1623636600

In this video, we will see how to code deletion in a binary search tree.

#c-programming #c #programming

1625122494

Searching for an element’s presence in a list is usually done using linear search and binary search. Linear search is time-consuming and memory expensive but is the simplest way to search for an element. On the other hand, Binary search is effective mainly due to the reduction of list dimension with each recursive function call or iteration. A practical implementation of binary search is autocompletion.

The objective of this project is to create a simple python program to implement binary search. It can be implemented in two ways: recursive (function calls) and iterative.

The project uses loops and functions to implement the search function. Hence good knowledge of python loops and function calls is sufficient to understand the code flow.

#python tutorials #binary search python #binary search python program #iterative binary search python #recursive binary search python

1615458581

Today we will be discussing one of the leetcode’s October daily coding challenge questions.

Before I jump into the problem, we need to understand a few features of Binary Search Tree.

A binary search tree is a tree-like data structure with added features as follows.

The subtree to the left side of the root node will always be lesser than the root node.

The subtree to the right side of the root node will always be greater than the root node.

The left and right subtree of the root node must be a Binary Search Tree

#python #coding #leetcode-medium #binary-search-tree #binary search

1596650400

Given an array **A[ ]** consisting of **N** distinct integers, the task is to find the number of elements which are **strictly greater **than all the elements preceding it and strictly greater than at least **K** elements on its right.

**Examples:**

_ A[] = {2, 5, 1, 7, 3, 4, 0}, K = 3_Input:

_ 2_Output:

Explanation:

The only array elements satisfying the given conditions are:

*5**: Greater than all elements on its left {2} and at least K(= 3) elements on its right {1, 3, 4, 0}**7**: Greater than all elements on its left {2, 5, 1} and at least K(= 3) elements on its right {3, 4, 0}*

Therefore, the count is 2.

_ A[] = {11, 2, 4, 7, 5, 9, 6, 3}, K = 2_Input:

_ 1_Output:

**Naive Approach:**

The simplest approach to solve the problem is to traverse the array and for each element, traverse all the elements on its left and check if all of them are smaller than it or not and traverse all elements on its right to check if at least **K** elements are smaller than it or not. For every element satisfying the conditions, increase **count**. Finally, print the value of **count**.

** Time Complexity:**_ O(N2)_

** Auxiliary Space:**_ O(1)_

**Efficient Approach:**

The above approach can be further optimized by using Self-Balancing BST. Follow the steps below:

- Traverse the array from right to left and insert all elements one by one in an AVL Tree
- Using the AVL Tree generate an array
**countSmaller[]**which contains the count of smaller elements on the right of every array element. - Traverse the array and for every
**ith element**, check if it is the maximum obtained so far and**countSmaller[i]**is greater than or equal to**K**. - If so, increase
**count**. - Print the final value of
**count**as the answer.

Below is the implementation of the above approach:

