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 elements 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 – 

C++

// 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 */
Node* insert(Node* root, int val, int& suc)
{
    /* If the tree is empty, return a new node */
    if (!root)
        return newNode(val);
    // go to right subtree
    if (val >= root->data)
        root->right = insert(root->right, val, suc);
    // If key is smaller than root's key, go to left
    // subtree and set successor as current node
    else {
        suc = root->data;
        root->left = insert(root->left, val, suc);
    }
    return root;
}

// Function to replace every element with the
// least greater element on its right
void replace(int arr[], int n)
{
    Node* root = nullptr;
    // start from right to left
    for (int i = n - 1; i >= 0; i--) {
        int suc = -1;
        // insert current element into BST and
        // find its inorder successor
        root = insert(root, arr[i], suc);
        arr[i] = suc;
    }
}

// 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;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// 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

Python3

# Python3 program to replace every element
# with the least greater element on its right

# A binary Tree node


class Node:

    def __init__(self, d):

        self.data = d
        self.left = None
        self.right = None

# A utility function to insert a new node with
# given data in BST and find its successor


def insert(node, data):

    global succ

    # If the tree is empty, return a new node
    root = node

    if (node == None):
        return Node(data)

    # If key is smaller than root's key, go to left
    # subtree and set successor as current node
    if (data < node.data):

        # print("1")
        succ = node
        root.left = insert(node.left, data)

    # Go to right subtree
    elif (data > node.data):
        root.right = insert(node.right, data)

    return root

# Function to replace every element with the
# least greater element on its right


def replace(arr, n):

    global succ
    root = None

    # Start from right to left
    for i in range(n - 1, -1, -1):
        succ = None

        # Insert current element into BST and
        # find its inorder successor
        root = insert(root, arr[i])

        # Replace element by its inorder
        # successor in BST
        if (succ):
            arr[i] = succ.data

        # No inorder successor
        else:
            arr[i] = -1

    return arr


# Driver code
if __name__ == '__main__':

    arr = [8, 58, 71, 18, 31, 32, 63,
           92, 43, 3, 91, 93, 25, 80, 28]
    n = len(arr)
    succ = None

    arr = replace(arr, n)

    print(*arr)

# This code is contributed by mohit kumar 29

C#

// C# program to replace every element with
// the least greater element on its right
using System;

class BinarySearchTree {

    // A binary Tree node
    public class Node {
        public int data;
        public Node left, right;

        public Node(int d)
        {
            data = d;
            left = right = null;
        }
    }

    // Root of BST
    public static Node root;
    public static Node succ;

    // Constructor
    public BinarySearchTree()
    {
        root = null;
        succ = null;
    }

    // A utility function to insert a new node with
    // given data in BST and find its successor
    public static 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
    public 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 = BinarySearchTree.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
    static public void Main()
    {
        int[] arr = { 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++) {
            Console.Write(arr[i] + " ");
        }
    }
}

// This code is contributed by rag2127

Javascript

<script>

// Javascript program to
// replace every element with
// the least greater element
// on its right

    // A binary Tree node
    class Node{
        constructor(d)
        {
            this.data=d;
            this.left=this.right=null;
        }
    }

    // Root of BST
    let root=null;
    let succ=null;

    // A utility function to insert a new node with
    // given data in BST and find its successor
    function insert(node,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
    function replace(arr,n)
    {
        // start from right to left
    for(let i = n - 1; i >= 0; i--)
    {
        succ = null;

        // Insert current element into BST and
        // find its inorder successor
        root = 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
    let arr=[8, 58, 71, 18, 31,
             32, 63, 92, 43, 3,
             91, 93, 25, 80, 28 ];
       let n = arr.length;
    replace(arr, n);
    for(let i = 0; i < n; i++)
        document.write(arr[i] + " ");


// This code is contributed by unknown2108

</script>

Output

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

Time complexity: 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.
Auxiliary Space: O(h), Here h is the height of the BST and the extra space is used in recursion call stack.

