LeetCode 3254: Find the Power of K-Size Subarrays I Solution

Master LeetCode problem 3254 (Find the Power of K-Size Subarrays I), 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.

3254. Find the Power of K-Size Subarrays I

Medium

Array Sliding Window

Problem Explanation

Explanation:

To solve this problem, we can iterate through the array nums and for each subarray of size k starting at index i, we check if the subarray is consecutive and sorted in ascending order. If it is, we calculate the power of the subarray as the maximum element; otherwise, the power is -1.

We can achieve this by maintaining a deque (double-ended queue) to store the indices of elements in the current subarray. As we iterate through the array nums, we keep track of the maximum element within the current subarray using the deque.

At each step, we compare the current element with the maximum element from the front of the deque (which represents the maximum element in the current subarray). If the current element is greater, we remove all smaller elements from the back of the deque until we find an element greater than the current element. This ensures that the deque always contains indices in decreasing order of element values.

The power for a subarray starting at index i is the element at the front of the deque, or -1 if the subarray is not consecutive and sorted.

The time complexity of this approach is O(n) where n is the length of nums, and the space complexity is O(k) for the deque.

Solution Code

import java.util.ArrayDeque;

class Solution {
    public int[] findPower(int[] nums, int k) {
        int n = nums.length;
        int[] result = new int[n - k + 1];
        ArrayDeque<Integer> deque = new ArrayDeque<>();

        for (int i = 0; i < n; i++) {
            if (!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) {
                result[i - k + 1] = deque.peekFirst() == i - k + 1 ? nums[deque.peekFirst()] : -1;
            }
        }

        return result;
    }
}

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