# Number of elements smaller than root using preorder traversal of a BST

Given a preorder traversal of a BST. The task is to find the number of elements less than root.

Examples:

```Input: preorder[] = {3, 2, 1, 0, 5, 4, 6}
Output: 3

Input: preorder[] = {5, 4, 3, 2, 1}
Output: 4
```

For a binary search tree, a preorder traversal is of the form:

root, { elements in left subtree of root }, { elements in right subtree of root }

Simple approach:

1. Traverse the given preorder.
2. Check if the current element is greater than root.
3. If yes then return the indexOfCurrentElement – 1 as the no. elements smaller than root will be all the elements occurs before the currentelement except root.

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the first index of the element ` `// that is greater than the root ` `int` `findLargestIndex(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `i, root = arr[0]; ` ` `  `    ``// Traverse the given preorder ` `    ``for``(i = 0; i < n-1; i++) ` `    ``{ ` `        ``// Check if the number is greater than root ` `        ``// If yes then return that index-1 ` `        ``if``(arr[i] > root) ` `           ``return` `i-1; ` `    ``} ` ` `  `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `preorder[] = {3, 2, 1, 0, 5, 4, 6}; ` `    ``int` `n = ``sizeof``(preorder) / ``sizeof``(preorder[0]); ` ` `  `     ``cout << findLargestIndex(preorder, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of ` `// above approach ` ` `  `class` `GFG ` `{ ` `// Function to find the first  ` `// index of the element that  ` `// is greater than the root ` `static` `int` `findLargestIndex(``int` `arr[],  ` `                            ``int` `n) ` `{ ` `    ``int` `i, root = arr[``0``]; ` ` `  `    ``// Traverse the given preorder ` `    ``for``(i = ``0``; i < n - ``1``; i++) ` `    ``{ ` `        ``// Check if the number is ` `        ``// greater than root ` `        ``// If yes then return ` `        ``// that index-1 ` `        ``if``(arr[i] > root) ` `        ``return` `i-``1``; ` `    ``} ` `    ``return` `0``; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String ags[]) ` `{ ` `    ``int` `preorder[] = {``3``, ``2``, ``1``, ``0``, ``5``, ``4``, ``6``}; ` `    ``int` `n = preorder.length; ` ` `  `    ``System.out.println(findLargestIndex(preorder, n)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by Subhadeep Gupta `

## Python3

 `# Python3 implementation of above approach ` ` `  `# Function to find the first index of  ` `# the element that is greater than the root ` `def` `findLargestIndex(arr, n): ` ` `  `    ``i, root ``=` `arr[``0``], arr[``0``]; ` ` `  `    ``# Traverse the given preorder ` `    ``for` `i ``in` `range``(``0``, n ``-` `1``): ` `         `  `        ``# Check if the number is greater than  ` `        ``# root. If yes then return that index-1 ` `        ``if``(arr[i] > root): ` `            ``return` `i ``-` `1``; ` ` `  `# Driver Code ` `preorder``=` `[``3``, ``2``, ``1``, ``0``, ``5``, ``4``, ``6``]; ` `n ``=` `len``(preorder) ` ` `  `print``(findLargestIndex(preorder, n)); ` ` `  `# This code is contributed  ` `# by Akanksha Rai `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to find the first  ` `// index of the element that ` `// is greater than the root ` `static` `int` `findLargestIndex(``int` `[]arr,  ` `                            ``int` `n) ` `{ ` `    ``int` `i, root = arr[0]; ` ` `  `    ``// Traverse the given preorder ` `    ``for``(i = 0; i < n - 1; i++) ` `    ``{ ` `        ``// Check if the number is  ` `        ``// greater than root. If yes  ` `        ``// then return that index-1 ` `        ``if``(arr[i] > root) ` `        ``return` `i - 1; ` `    ``} ` `    ``return` `0; ` `} ` ` `  `// Driver Code ` `static` `public` `void` `Main() ` `{ ` `    ``int` `[]preorder = {3, 2, 1, 0, 5, 4, 6}; ` `    ``int` `n = preorder.Length; ` ` `  `    ``Console.WriteLine(findLargestIndex(preorder, n)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by Subhadeep Gupta `

## PHP

 ` ``\$root``)  ` `        ``return` `\$i` `- 1;  ` `    ``}  ` ` `  `}  ` ` `  `// Driver Code  ` `\$preorder` `= ``array``(3, 2, 1, 0, 5, 4, 6);  ` `\$n` `= ``count``(``\$preorder``); ` `echo` `findLargestIndex(``\$preorder``, ``\$n``);  ` ` `  `// This code is contributed  ` `// by 29AjayKumar ` `?> `

Output:

```3
```

Time complexity: O(n)

Efficient approach (Using Binary Search): Here the idea is to make use of an extended form of binary search. The steps are as follows:

1. Go to mid. Check if the element at mid is greater than root. If yes then we recurse on the left half of array.
2. Else if the element at mid is lesser than root and element at mid+1 is greater than root we return mid as our answer.
3. Else we recurse on the right half of array to repeat the above steps.

Below is the implementation of the above idea.

