Category: Python Coding Test

Hello! In this lecture, we will cover an algorithm problem to calculate Ichin numbers. An Ichin number is a number made up of 0s and 1s that does not have two consecutive 1s. For example, 3-digit Ichin numbers are 000, 001, 010, 100, 101, 110, which totals to 6.

Problem Description

Given an integer N, we will solve the problem of finding all N-digit Ichin numbers and outputting their count.

Problem

Input N and output the count of all N-digit Ichin numbers.

Example Input

N = 3

Example Output

6

Approach to the Problem

There are two approaches to solve this problem. The first method uses recursion with DFS (Depth-First Search), and the second method uses Dynamic Programming. Let’s take a detailed look at each method.

1. Method Using Recursive DFS

The recursive method follows these two rules to generate Ichin numbers:

• If the current digit is 0, we can place either 0 or 1 in the next position.
• If the current digit is 1, we can only place 0 in the next position.

According to these rules, we can write a recursive function to generate Ichin numbers. Here is the implementation:

def count_ichin(N, current_sequence, last_digit):
    if len(current_sequence) == N:
        return 1

    count = 0
    if last_digit == 0:
        count += count_ichin(N, current_sequence + '0', 0)
        count += count_ichin(N, current_sequence + '1', 1)
    else:
        count += count_ichin(N, current_sequence + '0', 0)

    return count

N = 3
result = count_ichin(N, '', 0)
print(result)

The above code defines a recursive function count_ichin() to generate Ichin numbers, passing N and an empty string as initial values. The last digit starts as 0.

2. Method Using Dynamic Programming

When calculating Ichin numbers, using dynamic programming allows for a more efficient solution through memorization. The count of Ichin numbers I(n) can be defined with the following recurrence relation:

I(n) = I(n-1) + I(n-2)

The meaning of this equation is as follows:

• If you place 0 in the n-1 position: The count of Ichin numbers is I(n-1).
• If you place 10 in the n-2 position: The count of Ichin numbers is I(n-2).

Now we will implement dynamic programming based on this recurrence relation:

def find_ichin_count(N):
    if N == 1:
        return 1
    elif N == 2:
        return 1

    dp = [0] * (N + 1)
    dp[1] = 1
    dp[2] = 1

    for i in range(3, N + 1):
        dp[i] = dp[i - 1] + dp[i - 2]

    return dp[N]

N = 3
result = find_ichin_count(N)
print(result)

With the above code, the dynamic programming approach to finding Ichin numbers has been efficiently implemented.

Comparison and Selection

The recursive method is easy to understand but may be inefficient for large N values. In contrast, the dynamic programming method uses memory to reuse previous computation results, making it more performant. Generally, it is advisable to use dynamic programming for larger N values.

Conclusion

In this lecture, we discussed the problem of finding Ichin numbers. We learned to calculate Ichin numbers using both recursive and dynamic programming methods. I hope this problem helps you enhance your algorithmic problem-solving skills.

Thank you!

A binary tree is one of the fundamental data structures in computer science and algorithms, playing a crucial role in many problems. Understanding binary trees and the ability to solve problems involving them are highly valued in coding interviews. In this article, we will select one problem related to binary trees and take a detailed look at the problem description and the solution process.

Problem: Maximum Depth of a Binary Tree

Write a function to find the maximum depth of a given binary tree. The depth of a binary tree is the number of nodes along the longest path from the root node down to the farthest leaf node. For example, let’s assume we have a binary tree as follows.

      1
     / \
    2   3
   / \
  4   5

In this case, the maximum depth of the binary tree is 3 (node 1 → node 2 → node 4 or node 5). The signature of the function is as follows:

def maxDepth(root: TreeNode) -> int:

Problem Definition

The input parameter root given as input is the root node of the binary tree. This node is defined as an instance of the TreeNode class, which has pointers pointing to its left and right child nodes. If the binary tree is empty, the depth is 0.

Input Example

      1
     / \
    2   3
   / \
  4   5

When calling maxDepth(root), the return value should be 3.

Output Example

3

Problem Solving Approach

A Depth-First Search (DFS) approach can be used to solve this problem. By using the DFS method to traverse the nodes of the tree, we can recursively calculate the depth from each node to its leaf nodes.

Step 1: Define the TreeNode Class

First, we need to write the TreeNode class that defines the nodes of the binary tree. Each node has a value and pointers to its left and right children.

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

Step 2: Define the Recursive Function for Maximum Depth

We will define a recursive function to calculate the maximum depth. We will use recursive calls to determine the depth of each subtree and select the maximum value among them.

def maxDepth(root: TreeNode) -> int:
    # Base case: If the node is None, the depth is 0
    if not root:
        return 0
    # Calculate the depth of the left and right subtrees
    left_depth = maxDepth(root.left)
    right_depth = maxDepth(root.right)
    # Return the maximum depth including the current node
    return max(left_depth, right_depth) + 1

Step 3: Implement the Final Function

We have now implemented the complete maxDepth function. This function returns the ‘maximum depth’ of the given binary tree.