Another Approach:

We can use the Next Greater Element using stack algorithm to solve this problem in O(Nlog(N)) time and O(N) space.

Algorithm:

1. First, we take an array of pairs namely temp, and store each element and its index in this array,i.e. temp[i] will be storing {arr[i],i}.
2. Sort the array according to the array elements.
3. Now get the next greater index for each and every index of the temp array in an array namely index by using Next Greater Element using stack.
4. Now index[i] stores the index of the next least greater element of the element temp[i].first and if index[i] is -1, then it means that there is no least greater element of the element temp[i].second at its right side.
5. Now take a result array where result[i] will be equal to a[indexes[temp[i].second]] if index[i] is not -1 otherwise result[i] will be equal to -1.

Below is the implementation of the above approach

C++

#include <bits/stdc++.h>
using namespace std;
// function to get the next least greater index for each and
// every temp.second of the temp array using stack this
// function is similar to the Next Greater element for each
// and every element of an array using stack difference is
// we are finding the next greater index not value and the
// indexes are stored in the temp[i].second for all i
vector<int> nextGreaterIndex(vector<pair<int, int> >& temp)
{
    int n = temp.size();
    // initially result[i] for all i is -1
    vector<int> res(n, -1);
    stack<int> stack;
    for (int i = 0; i < n; i++) {
        // if the stack is empty or this index is smaller
        // than the index stored at top of the stack then we
        // push this index to the stack
        if (stack.empty() || temp[i].second < stack.top())
            stack.push(
                temp[i].second); // notice temp[i].second is
                                 // the index
        // else this index (i.e. temp[i].second) is greater
        // than the index stored at top of the stack we pop
        // all the indexes stored at stack's top and for all
        // these indexes we make this index i.e.
        // temp[i].second as their next greater index
        else {
            while (!stack.empty()
                   && temp[i].second > stack.top()) {
                res[stack.top()] = temp[i].second;
                stack.pop();
            }
            // after that push the current index to the
            // stack
            stack.push(temp[i].second);
        }
    }
    // now res will store the next least greater indexes for
    // each and every indexes stored at temp[i].second for
    // all i
    return res;
}
// now we will be using above function for finding the next
// greater index for each and every indexes stored at
// temp[i].second
vector<int> replaceByLeastGreaterUsingStack(int arr[],
                                            int n)
{
    // first of all in temp we store the pairs of {arr[i].i}
    vector<pair<int, int> > temp;
    for (int i = 0; i < n; i++) {
        temp.push_back({ arr[i], i });
    }
    // we sort the temp according to the first of the pair
    // i.e value
    sort(temp.begin(), temp.end(),
         [](const pair<int, int>& a,
            const pair<int, int>& b) {
             if (a.first == b.first)
                 return a.second > b.second;
             return a.first < b.first;
         });
    // now indexes vector will store the next greater index
    // for each temp[i].second index
    vector<int> indexes = nextGreaterIndex(temp);
    // we initialize a result vector with all -1
    vector<int> res(n, -1);
    for (int i = 0; i < n; i++) {
        // now if there is no next greater index after the
        // index temp[i].second the result will be -1
        // otherwise the result will be the element of the
        // array arr at index indexes[temp[i].second]
        if (indexes[temp[i].second] != -1)
            res[temp[i].second]
                = arr[indexes[temp[i].second]];
    }
    // return the res which will store the least greater
    // element of each and every element in the array at its
    // right side
    return res;
}
// driver code
int main()
{
    int arr[] = { 8,  58, 71, 18, 31, 32, 63, 92,
                  43, 3,  91, 93, 25, 80, 28 };
    int n = sizeof(arr) / sizeof(int);
    auto res = replaceByLeastGreaterUsingStack(arr, n);
    cout << "Least Greater elements on the right side are "
         << endl;
    for (int i : res)
        cout << i << ' ';
    cout << endl;
    return 0;
} // this code is contributed by Dipti Prakash Sinha

