Replace every element with the least greater element on its right - GeeksforGeeks

18 Apr, 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++
Java
Python3
C#
Javascript
// C++ program to replace every element with the// least greater element on its right#include <bits/stdc++.h>using namespace std; // A binary Tree nodestruct Node {    int data;    Node *left, *right;}; // A utility function to create a new BST nodeNode* 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 rightvoid 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 functionsint 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 program to replace every element with// the least greater element on its rightimport java.io.*; class BinarySearchTree{     // A binary Tree nodeclass Node{    int data;    Node left, right;     Node(int d)    {        data = d;        left = right = null;    }} // Root of BSTstatic Node root;static Node succ; // ConstructorBinarySearchTree(){    root = null;    succ = null;} // A utility function to insert a new node with// given data in BST and find its successorNode 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 rightstatic 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 codepublic 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 program to replace every element# with the least greater element on its right # A binary Tree nodeclass 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 successordef 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 rightdef 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 codeif __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# program to replace every element with// the least greater element on its rightusing System; class BinarySearchTree{     // A binary Tree nodepublic class Node{    public int data;    public Node left, right;         public Node(int d)    {        data = d;        left = right = null;    }} // Root of BSTpublic static Node root;public static Node succ; // Constructorpublic BinarySearchTree(){    root = null;    succ = null;} // A utility function to insert a new node with// given data in BST and find its successorpublic 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 rightpublic 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 codestatic 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
<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>
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:
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}.Sort the array according to the array elements.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.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.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.
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}.
Sort the array according to the array elements.
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.
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.
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++
Python3
#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 ivector<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].secondvector<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());    // 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 codeint 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
# 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 idef 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
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;}
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.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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