Given an array of integers, replace every element with the least greater element on its right side in the array. If there are no greater element on 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(n^{2}).

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 exists), we replace it by -1.

Below is C++ implementation of above idea –

// C++ program to replace every element with the // least greater element on its right #include <iostream> using namespace std; // A binary Tree node struct Node { int data; Node *left, *right; }; // A utility function to create a new BST node Node* newNode(int item) { Node* temp = new Node; temp->data = item; temp->left = temp->right = NULL; return temp; } /* A utility function to insert a new node with given data in BST and find its successor */ void insert(Node*& node, int data, Node*& succ) { /* If the tree is empty, return a new node */ if (node == NULL) node = newNode(data); // If key is smaller than root's key, go to left // subtree and set successor as current node if (data < node->data) { succ = node; insert(node->left, data, succ); } // go to right subtree else if (data > node->data) insert(node->right, data, succ); } // Function to replace every element with the // least greater element on its right void replace(int arr[], int n) { Node* root = NULL; // start from right to left for (int i = n - 1; i >= 0; i--) { Node* succ = NULL; // insert current element into BST and // find its inorder successor insert(root, arr[i], succ); // replace element by its inorder // successor in BST if (succ) arr[i] = succ->data; else // No inorder successor arr[i] = -1; } } // Driver Program to test above functions int main() { int arr[] = { 8, 58, 71, 18, 31, 32, 63, 92, 43, 3, 91, 93, 25, 80, 28 }; int n = sizeof(arr)/ sizeof(arr[0]); replace(arr, n); for (int i = 0; i < n; i++) cout << arr[i] << " "; return 0; }

Output:

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

**Worst case time complexity** of above solution is also O(n^{2}) as it uses BST. The worst case will happen when array is sorted in ascending or descending order. The complexity can easily be reduced to O(nlogn) by using balanced trees like red-black trees.

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

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