Java

// Java program for above approach

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Stack;

public class GFF {

    // function to get the next least greater index for each
    // and every temp.second of the temp array using stack
    // this function is similar to the Next Greater element
    // for each and every element of an array using stack
    // difference is we are finding the next greater index
    // not value and the indexes are stored in the
    // temp[i].second for all i
    static int[] nextGreaterIndex(ArrayList<int[]> temp)
    {
        int n = temp.size();
        // initially result[i] for all i is -1
        int[] res = new int[n];
        Arrays.fill(res, -1);
        Stack<Integer> stack = new Stack<Integer>();
        for (int i = 0; i < n; i++) {
            // if the stack is empty or this index is
            // smaller than the index stored at top of the
            // stack then we push this index to the stack
            if (stack.empty()
                || temp.get(i)[1] < stack.peek())
                stack.push(temp.get(
                    i)[1]); // notice temp[i].second is
                            // the index
            // else this index (i.e. temp[i].second) is
            // greater than the index stored at top of the
            // stack we pop all the indexes stored at
            // stack's top and for all these indexes we make
            // this index i.e. temp[i].second as their next
            // greater index
            else {
                while (!stack.empty()
                       && temp.get(i)[1] > stack.peek()) {
                    res[stack.peek()] = temp.get(i)[1];
                    stack.pop();
                }
                // after that push the current index to the
                // stack
                stack.push(temp.get(i)[1]);
            }
        }
        // now res will store the next least greater indexes
        // for each and every indexes stored at
        // temp[i].second for all i
        return res;
    }

    // now we will be using above function for finding the
    // next greater index for each and every indexes stored
    // at temp[i].second
    static int[] replaceByLeastGreaterUsingStack(int arr[],
                                                 int n)
    {
        // first of all in temp we store the pairs of
        // {arr[i].i}
        ArrayList<int[]> temp = new ArrayList<int[]>();
        for (int i = 0; i < n; i++) {
            temp.add(new int[] { arr[i], i });
        }
        // we sort the temp according to the first of the
        // pair i.e value
        Collections.sort(temp, (a, b) -> {
            if (a[0] == b[0])
                return b[1] - a[1];
            return a[0] - b[0];
        });

        // now indexes vector will store the next greater
        // index for each temp[i].second index
        int[] indexes = nextGreaterIndex(temp);
        // we initialize a result vector with all -1
        int[] res = new int[n];
        Arrays.fill(res, -1);
        for (int i = 0; i < n; i++) {
            // now if there is no next greater index after
            // the index temp[i].second the result will be
            // -1 otherwise the result will be the element
            // of the array arr at index
            // indexes[temp[i].second]
            if (indexes[temp.get(i)[1]] != -1)
                res[temp.get(i)[1]]
                    = arr[indexes[temp.get(i)[1]]];
        }
        // return the res which will store the least greater
        // element of each and every element in the array at
        // its right side
        return res;
    }

    // driver code
    public static void main(String[] args)
    {
        int arr[] = { 8,  58, 71, 18, 31, 32, 63, 92,
                      43, 3,  91, 93, 25, 80, 28 };
        int n = arr.length;
        int[] res = replaceByLeastGreaterUsingStack(arr, n);
        System.out.println(
            "Least Greater elements on the right side are ");
        for (int i : res)
            System.out.print(i + " ");
        System.out.println();
    }
}

// This code is contributed by Lovely Jain

Python3

# function to get the next least greater index for each and
# every temp[1] of the temp array using stack this
# function is similar to the Next Greater element for each
# and every element of an array using stack difference is
# we are finding the next greater index not value and the
# indexes are stored in the temp[i][1] for all i


def nextGreaterIndex(temp):

    n = len(temp)

