# 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 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]
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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(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.

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