3612. Adjacent Increasing Subarrays Detection I¶

Difficulty: Easy

3612. 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.

A subarray is a contiguous non-empty sequence of elements within an array.

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 {
    public boolean hasIncreasingSubarrays(List<Integer> nums, int k) {
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            if ((i + 2 * k <= n) && check(nums, i , i + 2 * k, k)) return true;
        }
        return false;
    }
    private boolean check(List<Integer> nums, int start, int end, int k) {
        int n = nums.size();
        int prev = -2000;        
        for (int i = start; i < start + k; i++) {
            if (prev == -2000) {
                prev = nums.get(i);
            }
            else {
                if (nums.get(i) <= prev) return false;
                prev = nums.get(i);
            }
        }
        prev = -2000;
        for (int i = start + k; i < end; i++) {
            System.out.println(nums.get(i));
            if (prev == -2000) {
                prev = nums.get(i);
            }
            else {
                if (nums.get(i) <= prev) return false;
                prev = nums.get(i);
            }
        }
        return true;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here