Rearrange Sequence Problem

Rearrange Sequence Problem – Largest Subarray that Can Be Rearranged to Form a Contiguous Sequence

Problem Statement:

You are given an array of size N containing integers which may not be unique. Your task is to find the size of the largest subarray that can be rearranged to form a strictly contiguous sequence.

A contiguous sequence is a set of numbers that are in consecutive order (e.g., [1, 2, 3, 4]).

Example 1:

Input:

5
4 3 3 1 1

Output:

2

Explanation:
The largest subarray that can be rearranged to form a contiguous sequence here is {4, 3}, which can be rearranged to {3, 4}.

Best Technique – Brute Force Approach

This problem can be solved in several ways. Let's start by breaking down the brute force approach, which works by checking all possible subarrays of the given array. It’s not the most efficient solution, but it’s easy to understand and implement.

The key idea is:

  • For every possible subarray, check if it can be rearranged into a contiguous sequence.

  • A contiguous sequence has the property that max - min + 1 == length of the subarray.

Different Approaches:

Approach 1: Brute Force (Checking all subarrays)

This approach checks all possible subarrays of the given array. For each subarray, we compute its min and max values, and check if the difference between them plus 1 is equal to the subarray length. If it is, the subarray can be rearranged into a contiguous sequence.

import java.util.*;

public class RearrangeSequence {

    // Helper function to check if a subarray can form a contiguous sequence
    public static boolean isContiguous(List<Integer> arr) {
        int max = Collections.max(arr);
        int min = Collections.min(arr);
        return max - min + 1 == arr.size();
    }

    // Brute force solution to find the largest contiguous subarray
    public static int bruteForceLargestContiguousSubarray(int[] arr) {
        int n = arr.length;
        int maxLen = 0;

        // Check all subarrays
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                List<Integer> subarray = new ArrayList<>();
                for (int k = i; k <= j; k++) {
                    subarray.add(arr[k]);
                }
                if (isContiguous(subarray)) {
                    maxLen = Math.max(maxLen, subarray.size());
                }
            }
        }
        return maxLen;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt(); // Number of test cases
        while (T-- > 0) {
            int N = sc.nextInt(); // Size of the array
            int[] arr = new int[N];
            for (int i = 0; i < N; i++) {
                arr[i] = sc.nextInt(); // Input array
            }
            System.out.println(bruteForceLargestContiguousSubarray(arr)); // Output the result for each test case
        }
    }
}

Approach 2: Optimized Approach (Using Sorting)

We can optimize the brute force approach by sorting the subarray first and then checking whether the array forms a contiguous sequence. If the subarray after sorting has max - min + 1 == length, then it's a contiguous sequence.

import java.util.*;

public class RearrangeSequence {

    // Helper function to check if a subarray can form a contiguous sequence
    public static boolean isContiguous(List<Integer> arr) {
        Collections.sort(arr);  // Sort the subarray
        int max = arr.get(arr.size() - 1);
        int min = arr.get(0);
        return max - min + 1 == arr.size();
    }

    // Optimized approach to find the largest contiguous subarray
    public static int optimizedLargestContiguousSubarray(int[] arr) {
        int n = arr.length;
        int maxLen = 0;

        // Check all subarrays
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                List<Integer> subarray = new ArrayList<>();
                for (int k = i; k <= j; k++) {
                    subarray.add(arr[k]);
                }
                if (isContiguous(subarray)) {
                    maxLen = Math.max(maxLen, subarray.size());
                }
            }
        }
        return maxLen;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt(); // Number of test cases
        while (T-- > 0) {
            int N = sc.nextInt(); // Size of the array
            int[] arr = new int[N];
            for (int i = 0; i < N; i++) {
                arr[i] = sc.nextInt(); // Input array
            }
            System.out.println(optimizedLargestContiguousSubarray(arr)); // Output the result for each test case
        }
    }
}

Approach 3: Efficient Approach (Using HashSet)

In this approach, we use a HashSet to check if the elements in the subarray can form a contiguous sequence without sorting. The idea is to keep track of unique elements, and then check if the difference between the max and min elements is equal to the length of the subarray minus 1.

import java.util.*;

public class RearrangeSequence {

    // Helper function to check if a subarray can form a contiguous sequence
    public static boolean isContiguous(List<Integer> arr) {
        Set<Integer> uniqueElements = new HashSet<>(arr);
        int max = Collections.max(arr);
        int min = Collections.min(arr);
        return max - min + 1 == uniqueElements.size();
    }

    // Efficient approach to find the largest contiguous subarray
    public static int efficientLargestContiguousSubarray(int[] arr) {
        int n = arr.length;
        int maxLen = 0;

        // Check all subarrays
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                List<Integer> subarray = new ArrayList<>();
                for (int k = i; k <= j; k++) {
                    subarray.add(arr[k]);
                }
                if (isContiguous(subarray)) {
                    maxLen = Math.max(maxLen, subarray.size());
                }
            }
        }
        return maxLen;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt(); // Number of test cases
        while (T-- > 0) {
            int N = sc.nextInt(); // Size of the array
            int[] arr = new int[N];
            for (int i = 0; i < N; i++) {
                arr[i] = sc.nextInt(); // Input array
            }
            System.out.println(efficientLargestContiguousSubarray(arr)); // Output the result for each test case
        }
    }
}

⚠️ Common Pitfalls:

  1. Off-by-one Errors: Ensure you account for zero-based indexing when defining your subarrays.

  2. Efficiency Issues: While the brute force approach works fine for small arrays, it has a time complexity of O(N^3) and is inefficient for large arrays (like N > 1000). It's crucial to optimize the solution for larger inputs.

  3. Edge Cases: Don’t forget to handle cases where there might be duplicate numbers, arrays with all the same values, or arrays where no valid subarray exists.

📊 Time and Space Complexity:

Approach               Time Complexity     Space Complexity
Brute Force                   O(N^3)         O(N)
Optimized Sorting Approach                   O(N^2 * log N)         O(N)
Efficient HashSet Approach                   O(N^2)         O(N)

🌍 Real-World Applications:

  1. Data Preprocessing: Finding contiguous subarrays in data could be useful for problems in data analysis or machine learning tasks where certain sequences or patterns need to be extracted or cleaned.

  2. Sequence Matching: This problem could be applied to problems where you need to find subsequences that match a certain property, such as genetic sequences or financial data.

  3. Optimization Problems: Similar problems can arise in optimization tasks where subarrays or subsequences need to be transformed or rearranged.


I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

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