Step 4: Analyze Time Complexity and Space Complexity

The time complexity of this algorithm is O(n), where n is the number of nodes in the binary tree. The time complexity is proportional to the size of the tree, as we visit each node once. The space complexity is O(h), where h is the height of the tree. In the worst case, the space complexity can be O(n), while in a balanced binary tree, it will be O(log n).

Test Cases

Let’s write some test cases to validate the function we created.

# Test cases 
def test_maxDepth():
    # Test case 1
    root1 = TreeNode(1, TreeNode(2, TreeNode(4), TreeNode(5)), TreeNode(3))
    assert maxDepth(root1) == 3, "Test case 1 failed"
    
    # Test case 2
    root2 = TreeNode(1)
    assert maxDepth(root2) == 1, "Test case 2 failed"
    
    # Test case 3: Empty tree
    root3 = None
    assert maxDepth(root3) == 0, "Test case 3 failed"
    
    # Test case 4
    root4 = TreeNode(1, TreeNode(2))
    assert maxDepth(root4) == 2, "Test case 4 failed"
    
    print("All test cases passed!")

test_maxDepth()

Conclusion

In this article, we introduced the problem of finding the maximum depth of a binary tree and explained the solution method in detail. There are many diverse problems related to binary trees, so it is important to practice by encountering various challenges. Problems like the Algorithm Challenge can help improve your skills further. Understanding the concept of binary trees and the basic principles of DFS traversal is greatly beneficial in coding tests. I hope you will continue to solve various algorithm problems to enhance your abilities.

1. What is Binary Search?

Binary Search is a highly efficient search algorithm used to find a specific value in a sorted array. This algorithm works by dividing the given list in half to search for the desired value, making it much faster than the typical Linear Search.

The key idea behind binary search is to take advantage of the fact that the list is sorted. The algorithm operates through the following steps:

1. Find the middle index of the list.
2. Check if the middle element matches the value you are looking for.
3. If they do not match, adjust the search range based on the comparison between the middle element and the target value. If it is smaller than the middle element, search the left half; if larger, search the right half.
4. Repeat this process until the target value is found.

2. Time Complexity of Binary Search

The time complexity of the binary search algorithm is O(log n). This is because it halves the search space at each step when the size of the list to be searched is n. As a result, binary search works efficiently even with very large data sets.

3. Problem: Find the Index of a Specific Value

Problem Description

Given a sorted integer array arr and an integer target, write a binary search function that returns the index of target. If the target does not exist, it should return -1.

Input

• The first line contains the size of the array n. (1 ≤ n ≤ 10^5)
• The second line contains n integers separated by spaces.
• The third line contains the target value target. (-10^9 ≤ target ≤ 10^9)

Output

Print the index of target. A value of -1 indicates that the target does not exist.

4. Example Input for the Problem

                5
                1 2 3 4 5
                3
            

Example Output

2

5. Problem Solving Process

This section describes the necessary steps to solve the problem. We will go through each step to find the target value in the given array using the binary search algorithm.

5.1 Implementation of the Algorithm

First, let’s define the basic structure needed to implement binary search. The function will take the array and the target value as arguments and return the index or -1. Now, let’s write the code.

                def binary_search(arr, target):
                    left, right = 0, len(arr) - 1
                    
                    while left <= right:
                        mid = (left + right) // 2
                        
                        if arr[mid] == target:
                            return mid
                        elif arr[mid] < target:
                            left = mid + 1
                        else:
                            right = mid - 1
                    return -1
            

5.2 Code Explanation

The above code is a simple implementation of the binary search algorithm. left and right indicate the current search range. Initially, left is 0 and right is the last index of the array.

while left <= right: The condition runs while left is less than or equal to right. It calculates the middle value and stores it in mid, and adjusts the range based on comparisons with that value.

5.3 Input Handling and Output

Next, let's add the part that handles input and calls the binary_search function to print the result.

                n = int(input())
                arr = list(map(int, input().split()))
                target = int(input())
                
                result = binary_search(arr, target)
                print(result)
            

6. Code Optimization

While the above code performs basic binary search, there are ways to optimize it further. Particularly in Python, a more convenient method can be used to calculate the middle index of the list. To simplify the process, it is advisable to use Python's integer division instead of directly adding values to calculate mid.

                def binary_search_optimized(arr, target):
                    left, right = 0, len(arr) - 1
                    
                    while left <= right:
                        mid = left + (right - left) // 2
                        if arr[mid] == target:
                            return mid
                        elif arr[mid] < target:
                            left = mid + 1
                        else:
                            right = mid - 1
                    return -1
            

By calculating mid this way, it helps to prevent overflow in Python's memory. This makes the calculation of the middle value safer and more efficient.