- C++

`// C++ Program to implement`

`// the above appraoch`

`#include <bits/stdc++.h>`

`**using**`

`**namespace**`

`std;`

`// Structure of an AVL Tree Node`

`**struct**`

`node {`

`**int**`

`key;`

`**struct**`

`node* left;`

`**struct**`

`node* right;`

`**int**`

`height;`

`// Size of the tree rooted`

`// with this node`

`**int**`

`size;`

`};`

`// Utility function to get maximum`

`// of two integers`

`**int**`

`max(``**int**`

`a,`

`**int**`

`b);`

`// Utility function to get height`

`// of the tree rooted with N`

`**int**`

`height(``**struct**`

`node* N)`

`{`

`**if**`

`(N == NULL)`

`**return**`

`0;`

`**return**`

`N->height;`

`}`

`// Utility function to find size of`

`// the tree rooted with N`

`**int**`

`size(``**struct**`

`node* N)`

`{`

`**if**`

`(N == NULL)`

`**return**`

`0;`

`**return**`

`N->size;`

`}`

`// Utility function to get maximum`

`// of two integers`

`**int**`

`max(``**int**`

`a,`

`**int**`

`b)`

`{`

`**return**`

`(a > b) ? a : b;`

`}`

`// Helper function to allocates a`

`// new node with the given key`

`**struct**`

`node* newNode(``**int**`

`key)`

`{`

`**struct**`

`node* node`

`= (``**struct**`

`node*)`

`**malloc**``(``**sizeof**``(``**struct**`

`node));`

`node->key = key;`

`node->left = NULL;`

`node->right = NULL;`

`node->height = 1;`

`node->size = 1;`

`**return**`

`(node);`

`}`

`// Utility function to right rotate`

`// subtree rooted with y`

`**struct**`

`node* rightRotate(``**struct**`

`node* y)`

`{`

`**struct**`

`node* x = y->left;`

`**struct**`

`node* T2 = x->right;`

`// Perform rotation`

`x->right = y;`

`y->left = T2;`

`// Update heights`

`y->height = max(height(y->left),`

`height(y->right))`

`+ 1;`

`x->height = max(height(x->left),`

`height(x->right))`

`+ 1;`

`// Update sizes`

`y->size = size(y->left)`

`+ size(y->right) + 1;`

`x->size = size(x->left)`

`+ size(x->right) + 1;`

`// Return new root`

`**return**`

`x;`

`}`

`// Utility function to left rotate`

`// subtree rooted with x`

`**struct**`

`node* leftRotate(``**struct**`

`node* x)`

`{`

`**struct**`

`node* y = x->right;`

`**struct**`

`node* T2 = y->left;`

`// Perform rotation`

`y->left = x;`

`x->right = T2;`

`// Update heights`

`x->height = max(height(x->left),`

`height(x->right))`

`+ 1;`

`y->height = max(height(y->left),`

`height(y->right))`

`+ 1;`

`// Update sizes`

`x->size = size(x->left)`

`+ size(x->right) + 1;`

`y->size = size(y->left)`

`+ size(y->right) + 1;`

`// Return new root`

`**return**`

`y;`

`}`

`// Function to obtain Balance factor`

`// of node N`

`**int**`

`getBalance(``**struct**`

`node* N)`

`{`

`**if**`

`(N == NULL)`

`**return**`

`0;`

`**return**`

`height(N->left)`

`- height(N->right);`

`}`

`// Function to insert a new key to the`

`// tree rooted with node`

`**struct**`

`node* insert(``**struct**`

`node* node,`

`**int**`

`key,`

`**int**``* count)`

`{`

`// Perform the normal BST rotation`

`**if**`

`(node == NULL)`

`**return**`

`(newNode(key));`

`**if**`

`(key < node->key)`

`node->left`

`= insert(node->left, key, count);`

`**else**`

`{`

`node->right`

`= insert(node->right, key, count);`

`// Update count of smaller elements`

`*count = *count + size(node->left) + 1;`

`}`

`// Update height and size of the ancestor`

`node->height = max(height(node->left),`

`height(node->right))`

`+ 1;`

`node->size = size(node->left)`

`+ size(node->right) + 1;`

`// Get the balance factor of the ancestor`

`**int**`

`balance = getBalance(node);`

`// Left Left Case`

`**if**`

`(balance > 1 && key < node->left->key)`

`**return**`

`rightRotate(node);`

`// Right Right Case`

`**if**`

`(balance < -1 && key > node->right->key)`

`**return**`

`leftRotate(node);`

`// Left Right Case`

`**if**`

`(balance > 1 && key > node->left->key) {`

`node->left = leftRotate(node->left);`

`**return**`

`rightRotate(node);`

`}`

`// Right Left Case`

`**if**`

`(balance < -1 && key < node->right->key) {`

`node->right = rightRotate(node->right);`

`**return**`

`leftRotate(node);`

`}`

`**return**`

`node;`

`}`

`// Function to generate an array which contains`

`// count of smaller elements on the right`

`**void**`

`constructLowerArray(``**int**`

`arr[],`

`**int**`

`countSmaller[],`

`**int**`

`n)`

`{`

`**int**`

`i, j;`

`**struct**`

`node* root = NULL;`

`**for**`

`(i = 0; i < n; i++)`

`countSmaller[i] = 0;`

`// Insert all elements in the AVL Tree`

`// and get the count of smaller elements`

`**for**`

`(i = n - 1; i >= 0; i--) {`

`root = insert(root, arr[i],`

`&countSmaller[i]);`

`}`

`}`

`// Function to find the number`

`// of elements which are greater`

`// than all elements on its left`

`// and K elements on its right`

`**int**`

`countElements(``**int**`

`A[],`

`**int**`

`n,`

`**int**`

`K)`

`{`

`**int**`

`count = 0;`

`// Stores the count of smaller`

`// elements on its right`

`**int**``* countSmaller`

`= (``**int**``*)``**malloc**``(``**sizeof**``(``**int**``) * n);`

`constructLowerArray(A, countSmaller, n);`

`**int**`

`maxi = INT_MIN;`

`**for**`

`(``**int**`

`i = 0; i <= (n - K - 1); i++) {`

`**if**`

`(A[i] > maxi && countSmaller[i] >= K) {`

`count++;`

`maxi = A[i];`

`}`

`}`

`**return**`

`count;`

`}`

`// Driver Code`

`**int**`

`main()`

`{`

`**int**`

`A[] = { 2, 5, 1, 7, 3, 4, 0 };`

`**int**`

`n =`

`**sizeof**``(A) /`

`**sizeof**``(``**int**``);`

`**int**`

`K = 3;`

`cout << countElements(A, n, K);`

`**return**`

`0;`

`}`

**Output:**

```
2
```

**_Time Complexity: _***O(NlogN)*

**_Auxiliary Space: _***O(N)*

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

#advanced data structure #arrays #binary search tree #mathematical #searching #avl-tree #balanced binary search trees #rotation

1604008800

Static code analysis refers to the technique of approximating the runtime behavior of a program. In other words, it is the process of predicting the output of a program *without* actually executing it.

