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3254. Find the Power of K-Size Subarrays I

Description

You are given an array of integers nums of length n and a positive integer k.

The power of an array is defined as:

  • Its maximum element if all of its elements are consecutive and sorted in ascending order.
  • -1 otherwise.

You need to find the power of all subarrays of nums of size k.

Return an integer array results of size n - k + 1, where results[i] is the power of nums[i..(i + k - 1)].

 

Example 1:

Input: nums = [1,2,3,4,3,2,5], k = 3

Output: [3,4,-1,-1,-1]

Explanation:

There are 5 subarrays of nums of size 3:

  • [1, 2, 3] with the maximum element 3.
  • [2, 3, 4] with the maximum element 4.
  • [3, 4, 3] whose elements are not consecutive.
  • [4, 3, 2] whose elements are not sorted.
  • [3, 2, 5] whose elements are not consecutive.

Example 2:

Input: nums = [2,2,2,2,2], k = 4

Output: [-1,-1]

Example 3:

Input: nums = [3,2,3,2,3,2], k = 2

Output: [-1,3,-1,3,-1]

 

Constraints:

  • 1 <= n == nums.length <= 500
  • 1 <= nums[i] <= 105
  • 1 <= k <= n

Solutions

Solution 1: Recursion

Thinking

A window's power is its maximum iff the entries are consecutive increasing, otherwise \(-1\). \(n\le 500\) would allow rescanning each window, but consecutive windows share the same run.

Let \(f[i]\) be the consecutive-increasing length ending at \(i\). Then \(f[i]\ge k\) iff \([i-k+1,i]\) is valid, and the power is \(\textit{nums}[i]\). One recurrence, then emit by right end.

We define an array \(f\), where \(f[i]\) represents the length of the continuous increasing subsequence ending at the \(i\)-th element. Initially, \(f[i] = 1\).

Next, we traverse the array \(\textit{nums}\) to calculate the values of the array \(f\). If \(nums[i] = nums[i - 1] + 1\), then \(f[i] = f[i - 1] + 1\); otherwise, \(f[i] = 1\).

Then, we traverse the array \(f\) in the range \([k - 1, n)\). If \(f[i] \ge k\), we add \(\textit{nums}[i]\) to the answer array; otherwise, we add \(-1\).

After the traversal, we return the answer array.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) represents the length of the array \(\textit{nums}\).

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class Solution:
    def resultsArray(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        f = [1] * n
        for i in range(1, n):
            if nums[i] == nums[i - 1] + 1:
                f[i] = f[i - 1] + 1
        return [nums[i] if f[i] >= k else -1 for i in range(k - 1, n)]

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class Solution {
    public int[] resultsArray(int[] nums, int k) {
        int n = nums.length;
        int[] f = new int[n];
        Arrays.fill(f, 1);
        for (int i = 1; i < n; ++i) {
            if (nums[i] == nums[i - 1] + 1) {
                f[i] = f[i - 1] + 1;
            }
        }
        int[] ans = new int[n - k + 1];
        for (int i = k - 1; i < n; ++i) {
            ans[i - k + 1] = f[i] >= k ? nums[i] : -1;
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> resultsArray(vector<int>& nums, int k) {
        int n = nums.size();
        int f[n];
        f[0] = 1;
        for (int i = 1; i < n; ++i) {
            f[i] = nums[i] == nums[i - 1] + 1 ? f[i - 1] + 1 : 1;
        }
        vector<int> ans;
        for (int i = k - 1; i < n; ++i) {
            ans.push_back(f[i] >= k ? nums[i] : -1);
        }
        return ans;
    }
};

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func resultsArray(nums []int, k int) (ans []int) {
    n := len(nums)
    f := make([]int, n)
    f[0] = 1
    for i := 1; i < n; i++ {
        if nums[i] == nums[i-1]+1 {
            f[i] = f[i-1] + 1
        } else {
            f[i] = 1
        }
    }
    for i := k - 1; i < n; i++ {
        if f[i] >= k {
            ans = append(ans, nums[i])
        } else {
            ans = append(ans, -1)
        }
    }
    return
}

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function resultsArray(nums: number[], k: number): number[] {
    const n = nums.length;
    const f: number[] = Array(n).fill(1);
    for (let i = 1; i < n; ++i) {
        if (nums[i] === nums[i - 1] + 1) {
            f[i] = f[i - 1] + 1;
        }
    }
    const ans: number[] = [];
    for (let i = k - 1; i < n; ++i) {
        ans.push(f[i] >= k ? nums[i] : -1);
    }
    return ans;
}

Solution 2: Two Pointers

Thinking

Solution 1 stores an \(O(n)\) array \(f\). The run only needs its left end \(j\): reset \(j\) to \(i\) when the adjacent difference is not \(1\). The window is valid iff its left index is still \(\ge j\). A rolling pointer uses \(O(1)\) extra space.

Let pointer \(j\) be the start of the current segment whose adjacent differences are all exactly \(1\). Traverse the array from left to right: if \(i > 0\) and \(\textit{nums}[i] \neq \textit{nums}[i - 1] + 1\), update \(j\) to \(i\).

When \(i \ge k - 1\), the current window is \([i - k + 1,\ i]\). If \(i - k + 1 < j\), some adjacent pair in the window differs by more than \(1\), so the power is \(-1\); otherwise the window is consecutive integers in increasing order, and the power is the maximum value \(\textit{nums}[i]\).

The time complexity is \(O(n)\), and the space complexity is \(O(1)\). Here, \(n\) is the length of \(\textit{nums}\).

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export function resultsArray(nums: number[], k: number): number[] {
    const n = nums.length;
    const ans: number[] = [];

    for (let i = 0, j = 0; i < n; i++) {
        if (i && nums[i - 1] + 1 !== nums[i]) j = i;
        if (i >= k - 1) {
            ans.push(i - k + 1 < j ? -1 : nums[i]);
        }
    }

    return ans;
}

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export function resultsArray(nums, k) {
    const n = nums.length;
    const ans = [];

    for (let i = 0, j = 0; i < n; i++) {
        if (i && nums[i - 1] + 1 !== nums[i]) j = i;
        if (i >= k - 1) {
            ans.push(i - k + 1 < j ? -1 : nums[i]);
        }
    }

    return ans;
}

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