# 14.10.2025 [3349. Adjacent Increasing Subarrays Detection I]

Any increasing consequent windows size k #easy

14.10.2025

3349. Adjacent Increasing Subarrays Detection I medium blog post substack youtube

https://dmitrysamoylenko.com/2023/07/14/leetcode_daily.html

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https://t.me/leetcode_daily_unstoppable/1142

Problem TLDR

Any increasing consequent windows size k #easy

Intuition

Just brute-force every index. O(1) memory solution: count increasings, keep previous and current, check

Approach

• corner case: single 2k increasing chunk

Complexity

• Time complexity: $$O(n)$$

• Space complexity: $$O(n)$$, can be O(1)

Code

// 379ms
    fun hasIncreasingSubarrays(n: List<Int>, k: Int) = n.indices
        .any { i -> fun List<Int>.g()=this==sorted()&&toSet().size==k;
            n.slice(i..<min(n.size,i+k)).g() && 
            n.slice(min(n.size-1,i+k)..<min(n.size,i+k+k)).g()
        }

// 5ms
    pub fn has_increasing_subarrays(n: Vec<i32>, k: i32) -> bool {
        once(0).chain(n.chunk_by(|a,b| a < b).map(|c|c.len() as i32))
        .collect::<Vec<_>>().windows(2).any(|c| c[0].min(c[1]).max(c[1]/2) >= k)
    }