LeetCode 3255: Find the Power of K-Size Subarrays II Solution

Master LeetCode problem 3255 (Find the Power of K-Size Subarrays II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3255. Find the Power of K-Size Subarrays II

Medium

Array Sliding Window

Problem Explanation

Explanation:

To solve this problem, we can iterate through the array nums and maintain a deque that stores the indices of elements in descending order. We will only keep the indices that are within the sliding window of size k. The front of the deque will always point to the index of the maximum element within the current window.

At each step, we will check if the current index is still within the sliding window. If it is not, we will remove it from the deque. Next, we will update our result array based on the maximum element within the current window.

Algorithm:

Initialize an empty deque to store indices and an empty result array of size n - k + 1.
Iterate through the array nums from index 0 to n.
- Remove indices from the front of the deque that are outside the current window.
- Check if the current index is still within the window, if not, remove it.
- Add the current index to the back of the deque.
- If the current index is greater than or equal to k - 1 (i.e., the window size has reached k), update the result array based on the maximum element within the current window.
Return the result array.

Time Complexity:

The time complexity of this algorithm is O(n) where n is the size of the input array nums.

Space Complexity:

The space complexity of this algorithm is O(n) where n is the size of the input array nums.

Solution Code

class Solution {
    public int[] findPower(int[] nums, int k) {
        int n = nums.length;
        Deque<Integer> deque = new ArrayDeque<>();
        int[] results = new int[n - k + 1];
        
        for (int i = 0; i < n; i++) {
            while (!deque.isEmpty() && deque.peekFirst() < i - k + 1) {
                deque.pollFirst();
            }
            
            while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
                deque.pollLast();
            }
            
            deque.offerLast(i);
            
            if (i >= k - 1) {
                results[i - k + 1] = nums[deque.peekFirst()];
            }
        }
        
        return results;
    }
}

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