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count the smaller elements ` `int` `findLargestIndex(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `root = arr[0], lb = 0, ub = n-1; ` `    ``while``(lb < ub) ` `    ``{ ` ` `  `        ``int` `mid = (lb + ub)/2; ` ` `  `        ``// Check if the element at mid ` `        ``// is greater than root. ` `        ``if``(arr[mid] > root) ` `            ``ub = mid - 1; ` `        ``else` `        ``{ ` `          ``// if the element at mid is lesser  ` `          ``//  than root and element at mid+1  ` `          ``// is greater ` `           ``if``(arr[mid + 1] > root) ` `              ``return` `mid; ` `            ``else` `lb = mid + 1;  ` `        ``} ` `     ``} ` `     ``return` `lb; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `preorder[] = {3, 2, 1, 0, 5, 4, 6}; ` `    ``int` `n = ``sizeof``(preorder) / ``sizeof``(preorder[0]); ` ` `  `     ``cout << findLargestIndex(preorder, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation  ` `// of above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to count the ` `// smaller elements ` `static` `int` `findLargestIndex(``int` `arr[],  ` `                            ``int` `n) ` `{ ` `    ``int` `root = arr[``0``], ` `        ``lb = ``0``, ub = n - ``1``; ` `    ``while``(lb < ub) ` `    ``{ ` ` `  `        ``int` `mid = (lb + ub) / ``2``; ` ` `  `        ``// Check if the element at  ` `        ``// mid is greater than root. ` `        ``if``(arr[mid] > root) ` `            ``ub = mid - ``1``; ` `        ``else` `        ``{ ` `             `  `            ``// if the element at mid is  ` `            ``// lesser than root and  ` `            ``// element at mid+1 is greater ` `            ``if``(arr[mid + ``1``] > root) ` `                ``return` `mid; ` `            ``else` `lb = mid + ``1``;  ` `        ``} ` `    ``} ` `    ``return` `lb; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `preorder[] = {``3``, ``2``, ``1``, ``0``, ``5``, ``4``, ``6``}; ` `    ``int` `n = preorder.length; ` ` `  `    ``System.out.println( ` `           ``findLargestIndex(preorder, n)); ` `} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python3 implementation of above approach ` ` `  `# Function to count the smaller elements ` `def` `findLargestIndex(arr, n): ` `    ``root ``=` `arr[``0``]; ` `    ``lb ``=` `0``; ` `    ``ub ``=` `n ``-` `1``; ` `    ``while``(lb < ub): ` ` `  `        ``mid ``=` `(lb ``+` `ub) ``/``/` `2``; ` `         `  `        ``# Check if the element at mid ` `        ``# is greater than root. ` `        ``if``(arr[mid] > root): ` `            ``ub ``=` `mid ``-` `1``; ` `        ``else``: ` `             `  `            ``# if the element at mid is lesser  ` `            ``# than root and element at mid+1  ` `            ``# is greater ` `            ``if``(arr[mid ``+` `1``] > root): ` `                ``return` `mid; ` `            ``else``: ` `                ``lb ``=` `mid ``+` `1``;  ` `    ``return` `lb; ` ` `  `# Driver Code ` `preorder ``=` `[``3``, ``2``, ``1``, ``0``, ``5``, ``4``, ``6``]; ` `n ``=` `len``(preorder); ` ` `  `print``(findLargestIndex(preorder, n)); ` ` `  `# This code is contributed by mits `

## C#

 `// C# implementation  ` `// of above approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `// Function to count the  ` `// smaller elements  ` `static` `int` `findLargestIndex(``int` `[]arr,  ` `                            ``int` `n)  ` `{  ` `    ``int` `root = arr[0],  ` `        ``lb = 0, ub = n - 1;  ` `    ``while``(lb < ub)  ` `    ``{  ` ` `  `        ``int` `mid = (lb + ub) / 2;  ` ` `  `        ``// Check if the element at  ` `        ``// mid is greater than root.  ` `        ``if``(arr[mid] > root)  ` `            ``ub = mid - 1;  ` `        ``else` `        ``{  ` `             `  `            ``// if the element at mid is  ` `            ``// lesser than root and  ` `            ``// element at mid+1 is greater  ` `            ``if``(arr[mid + 1] > root)  ` `                ``return` `mid;  ` `            ``else` `lb = mid + 1;  ` `        ``}  ` `    ``}  ` `    ``return` `lb;  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `Main(String []args)  ` `{  ` `    ``int` `[]preorder = {3, 2, 1, 0, 5, 4, 6};  ` `    ``int` `n = preorder.Length;  ` ` `  `    ``Console.WriteLine(  ` `        ``findLargestIndex(preorder, n));  ` `}  ` `}  ` ` `  `// This code contributed by Rajput-Ji `

## PHP

 ` ``\$root``) ` `            ``\$ub` `= ``\$mid` `- 1; ` `        ``else` `        ``{ ` `            ``// if the element at mid is lesser  ` `            ``// than root and element at mid+1  ` `            ``// is greater ` `            ``if``(``\$arr``[``\$mid` `+ 1] > ``\$root``) ` `                ``return` `\$mid``; ` `            ``else` `                ``\$lb` `= ``\$mid` `+ 1;  ` `        ``} ` `    ``} ` `    ``return` `\$lb``; ` `} ` ` `  `// Driver Code ` `\$preorder` `= ``array``(3, 2, 1, 0, 5, 4, 6); ` `\$n` `= ``count``(``\$preorder``); ` ` `  `echo` `findLargestIndex(``\$preorder``, ``\$n``); ` ` `  `// This code is contributed by mits ` `?> `

Output:

```3
```

Time Complexity: O(logn)

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