Lately, however, the term “Static Code Analysis” is more commonly used to refer to one of the applications of this technique rather than the technique itself — **program comprehension** — understanding the program and detecting issues in it (anything from syntax errors to type mismatches, performance hogs likely bugs, security loopholes, etc.). This is the usage we’d be referring to throughout this post.

“The refinement of techniques for the prompt discovery of error serves as well as any other as a hallmark of what we mean by science.”

- J. Robert Oppenheimer

We cover a lot of ground in this post. The aim is to build an understanding of static code analysis and to equip you with the basic theory, and the right tools so that you can write analyzers on your own.

We start our journey with laying down the essential parts of the pipeline which a compiler follows to *understand* what a piece of code does. We learn where to tap points in this pipeline to plug in our analyzers and extract meaningful information. In the latter half, we get our feet wet, and write four such static analyzers, completely from scratch, in Python.

Note that although the ideas here are discussed in light of Python, static code analyzers across all programming languages are carved out along similar lines. We chose Python because of the availability of an easy to use `ast`

module, and wide adoption of the language itself.

Before a computer can finally *“understand”* and execute a piece of code, it goes through a series of complicated transformations:

As you can see in the diagram (go ahead, zoom it!), the static analyzers feed on the output of these stages. To be able to better understand the static analysis techniques, let’s look at each of these steps in some more detail:

The first thing that a compiler does when trying to understand a piece of code is to break it down into smaller chunks, also known as tokens. Tokens are akin to what words are in a language.

A token might consist of either a single character, like `(`

, or literals (like integers, strings, e.g., `7`

, `Bob`

, etc.), or reserved keywords of that language (e.g, `def`

in Python). Characters which do not contribute towards the semantics of a program, like trailing whitespace, comments, etc. are often discarded by the scanner.

Python provides the `tokenize`

module in its standard library to let you play around with tokens:

Python

1

```
import io
```

2

```
import tokenize
```

3

4

```
code = b"color = input('Enter your favourite color: ')"
```

5

6

```
for token in tokenize.tokenize(io.BytesIO(code).readline):
```

7

```
print(token)
```

Python

1

```
TokenInfo(type=62 (ENCODING), string='utf-8')
```

2

```
TokenInfo(type=1 (NAME), string='color')
```

3

```
TokenInfo(type=54 (OP), string='=')
```

4

```
TokenInfo(type=1 (NAME), string='input')
```

5

```
TokenInfo(type=54 (OP), string='(')
```

6

```
TokenInfo(type=3 (STRING), string="'Enter your favourite color: '")
```

7

```
TokenInfo(type=54 (OP), string=')')
```

8

```
TokenInfo(type=4 (NEWLINE), string='')
```

9

```
TokenInfo(type=0 (ENDMARKER), string='')
```

(Note that for the sake of readability, I’ve omitted a few columns from the result above — metadata like starting index, ending index, a copy of the line on which a token occurs, etc.)

#code quality #code review #static analysis #static code analysis #code analysis #static analysis tools #code review tips #static code analyzer #static code analysis tool #static analyzer

1599043260

Given a **Binary Tree** and an integer **D**, the task is to check if the distance between all pairs of same node values in the Tree is ? **D** or not. If found to be true, then print **Yes**. Otherwise, print **No**.

**Examples:**

_ D = 7 _Input:

```
1
/ \
2 3
/ \ / \
4 3 4 4
```

_ Yes _Output:

Explanation:

_The repeated value of nodes are 3 and 4. _

_The distance between the two nodes valued 3, is 3. _

_The maximum distance between any pair of nodes valued 4 is 4. _

Therefore, none of the distances exceed 7

_ D = 1 _Input:

```
3
/ \
3 3
\
3
```

_ No _Output:

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

**Approach: **

The idea is to observe that the problem is similar to finding the distance between two nodes of a tree. But there can be multiple pairs of nodes for which we have to find the distance. Follow the steps below:

- Perform the Post Order Traversal of the given tree and find the distance between the repeated pairs of nodes.
- Find the nodes that are repeated in the tree using unordered_map.
- For each repeated node of a particular value, find the maximum possible distance between any pair.
- If that distance is >
**D**, print “No”. - If no such node value is found having a pair containing that value, exceeding **D, **then print “Yes”.

#greedy #recursion #searching #tree #binary tree #frequency-counting #postorder traversal #tree-traversal