    # initially result[i] for all i is -1
    res = [-1 for i in range(n)]
    stack = []
    for i in range(n):

        # if the stack is empty or this index is smaller
        # than the index stored at top of the stack then we
        # append this index to the stack
        if (len(stack) == 0 or temp[i][1] < stack[-1]):
            stack.append(temp[i][1])  # notice temp[i][1] is
            # the index
        # else this index (i.e. temp[i][1]) is greater
        # than the index stored at top of the stack we pop
        # all the indexes stored at stack's top and for all
        # these indexes we make this index i.e.
        # temp[i][1] as their next greater index
        else:
            while (len(stack) > 0 and temp[i][1] > stack[-1]):
                res[stack[-1]] = temp[i][1]
                stack.pop()

            # after that append the current index to the stack
            stack.append(temp[i][1])

    # now res will store the next least greater indexes for
    # each and every indexes stored at temp[i][1] for
    # all i
    return res

# now we will be using above function for finding the next
# greater index for each and every indexes stored at
# temp[i][1]


def replaceByLeastGreaterUsingStack(arr, n):

    # first of all in temp we store the pairs of {arr[i].i}
    temp = []
    for i in range(n):
        temp.append([arr[i], i])

    # we sort the temp according to the first of the pair
    # i.e value
    temp.sort()

    # now indexes vector will store the next greater index
    # for each temp[i][1] index
    indexes = nextGreaterIndex(temp)

    # we initialize a result vector with all -1
    res = [-1 for i in range(n)]
    for i in range(n):

        # now if there is no next greater index after the
        # index temp[i][1] the result will be -1
        # otherwise the result will be the element of the
        # array arr at index indexes[temp[i][1]]
        if (indexes[temp[i][1]] != -1):
            res[temp[i][1]] = arr[indexes[temp[i][1]]]

    # return the res which will store the least greater
    # element of each and every element in the array at its
    # right side
    return res

# driver code


arr = [8,  58, 71, 18, 31, 32, 63, 92, 43, 3,  91, 93, 25, 80, 28]
n = len(arr)
res = replaceByLeastGreaterUsingStack(arr, n)
print("Least Greater elements on the right side are ")
for i in res:
    print(i, end=' ')
print()

# this code is contributed by shinjanpatra

C#

using System;
using System.Collections.Generic;
using System.Linq;

class GFG {

    // function to get the next least greater index for each
    // and every temp.second of the temp array using stack
    // this function is similar to the Next Greater element
    // for each and every element of an array using stack
    // difference is we are finding the next greater index
    // not value and the indexes are stored in the
    // temp[i].second for all i
    static int[] nextGreaterIndex(List<int[]> temp)
    {
        int n = temp.Count();
        // initially result[i] for all i is -1
        int[] res = new int[n];

        for (int i = 0; i < n; i++) {
            res[i] = -1;
        }

        Stack<int> stack = new Stack<int>();

        for (int i = 0; i < n; i++) {
            // if the stack is empty or this index is
            // smaller than the index stored at top of the
            // stack then we push this index to the stack
            if (stack.Count() == 0
                || temp[i][1] < stack.Peek()) {
                stack.Push(temp[i][1]); // notice temp[i][1]
                                        // is the index
            }

            // else this index (i.e. temp[i][1]) is
            // greater than the index stored at top of the
            // stack we pop all the indexes stored at
            // stack's top and for all these indexes we make
            // this index i.e. temp[i][1] as their next
            // greater index
            else {
                while (stack.Count() != 0
                       && temp[i][1] > stack.Peek()) {
                    res[stack.Peek()] = temp[i][1];
                    stack.Pop();
                }
                // after that push the current index to the
                // stack
                stack.Push(temp[i][1]);
            }
        }
        // now res will store the next least greater indexes
        // for each and every indexes stored at
        // temp[i][1] for all i
        return res;
    }

    // now we will be using above function for finding the
    // next greater index for each and every indexes stored
    // at temp[i][1]
    static int[] replaceByLeastGreaterUsingStack(int[] arr,
                                                 int n)
    {
        // first of all in temp we store the pairs of
        // {arr[i].i}
        List<int[]> temp = new List<int[]>();
        for (int i = 0; i < n; i++) {
            temp.Add(new int[] { arr[i], i });
        }

