LeetCode #3254 — MEDIUM

Find the Power of K-Size Subarrays I

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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)].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Sliding Window

Example 1

[1,2,3,4,3,2,5]
3

Example 2

[2,2,2,2,2]
4

Example 3

[3,2,3,2,3,2]
2

Core Insight

What unlocks the optimal approach

Can we use a brute force solution with nested loops and HashSet?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3254: Find the Power of K-Size Subarrays I
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;
    }
}

// Accepted solution for LeetCode #3254: Find the Power of K-Size Subarrays I
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
}

# Accepted solution for LeetCode #3254: Find the Power of K-Size Subarrays I
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)]

// Accepted solution for LeetCode #3254: Find the Power of K-Size Subarrays I
/**
 * [3254] Find the Power of K-Size Subarrays I
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn results_array(nums: Vec<i32>, k: i32) -> Vec<i32> {
        let k = k as usize;
        let n = nums.len();

        let mut count = 0;
        for i in 0..(k - 1) {
            if nums[i] + 1 == nums[i + 1] {
                count += 1;
            }
        }

        let mut result = Vec::with_capacity(n - k + 1);
        result.push(if count == k - 1 { nums[k - 1] } else { -1 });

        for i in k..n {
            if nums[i - 1] + 1 == nums[i] {
                count += 1;
            }

            if nums[i - k] + 1 == nums[i - k + 1] {
                count -= 1;
            }

            result.push(if count == k - 1 { nums[i] } else { -1 });
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3254() {
        assert_eq!(vec![1, 2, 3], Solution::results_array(vec![1, 2, 3], 1));
        assert_eq!(
            vec![3, 4, -1, -1, -1],
            Solution::results_array(vec![1, 2, 3, 4, 3, 2, 5], 3)
        );
        assert_eq!(
            vec![-1, -1],
            Solution::results_array(vec![2, 2, 2, 2, 2], 4)
        );
        assert_eq!(
            vec![-1, 3, -1, 3, -1],
            Solution::results_array(vec![3, 2, 3, 2, 3, 2], 2)
        );
    }
}

// Accepted solution for LeetCode #3254: Find the Power of K-Size Subarrays I
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.