🚀Day 1. K Sized Subarray Maximum 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given an array arr[] of integers and an integer k, your task is to find the maximum value for each contiguous subarray of size k. The output should be an array of maximum values corresponding to each contiguous subarray.

🔍 Example Walkthrough:

Example 1:

Input:

arr[] = [1, 2, 3, 1, 4, 5, 2, 3, 6] k = 3

Output:

[3, 3, 4, 5, 5, 5, 6]

Explanation:

1st window: [1, 2, 3], max = 3 2nd window: [2, 3, 1], max = 3 3rd window: [3, 1, 4], max = 4 4th window: [1, 4, 5], max = 5 5th window: [4, 5, 2], max = 5 6th window: [5, 2, 3], max = 5 7th window: [2, 3, 6], max = 6

Example 2:

Input:

arr[] = [8, 5, 10, 7, 9, 4, 15, 12, 90, 13] k = 4

Output:

[10, 10, 10, 15, 15, 90, 90]

Explanation:

1st window: [8, 5, 10, 7], max = 10 2nd window: [5, 10, 7, 9], max = 10 3rd window: [10, 7, 9, 4], max = 10 4th window: [7, 9, 4, 15], max = 15 5th window: [9, 4, 15, 12], max = 15 6th window: [4, 15, 12, 90], max = 90 7th window: [15, 12, 90, 13], max = 90

Constraints

• $(1 \leq \text{arr.size()} \leq 10^6)$

• $(1 \leq k \leq \text{arr.size()})$

• $(0 \leq arr[i] \leq 10^9)$

🎯 My Approach:

Sliding Window Maximum using Deque

• We maintain a deque where each element is an index, and the values at those indices are in decreasing order.

• As we iterate through the array, we add new elements to the deque and remove elements that fall out of the current window.

• The maximum element of the current window will always be at the front of the deque.

• This allows us to efficiently compute the maximum for each window in constant time on average.

Algorithm Steps:

1. Initialize a deque to store indices.

2. Loop through the array.

3. Remove elements from the front of the deque if they are outside the window.

4. Remove elements from the back of the deque if they are smaller than the current element (ensuring decreasing order).

5. Add the current index to the deque.

6. Once the first window is complete, append the front element (maximum) to the result.

7. Continue until all windows are processed.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N), because each element is added and removed from the deque at most once.

• Expected Auxiliary Space Complexity: O(K), the size of the deque.

📝 Solution Code

Code (C++)

class Solution {
public:
    vector<int> maxOfSubarrays(const vector<int>& arr, int k) {
        vector<int> res;
        deque<int> dq;
        for (int i = 0; i < arr.size(); ++i) {
            if (!dq.empty() && dq.front() <= i - k) dq.pop_front();
            while (!dq.empty() && arr[dq.back()] < arr[i]) dq.pop_back();
            dq.push_back(i);
            if (i >= k - 1) res.push_back(arr[dq.front()]);
        }
        return res;
    }
};

⚡ Alternative Approaches

2️⃣ Using Priority Queue (O(N log K) Time, O(K) Space)

This approach uses a max-heap to maintain the maximum in each window.

class Solution {
public:
    vector<int> maxOfSubarrays(const vector<int>& arr, int k) {
        vector<int> res;
        priority_queue<pair<int, int>> pq;
        for (int i = 0; i < arr.size(); ++i) {
            pq.emplace(arr[i], i);
            if (i >= k - 1) {
                while (pq.top().second <= i - k) pq.pop();
                res.push_back(pq.top().first);
            }
        }
        return res;
    }
};

🔹 Pros: Simpler logic using a priority queue. 🔹 Cons: Slightly slower due to O(log K) insertion/deletion operations.

3️⃣ Using Multiset (O(N log K) Time, O(K) Space)

This approach leverages multiset for ordered window maximum.

class Solution {
public:
    vector<int> maxOfSubarrays(const vector<int>& arr, int k) {
        vector<int> res;
        multiset<int> window;
        for (int i = 0; i < arr.size(); ++i) {
            window.insert(arr[i]);
            if (i >= k - 1) {
                res.push_back(*window.rbegin());
                window.erase(window.find(arr[i - k + 1]));
            }
        }
        return res;
    }
};

🔹 Pros: Provides ordered elements inside the window. 🔹 Cons: O(log K) insertion and deletion cause performance overhead.

📊 Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	✅ Pros	⚠️ Cons
Deque (Optimal)	🟢 O(N)	🟢 O(K)	Fastest, minimal memory	None
Priority Queue	🟡 O(N log K)	🟡 O(K)	Simple to implement	Slower than deque
Multiset	🟡 O(N log K)	🟡 O(K)	Maintains order	Slower than deque

💡 Best Choice?

• ✅ For optimal performance: Deque-based approach (O(N) time, O(K) space).

• ✅ For priority order requirement: Multiset-based approach.

• ✅ For easier implementation: Priority queue approach.

Code (Java)

class Solution {
    public ArrayList<Integer> maxOfSubarrays(int[] arr, int k) {
        ArrayList<Integer> res = new ArrayList<>();
        Deque<Integer> dq = new ArrayDeque<>();
        for (int i = 0; i < arr.length; i++) {
            if (!dq.isEmpty() && dq.peekFirst() <= i - k) dq.pollFirst();
            while (!dq.isEmpty() && arr[dq.peekLast()] < arr[i]) dq.pollLast();
            dq.offerLast(i);
            if (i >= k - 1) res.add(arr[dq.peekFirst()]);
        }
        return res;
    }
}

Code (Python)

class Solution:
    def maxOfSubarrays(self, arr, k):
        res, dq = [], deque()
        for i in range(len(arr)):
            if dq and dq[0] <= i - k:
                dq.popleft()
            while dq and arr[dq[-1]] < arr[i]:
                dq.pop()
            dq.append(i)
            if i >= k - 1:
                res.append(arr[dq[0]])
        return res

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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