LeetCode #3349 — EASY

Adjacent Increasing Subarrays Detection I

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given an array nums of n integers and an integer k, determine whether there exist two adjacent subarrays of length k such that both subarrays are strictly increasing. Specifically, check if there are two subarrays starting at indices a and b (a < b), where:

Both subarrays nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing.
The subarrays must be adjacent, meaning b = a + k.

Return true if it is possible to find two such subarrays, and false otherwise.

Example 1:

Input: nums = [2,5,7,8,9,2,3,4,3,1], k = 3

Output: true

Explanation:

The subarray starting at index 2 is [7, 8, 9], which is strictly increasing.
The subarray starting at index 5 is [2, 3, 4], which is also strictly increasing.
These two subarrays are adjacent, so the result is true.

Example 2:

Input: nums = [1,2,3,4,4,4,4,5,6,7], k = 5

Output: false

Constraints:

2 <= nums.length <= 100
1 < 2 * k <= nums.length
-1000 <= nums[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: Given an array nums of n integers and an integer k, determine whether there exist two adjacent subarrays of length k such that both subarrays are strictly increasing. Specifically, check if there are two subarrays starting at indices a and b (a < b), where: Both subarrays nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing. The subarrays must be adjacent, meaning b = a + k. Return true if it is possible to find two such subarrays, and false otherwise.

Baseline thinking

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

Pattern signal: Array

Example 1

[2,5,7,8,9,2,3,4,3,1]
3

Example 2

[1,2,3,4,4,4,4,5,6,7]
5

Step 02

Core Insight

What unlocks the optimal approach

Store the longest decreasing subarray starting and ending at an index.

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 #3349: Adjacent Increasing Subarrays Detection I
class Solution {
    public boolean hasIncreasingSubarrays(List<Integer> nums, int k) {
        int mx = 0, pre = 0, cur = 0;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            ++cur;
            if (i == n - 1 || nums.get(i) >= nums.get(i + 1)) {
                mx = Math.max(mx, Math.max(cur / 2, Math.min(pre, cur)));
                pre = cur;
                cur = 0;
            }
        }
        return mx >= k;
    }
}

// Accepted solution for LeetCode #3349: Adjacent Increasing Subarrays Detection I
func hasIncreasingSubarrays(nums []int, k int) bool {
	mx, pre, cur := 0, 0, 0
	for i, x := range nums {
		cur++
		if i == len(nums)-1 || x >= nums[i+1] {
			mx = max(mx, max(cur/2, min(pre, cur)))
			pre, cur = cur, 0
		}
	}
	return mx >= k
}

# Accepted solution for LeetCode #3349: Adjacent Increasing Subarrays Detection I
class Solution:
    def hasIncreasingSubarrays(self, nums: List[int], k: int) -> bool:
        mx = pre = cur = 0
        for i, x in enumerate(nums):
            cur += 1
            if i == len(nums) - 1 or x >= nums[i + 1]:
                mx = max(mx, cur // 2, min(pre, cur))
                pre, cur = cur, 0
        return mx >= k

// Accepted solution for LeetCode #3349: Adjacent Increasing Subarrays Detection I
impl Solution {
    pub fn has_increasing_subarrays(nums: Vec<i32>, k: i32) -> bool {
        let n = nums.len();
        let (mut mx, mut pre, mut cur) = (0, 0, 0);

        for i in 0..n {
            cur += 1;
            if i == n - 1 || nums[i] >= nums[i + 1] {
                mx = mx.max(cur / 2).max(pre.min(cur));
                pre = cur;
                cur = 0;
            }
        }

        mx >= k
    }
}

// Accepted solution for LeetCode #3349: Adjacent Increasing Subarrays Detection I
function hasIncreasingSubarrays(nums: number[], k: number): boolean {
    let [mx, pre, cur] = [0, 0, 0];
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        ++cur;
        if (i === n - 1 || nums[i] >= nums[i + 1]) {
            mx = Math.max(mx, (cur / 2) | 0, Math.min(pre, cur));
            [pre, cur] = [cur, 0];
        }
    }
    return mx >= k;
}

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(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.