Skip to content
Related Articles

Related Articles

Replace every element with the least greater element on its right

View Discussion
Improve Article
Save Article
  • Difficulty Level : Hard
  • Last Updated : 04 Jul, 2022

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* node, int data, Node*& succ)
{
     
    /* If the tree is empty, return a new node */
    if (node == NULL)
        return 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;
        node->left = insert(node->left, data, succ);
    }
 
    // go to right subtree
    else if (data > node->data)
        node->right = insert(node->right, data, succ);
    return node;
}
 
// 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
        root = 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;
}

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 

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.

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 a[1] - b[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[] = { 858, 71, 18, 31, 32, 63, 92,
                      43, 391, 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 = [ 858, 71, 18, 31, 32, 63, 92, 43, 391, 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

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 

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 <vector>
#include <set>
 
using namespace std;
 
void solve(vector<int>& arr) {
    set<int> s;
    vector<int> ans(arr.size());
    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;
}

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 

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.


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!