        // we sort the temp according to the first of the
        // pair i.e value
        temp.Sort((a, b) = > {
            if (a[0] == b[0])
                return a[1] - b[1];
            return a[0] - b[0];
        });

        // now indexes vector will store the next greater
        // index for each temp[i][1] index
        int[] indexes = nextGreaterIndex(temp);
        // we initialize a result vector with all -1
        int[] res = new int[n];

        for (int i = 0; i < n; i++) {
            res[i] = -1;
        }

        for (int i = 0; i < n; i++) {
            // now if there is no next greater index after
            // the index temp[i][1] the result will be
            // -1 otherwise the result will be the element
            // of the array arr at index
            // indexes[temp[i][1]]
            if (indexes[temp[i][1]] != -1)
                res[temp[i][1]] = arr[indexes[temp[i][1]]];
        }
        // return the res which will store the least greater
        // element of each and every element in the array at
        // its right side
        return res;
    }

    // driver code
    public static void Main()
    {
        int[] arr
            = new int[] { 8,  58, 71, 18, 31, 32, 63, 92,
                          43, 3,  91, 93, 25, 80, 28 };
        int n = arr.Length;
        int[] res = replaceByLeastGreaterUsingStack(arr, n);

        Console.WriteLine(
            "Least Greater elements on the right side are ");

        foreach(var i in res) Console.Write(i + " ");
        Console.WriteLine();
    }
}

// This code is contributed by Tapesh (tapeshdua420)

Javascript

<script>

// function to get the next least greater index for each and
// every temp[1] of the temp array using stack this
// function is similar to the Next Greater element for each
// and every element of an array using stack difference is
// we are finding the next greater index not value and the
// indexes are stored in the temp[i][1] for all i

function mycmp(a,b){
    if(a[0] == b[0])
     return b[1] - a[1];
     return a[0] - b[0];
}

function nextGreaterIndex(temp)
{
    let n = temp.length;
    // initially result[i] for all i is -1
    let res = new Array(n).fill(-1);
    let stack = [];
    for (let i = 0; i < n; i++) {
        // if the stack is empty or this index is smaller
        // than the index stored at top of the stack then we
        // push this index to the stack
        if (stack.length == 0 || temp[i][1] < stack[stack.length-1])
            stack.push(temp[i][1]); // notice temp[i][1] is
                                 // the index
        // else this index (i.e. temp[i][1]) is greater
        // than the index stored at top of the stack we pop
        // all the indexes stored at stack's top and for all
        // these indexes we make this index i.e.
        // temp[i][1] as their next greater index
        else {
            while (stack.length > 0 && temp[i][1] > stack[stack.length-1]) {
                res[stack[stack.length-1]] = temp[i][1];
                stack.pop();
            }
            // after that push the current index to the stack
            stack.push(temp[i][1]);
        }
    }
    // now res will store the next least greater indexes for
    // each and every indexes stored at temp[i][1] for
    // all i
    return res;
}
// now we will be using above function for finding the next
// greater index for each and every indexes stored at
// temp[i][1]
function replaceByLeastGreaterUsingStack(arr,n)
{
    // first of all in temp we store the pairs of {arr[i].i}
    let temp = [];
    for (let i = 0; i < n; i++) {
        temp.push([arr[i], i]);
    }
    // we sort the temp according to the first of the pair
    // i.e value
    temp.sort(mycmp);
    // now indexes vector will store the next greater index
    // for each temp[i][1] index
    let indexes = nextGreaterIndex(temp);
    // we initialize a result vector with all -1
    let res = new Array(n).fill(-1);
    for (let i = 0; i < n; i++) {
        // now if there is no next greater index after the
        // index temp[i][1] the result will be -1
        // otherwise the result will be the element of the
        // array arr at index indexes[temp[i][1]]
        if (indexes[temp[i][1]] != -1)
            res[temp[i][1]]
                = arr[indexes[temp[i][1]]];
    }
    // return the res which will store the least greater
    // element of each and every element in the array at its
    // right side
    return res;
}
// driver code

let arr = [ 8,  58, 71, 18, 31, 32, 63, 92,
                  43, 3,  91, 93, 25, 80, 28 ];
let n = arr.length;
let res = replaceByLeastGreaterUsingStack(arr, n);
document.write("Least Greater elements on the right side are ","</br>");
for (let i of res)
    document.write(i,' ');
document.write("</br>");

