Find the power of k size subarrays i

Array Sliding Window

62.1% acceptance

Feb 25, 2026

671

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)].

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

impl Solution {
  pub fn results_array(nums: Vec<i32>, k: i32) -> Vec<i32> {
    let n = nums.len();
    let k = k as usize;
    // streak[i] = number of consecutive valid "nums[j+1]==nums[j]+1" edges ending at i
    // streak[0] = 0 (no left edge)
    let mut streak = vec![0usize; n];
    for i in 1..n {
      if nums[i] == nums[i - 1] + 1 {
        streak[i] = streak[i - 1] + 1;
      }
    }
    // window [l, l+k-1] is valid iff streak[l+k-1] >= k-1
    (0..=(n - k))
      .map(|l| {
        if streak[l + k - 1] >= k - 1 {
          nums[l + k - 1]
        } else {
          -1
        }
      })
      .collect()
  }
}