# 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 â€“

## C++

 // C++ program to replace every element with the // least greater element on its right #include 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



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 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 nextGreaterIndex(vector >& temp) {     int n = temp.size();     // initially result[i] for all i is -1     vector res(n, -1);     stack 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 replaceByLeastGreaterUsingStack(int arr[],                                             int n) {     // first of all in temp we store the pairs of {arr[i].i}     vector > 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& a,             const pair& 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 indexes = nextGreaterIndex(temp);     // we initialize a result vector with all -1     vector 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 temp)     {         int n = temp.size();         // initially result[i] for all i is -1         int[] res = new int[n];         Arrays.fill(res, -1);         Stack stack = new 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.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 temp = new ArrayList();         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 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 stack = new 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.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 temp = new List();         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



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 #include #include   using namespace std;   void solve(vector& arr) {     set 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& arr) {     for (int i : arr)         cout << i << " ";     cout << "\n"; }   int main() {     vector 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 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();         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).

#### 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.

## C++

 #include #include #include   using namespace std;   vector replace_with_least_greater(const vector& arr) {     int n = arr.size();     vector new_arr(n, -1);     for (int i = 0; i < n; i++) {         int min_greater = numeric_limits::max();         for (int j = i + 1; j < n; j++) {             if (arr[j] > arr[i] && arr[j] < min_greater) {                 min_greater = arr[j];             }         }         if (min_greater != numeric_limits::max()) {             new_arr[i] = min_greater;         }     }     return new_arr; }   int main() {     vector arr = {8, 58, 71, 18, 31, 32, 63, 92, 43, 3, 91, 93, 25, 80, 28};     vector result = replace_with_least_greater(arr);       // Print the result     for (int val : result) {         cout << val << " ";     }     cout << endl;       return 0; }

## Java

 import java.util.ArrayList; import java.util.List;   public class Main {       // Function to replace each element in the input list with the least greater element to its right.     public static List replaceWithLeastGreater(List arr) {         int n = arr.size();         List newArr = new ArrayList<>(n);                   for (int i = 0; i < n; i++) {             int minGreater = Integer.MAX_VALUE; // Initialize minGreater to a very large value.               // Iterate through elements to the right of the current element.             for (int j = i + 1; j < n; j++) {                 // Check if the current element is greater than the current minimum greater element                 // found so far and is smaller than any previously found minimum greater element.                 if (arr.get(j) > arr.get(i) && arr.get(j) < minGreater) {                     minGreater = arr.get(j); // Update minGreater with the new minimum greater element.                 }             }               if (minGreater != Integer.MAX_VALUE) {                 newArr.add(minGreater); // Add the minimum greater element to the new list.             } else {                 newArr.add(-1); // If no greater element was found, add -1 to the new list.             }         }           return newArr; // Return the list with replaced elements.     }       public static void main(String[] args) {         // Create an input list of integers.         List arr = List.of(8, 58, 71, 18, 31, 32, 63, 92, 43, 3, 91, 93, 25, 80, 28);                   // Call the replaceWithLeastGreater function to get the result.         List result = replaceWithLeastGreater(arr);           // Print the result         for (int val : result) {             System.out.print(val + " "); // Print each element in the result list.         }         System.out.println();     } }

## 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))

## C#

 using System; using System.Collections.Generic;   class Program {     // Function to replace each element with the least greater element on its right     static List ReplaceWithLeastGreater(List arr)     {         int n = arr.Count;         List newArr = new List(n);         for (int i = 0; i < n; i++)         {             int minGreater = int.MaxValue;             for (int j = i + 1; j < n; j++)             {                 // Check if the element at index j is greater than the current element                 // and if it is smaller than the current minimum greater value                 if (arr[j] > arr[i] && arr[j] < minGreater)                 {                     minGreater = arr[j];                 }             }             if (minGreater != int.MaxValue)             {                 // If a least greater element is found, add it to the new list                 newArr.Add(minGreater);             }             else             {                 // If no least greater element is found, add -1 to the new list                 newArr.Add(-1);             }         }         return newArr;     }       // Main method     static void Main()     {         List arr = new List { 8, 58, 71, 18, 31, 32, 63, 92, 43, 3, 91, 93, 25, 80, 28 };         List result = ReplaceWithLeastGreater(arr);           // Print the result         foreach (int val in result)         {             Console.Write(val + " ");         }         Console.WriteLine();     } }

## Javascript

 // Function to replace each element in the input // list with the least greater element to its right. function replaceWithLeastGreater(arr) {     const n = arr.length;     const newArr = [];       for (let i = 0; i < n; i++) {         let minGreater = Number.MAX_VALUE; // Initialize minGreater to a very large value.           // Iterate through elements to the right of the current element.         for (let j = i + 1; j < n; j++) {             // Check if the current element is greater than the current minimum greater element             // found so far and is smaller than any previously found minimum greater element.             if (arr[j] > arr[i] && arr[j] < minGreater) {                 minGreater = arr[j]; // Update minGreater with the new minimum greater element.             }         }           if (minGreater !== Number.MAX_VALUE) {             newArr.push(minGreater); // Add the minimum greater element to the new list.         } else {             newArr.push(-1); // If no greater element was found, add -1 to the new list.         }     }       return newArr; // Return the list with replaced elements. }   // Create an input list of integers. const arr = [8, 58, 71, 18, 31, 32, 63, 92, 43, 3, 91, 93, 25, 80, 28];   // Call the replaceWithLeastGreater function to get the result. const result = replaceWithLeastGreater(arr);   // Print the result for (const val of result) {     process.stdout.write(val + ' '); // Print each element in the result list. } process.stdout.write('\n');

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.

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