LeetCode in Kotlin

3349. Adjacent Increasing Subarrays Detection I

Easy

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

Solution

class Solution {
    fun hasIncreasingSubarrays(nums: List<Int>, k: Int): Boolean {
        val l = nums.size
        if (l < k * 2) {
            return false
        }
        for (i in 0..<l - 2 * k + 1) {
            if (check(i, k, nums) && check(i + k, k, nums)) {
                return true
            }
        }
        return false
    }

    private fun check(p: Int, k: Int, nums: List<Int>): Boolean {
        for (i in p..<p + k - 1) {
            if (nums[i] >= nums[i + 1]) {
                return false
            }
        }
        return true
    }
}

This site is open source. Improve this page.