    else if (data > node.data)
        node.right = insert(node.right, data);
          
    return node;
}
  
// Function to replace every element with the
// least greater element on its right
static void replace(int arr[], int n)
{
    BinarySearchTree tree = new BinarySearchTree();
  
    // start from right to left
    for(int i = n - 1; i >= 0; i--) 
    {
        succ = null;
          
        // Insert current element into BST and
        // find its inorder successor
        root = tree.insert(root, arr[i]);
  
        // Replace element by its inorder
        // successor in BST
        if (succ != null)
            arr[i] = succ.data;
              
        // No inorder successor
        else 
            arr[i] = -1;
    }
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = new int[] { 8, 58, 71, 18, 31
                            32, 63, 92, 43, 3,
                            91, 93, 25, 80, 28 };
    int n = arr.length;
  
    replace(arr, n);
  
    for(int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}
}
  
// The code is contributed by Tushar Bansal
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Output:

18 63 80 25 32 43 80 93 80 25 93 -1 28 -1 -1

The worst-case time complexity of the above solution is also O(n2) as it uses BST. The worst-case will happen when array is sorted in ascending or descending order. The complexity can easily be reduced to O(nlogn) by using balanced trees like red-black trees.
This article is contributed by Aditya Goel. 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.
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