// this code is contributed by shinjanpatra

</script>

Output

Least Greater elements on the right side are 
18 63 80 25 32 43 80 93 80 25 93 -1 28 -1 -1 

Time Complexity: O(N log N)

Space Complexity: O(N)

Another approach with set

A different way to think about the problem is listing our requirements and then thinking over it to find a solution. If we traverse the array from backwards, we need  a data structure(ds) to support:

1. Insert an element into our ds in sorted order (so at any point of time the elements in our ds are sorted)
2. Finding the upper bound of the current element (upper bound will give just greater element from our ds if present)

Carefully observing at our requirements, a set is what comes in mind. 

Why not multiset? Well we can use a multiset but there is no need to store an element more than once.

Let’s code our approach

Time and space complexity: We insert each element in our set and find upper bound for each element using a loop so its time complexity is O(n*log(n)). We are storing each element in our set so space complexity is O(n)

C++

#include <iostream>
#include <set>
#include <vector>

using namespace std;

void solve(vector<int>& arr)
{
    set<int> s;
    for (int i = arr.size() - 1; i >= 0;
         i--) { // traversing the array backwards
        s.insert(arr[i]); // inserting the element into set
        auto it
            = s.upper_bound(arr[i]); // finding upper bound
        if (it == s.end())
            arr[i] = -1; // if upper_bound does not exist
                         // then -1
        else
            arr[i] = *it; // if upper_bound exists, lets
                          // take it
    }
}

void printArray(vector<int>& arr)
{
    for (int i : arr)
        cout << i << " ";
    cout << "\n";
}

int main()
{
    vector<int> arr = { 8,  58, 71, 18, 31, 32, 63, 92,
                        43, 3,  91, 93, 25, 80, 28 };
    printArray(arr);
    solve(arr);
    printArray(arr);
    return 0;
}

Java

import java.util.*;

public class Main {
    public static void main(String[] args)
    {
        int[] arr = { 8,  58, 71, 18, 31, 32, 63, 92,
                      43, 3,  91, 93, 25, 80, 28 };
        printArray(arr);
        solve(arr);
        printArray(arr);
    }
    public static void solve(int[] arr)
    {
        TreeSet<Integer> s = new TreeSet<>();
        for (int i = arr.length - 1; i >= 0;
             i--) { // traversing the array backwards
            s.add(arr[i]); // inserting the element into set
            Integer it
                = s.higher(arr[i]); // finding upper bound
                                    // (higher in java)
            if (it == null)
                arr[i] = -1; // if upper_bound does not
                             // exist then -1
            else
                arr[i] = it; // if upper_bound exists, lets
                             // take it
        }
    }
    public static void printArray(int[] arr)
    {
        for (int i : arr)
            System.out.print(i + " ");
        System.out.println();
    }
}

// This code is contributed by Tapesh (tapeshdua420)

Python3

from typing import List
from bisect import bisect_right

def solve(arr: List[int]) -> List[int]:
    s = set()
    for i in range(len(arr) - 1, -1, -1):
        s.add(arr[i])
        upper_bound = bisect_right(sorted(s), arr[i])
        if upper_bound == len(s):
            arr[i] = -1
        else:
            arr[i] = sorted(s)[upper_bound]
    return arr

def print_array(arr: List[int]):
    print(*arr)

if __name__ == "__main__":
    arr = [8, 58, 71, 18, 31, 32, 63, 92, 43, 3, 91, 93, 25, 80, 28]
    print_array(arr)
    solve(arr)
    print_array(arr)

