Codeforces 677A – Vanya and Fence Solution in Java (Step-by-Step Explanation)

Introduction:

Codeforces 677A – Vanya and Fence is one of the most popular beginner-friendly competitive programming problems. If you are preparing for coding interviews or improving your problem-solving skills, this problem helps you understand basic loops and conditional logic.

In this article, we will break down the logic clearly and implement the solution in Java with a detailed explanation.

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Problem Summary

Vanya and his friends are walking in a straight line. Each friend has a certain height. There is a fence of height h.

Rules:

1. If a friend's height is less than or equal to h, they take 1 unit width.
2. If a friend's height is greater than h, they take 2 units width.

Your task is to calculate the total width required for all friends to pass through the fence without bending.

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Input Format

1. First line contains two integers:
1. n → number of friends
2. h → height of the fence
2. Second line contains n integers representing heights of friends

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Output

Print a single integer — the total width required.

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Understanding the Logic

Let’s break it down step by step:

Read the number of friends (n) and fence height (h).

1. Initialize a variable totalWidth = 0.

2. Loop through each friend's height:

   • If height ≤ h → add 1 to totalWidth

   • If height > h → add 2 to totalWidth

3. Print the totalWidth.

Why This Works

We simply count:

• 1 unit for shorter or equal height friends
• 2 units for taller friends

This problem mainly tests:

1. Looping
2. Conditional statements
3. Basic input handling

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Example Explanation

Input:

3 7
4 5 14

Step-by-step:

• Friend 1 → height 4 ≤ 7 → width = 1
• Friend 2 → height 5 ≤ 7 → width = 1
• Friend 3 → height 14 > 7 → width = 2

Total Width = 1 + 1 + 2 = 4

Output:

4

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Java Implementation

import java.util.Scanner;
 public class Main{
    public static void main(String args[]){
        // create constructor to take input from user using console
        Scanner sc = new Scanner(System.in);
        
        int noEle = sc.nextInt();
        int targetEle = sc.nextInt();
        // array to store the values
        int nums[] = new int[noEle];
        // store response
        int res =0;
        // take input from user and store the elements in an array
        for(int pos =0;pos<noEle;pos++){
  nums[pos] = sc.nextInt();
        }
        // check the logic explained below
        for(int num:nums){
                res = num <= targetEle ? res + 1 : res + 2;
        }
        System.out.println(res);
        // make sure to close the reader to prevent attacks
        sc.close();
    }
}

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Time and Space Complexity

• Time Complexity: O(n)
  (We loop through all friends once)

• Space Complexity: O(1)
  (Only one variable used for counting)

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Why This Problem Is Important for Beginners

This problem helps you:

1. Understand basic conditional logic
2. Practice input/output handling
3. Improve loop implementation
4. Build foundation for harder competitive programming problems

If you are new to competitive programming, this is a great starting point.

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Frequently Asked Questions

1. What is the difficulty level of Codeforces 677A?

It is considered a beginner-level problem, suitable for those starting competitive programming.

2. Which concepts are used in this problem?

Basic loops, conditional statements, and simple arithmetic.

3. Can this problem be solved in other languages?

Yes. It can be solved in C++, Python, Java, and other languages easily.

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Final Thoughts

Codeforces 677A – Vanya and Fence is a simple yet effective problem to strengthen your fundamentals. By understanding this logic, you prepare yourself for more advanced competitive programming challenges.

If you found this explanation helpful, try solving similar beginner-level problems to improve your coding skills consistently.