7. Variations of the Problem

The binary search algorithm can be applied to various variations beyond finding the index of a specific value. For example, problems include finding the leftmost (or rightmost) index in an array or finding the maximum (or minimum) value that satisfies a specific condition.

7.1 Example Problem: Finding the First Position

Let's solve the problem of finding the first position of a specific value in a given integer array. To do this, we will use binary search, but if the mid value equals the target value, we will continue searching to the left.

                def binary_search_first(arr, target):
                    left, right = 0, len(arr) - 1
                    result = -1  # Variable to store the result
                    
                    while left <= right:
                        mid = left + (right - left) // 2
                        if arr[mid] == target:
                            result = mid     # Store the current index
                            right = mid - 1  # Search left
                        elif arr[mid] < target:
                            left = mid + 1
                        else:
                            right = mid - 1
                    return result
            

8. Conclusion

Binary search is a highly efficient algorithm for finding a specific value in a sorted array. In this tutorial, we covered the basic concepts of binary search, its time complexity, algorithm implementation, optimization, and variations. By using binary search, you can effectively solve problems frequently encountered in coding tests. Continue to practice various problems and master techniques utilizing binary search to enhance your coding skills.

We hope you continue learning various algorithms through more tutorials and problem-solving sessions. Algorithm problems can improve your skills through repetitive practice, so maintain a persistent attitude toward challenges.

In this article, we will discuss the concept of Bipartite Graphs and the algorithmic problem for determining whether a graph is bipartite.
A bipartite graph is one that can be divided into two sets of nodes, and the key is to determine whether adjacent nodes belong to different sets.

Problem Description

We will solve the problem of determining whether a given graph, made up of vertices and edges, is bipartite.
Specifically, this problem will take the following input.

• First line: The number of vertices n and the number of edges m are given, separated by a space.
• Next m lines: Two vertices u and v connected by each edge are provided.

The output will be YES if the given graph is bipartite; otherwise, it will output NO.

Example Input 1

3 3
1 2
2 3
1 3

Example Output 1

NO

Example Input 2

3 2
1 2
2 3

Example Output 2

YES

Problem Solving Process

1. Understanding the Definition of a Bipartite Graph

A bipartite graph is one where, when each node is divided into two groups, no nodes within the same group are connected.
Such graphs can generally be identified through the possibility of bipartite coloring.

In other words, when coloring a node, the adjacent nodes should be colored with the opposite color, and if there are no nodes colored the same until the end, it is a bipartite graph.

2. Graph Representation Method

To represent the given graph, we will use an adjacency list. We maintain a list of vertices connected to each vertex.
In Python, we can easily construct the graph using a dictionary.

Python Code Example (Graph Construction)

def create_graph(n, edges):
    graph = {i: [] for i in range(1, n + 1)}
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)
    return graph

3. Coloring Using BFS or DFS

We can use either the BFS or DFS algorithm to traverse the graph. We will use the BFS method to determine if the graph is bipartite.

The basic idea of BFS is to color the starting node with an arbitrary color and proceed to color all adjacent nodes with the opposite color.
If any adjacent node is already colored and matches the current color we are trying to color it with, then it is not a bipartite graph.

Python Code Example (BFS Coloring)

from collections import deque

def is_bipartite(graph, n):
    color = {}
    for node in range(1, n + 1):
        if node not in color:
            queue = deque([node])
            color[node] = 0  # Color the starting node

            while queue:
                current = queue.popleft()

                for neighbor in graph[current]:
                    if neighbor not in color:
                        color[neighbor] = 1 - color[current]  # Color with opposite color
                        queue.append(neighbor)
                    elif color[neighbor] == color[current]:
                        # If the same color, then it is not a bipartite graph
                        return False
    return True

4. Implementing the Entire Program

Now we will integrate the graph construction and the bipartite determination logic to complete the entire program.

def main():
    n, m = map(int, input().split())
    edges = [tuple(map(int, input().split())) for _ in range(m)]

    graph = create_graph(n, edges)
    
    if is_bipartite(graph, n):
        print("YES")
    else:
        print("NO")

if __name__ == "__main__":
    main()

Conclusion

In this article, we explored the concept of bipartite graphs and the algorithmic problem of determining them.
We explained an efficient method to identify bipartite graphs through BFS and examined a more practical approach using Python code examples.

We plan to cover various algorithm topics in the future, so please continue to stay tuned. Thank you!