• Problem Statement
• Key Insight
• Code
• Conclusion

Codeforces Problem 1850D - Balanced Round

In this blog, we’ll explore the solution to D. Balanced Round, a fascinating problem from Codeforces. This post explains the problem, discusses our thought process, and presents the final solution in a simple and clear manner.

problem link: Codeforces Problem 1850D - Balanced Round from Codeforces.

Problem Statement

We are given a list of n problems, each with a difficulty level a[i]. A round is balanced if the absolute difference between the difficulties of any two consecutive problems is at most k.

To achieve this:

• We can remove some problems (possibly zero).
• We can rearrange the remaining problems in any order.

Our goal is to find the minimum number of problems to remove to make the round balanced.

Input Format

1. An integer t representing the number of test cases.
2. For each test case:
   • Two integers, n (number of problems) and k (maximum allowed difference).
   • An array a of size n, representing the difficulty levels.

Key Observations

1. Rearrangement Doesn't Matter: Since we can reorder the array, the problem boils down to finding a subset of problems such that the maximum difference between consecutive elements is ≤ k.
2. Sorted Array Insight: After sorting, consecutive elements have the smallest possible differences. The task reduces to finding the largest contiguous subarray in the sorted array where the difference between the smallest and largest element is ≤ k.

The number of problems to remove is:

problems to remove = n - maxSubarray

Our Solution

To solve the problem, we:

1. Sort the Array: Sorting ensures consecutive elements are as close in value as possible.
2. Find the Largest Valid Subarray: Use a loop to track the size of the largest subarray where all consecutive differences are ≤ k.
3. Calculate the Result: Subtract the size of the largest valid subarray from the total number of problems.

Example Walkthrough

Input:

2
6 3
1 5 7 9 13 15
5 4
1 10 15 20 25
    

Output:

3
3
    

Explanation:

• Test Case 1:
  • Sorted array: [1, 5, 7, 9, 13, 15]
  • Largest subarray with differences ≤ 3: [5, 7, 9] (size = 3).
  • Minimum removals: 6 - 3 = 3.
• Test Case 2:
  • Sorted array: [1, 10, 15, 20, 25]
  • Largest subarray with differences ≤ 4: [1] (size = 1).
  • Minimum removals: 5 - 1 = 4.

Code Implementations

Below are the implementations in Java and C++.

C++

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    int t;
    cin >> t;

    while (t--) {
        int n, k;
        cin >> n >> k;
        vector<int> arr(n);

        for (int i = 0; i < n; ++i) {
            cin >> arr[i];
        }

        sort(arr.begin(), arr.end());

        int maxSubarray = 1;
        int currentLength = 1;

        for (int i = 1; i < n; ++i) {
            if (arr[i] - arr[i - 1] <= k) {
                currentLength++;
            } else {
                maxSubarray = max(maxSubarray, currentLength);
                currentLength = 1;
            }
        }

        maxSubarray = max(maxSubarray, currentLength);
        cout << n - maxSubarray << endl;
    }

    return 0;
}
    
        

JAVA

import java.util.*;

public class BalancedRound {

    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        int t = scn.nextInt();  // Number of test cases

        while (t-- > 0) {
            int n = scn.nextInt();  // Length of the array
            int k = scn.nextInt();  // Maximum allowed difference
            int[] arr = new int[n];

            // Read array elements
            for (int i = 0; i < n; i++) {
                arr[i] = scn.nextInt();
            }

            // Sort the array
            Arrays.sort(arr);

            // Find the longest subarray with differences ≤ k
            int maxSubarray = 1; // Maximum valid subarray size
            int currentLength = 1; // Current valid subarray size

            for (int i = 1; i < n; i++) {
                if (arr[i] - arr[i - 1] <= k) {
                    currentLength++; // Extend the current subarray
                } else {
                    maxSubarray = Math.max(maxSubarray, currentLength);
                    currentLength = 1; // Reset for a new subarray
                }
            }

            // Ensure the maximum size is updated after the loop
            maxSubarray = Math.max(maxSubarray, currentLength);

            // The minimum elements to remove are n - maxSubarray
            System.out.println(n - maxSubarray);
        }

        scn.close();
    }
}
        

Complexity Analysis

• Sorting: Sorting the array takes O(n log n).
• Finding the Largest Subarray: A single pass through the array takes O(n).
• Total Complexity: For t test cases, the complexity is: O(T * n log n), where T is the number of test cases and n is the size of the largest array.

Conclusion

This solution provides a clean, efficient approach to solving D. Balanced Round. Sorting and the sliding window technique ensure the solution is both optimal and easy to implement. Let me know if you have questions or suggestions!