    # This code is contributed by vikranshirsath177.

C#

// Include namespace system
using System;
using System.Collections.Generic;

public class GFG {
    public static void Main(String[] args)
    {
        int[] arr = { 8,  58, 71, 18, 31, 32, 63, 92,
                      43, 3,  91, 93, 25, 80, 28 };
        GFG.printArray(arr);
        GFG.solve(arr);
        GFG.printArray(arr);
    }
    public static void solve(int[] arr)
    {
        var s = new SortedSet<int>();
        for (int i = arr.Length - 1; i >= 0; i--) {
            // traversing the array backwards
            s.Add(arr[i]);
            // inserting the element into set
            var it = -1;
            // finding upper bound

            foreach(int j in s)
            {
                if (j > arr[i]) {
                    it = j;
                    break;
                }
            }

            if (it == -1) {
                arr[i] = -1;
            }
            else {
                arr[i] = it;
            }
        }
    }
    public static void printArray(int[] arr)
    {
        foreach(int i in arr)
        {
            Console.Write(i.ToString() + " ");
        }
        Console.WriteLine();
    }
}

// This code is contributed by aadityaburujwale.

Javascript

function solve(arr) {
  let s = new Set();
  for (let i = arr.length - 1; i >= 0; i--) {
    // traversing the array backwards
    s.add(arr[i]);
    // inserting the element into set
    let it = -1;
    // finding upper bound

    for (let j of s) {
      if (j > arr[i]) {
        it = j;
        break;
      }
    }

    if (it == -1) {
      arr[i] = -1;
    } else {
      arr[i] = it;
    }
  }
}

function printArray(arr) {
  for (let i of arr) {
    console.log(i + " ");
  }
  console.log();
}

let arr = [8, 58, 71, 18, 31, 32, 63, 92, 43, 3, 91, 93, 25, 80, 28];
printArray(arr);
solve(arr);
printArray(arr);

Output

8 58 71 18 31 32 63 92 43 3 91 93 25 80 28 
18 63 80 25 32 43 80 93 80 25 93 -1 28 -1 -1 

Time complexity -The time complexity of the given program is O(n*logn), where n is the size of the input array.

The reason for this is that the program uses a set to store the unique elements of the input array, and for each element in the array, it performs a single insertion operation and a single upper_bound operation on the set. Both of these operations have a time complexity of O(logn) in the average case, and since they are performed n times, the overall time complexity is O(n*logn).

Space complexity-The space complexity of the program is also O(n), as it uses a set to store the unique elements of the input array. In the worst case, where all elements of the array are unique, the set will have to store n elements, leading to a space complexity of O(n).

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Approach#4:using  2 loops 

Algorithm

1. Initialize a new list with all -1 values to represent the elements for which there is no greater element on its right.
2. For each element in the input list:
a. Initialize a variable “min_greater” with a value of infinity.
b. Use another loop to find the minimum element in the input list that is greater than the current element.
c. If such an element is found, update the value of “min_greater” to be the minimum element found in step b.
d. If the value of “min_greater” is still infinity, it means there is no greater element on the right of the current element, so do nothing.
e. Otherwise, update the corresponding element in the new list with the value of “min_greater”.
5. Return the new list.

Python3

def replace_with_least_greater(arr):
    n = len(arr)
    new_arr = [-1] * n
    for i in range(n):
        min_greater = float('inf')
        for j in range(i+1, n):
            if arr[j] > arr[i] and arr[j] < min_greater:
                min_greater = arr[j]
        if min_greater != float('inf'):
            new_arr[i] = min_greater
    return new_arr
arr = [8, 58, 71, 18, 31, 32, 63, 92, 43, 3, 91, 93, 25, 80, 28]
print(replace_with_least_greater(arr))

Output

[18, 63, 80, 25, 32, 43, 80, 93, 80, 25, 93, -1, 28, -1, -1]

Time Complexity: O(n^2), where n is the length of the input array. This is because we use two nested loops to iterate over all pairs of elements in the input array.
Space Complexity: O(n), where n is the length of the input array. This is because we create a new list of the same length as the input array to store the output.