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3349. Adjacent Increasing Subarrays Detection I

Description

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

Solutions

Solution 1

  • 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;
    }
}


  • class Solution {
public:
    bool hasIncreasingSubarrays(vector<int>& 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[i] >= nums[i + 1]) {
                mx = max({mx, cur / 2, min(pre, cur)});
                pre = cur;
                cur = 0;
            }
        }
        return mx >= k;
    }
};


  • 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


  • 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
}


  • 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;
}


  • /**
 * @param {number[]} nums
 * @param {number} k
 * @return {boolean}
 */
var hasIncreasingSubarrays = function (nums, k) {
    const n = nums.length;
    let [mx, pre, cur] = [0, 0, 0];
    for (let i = 0; i < n; ++i) {
        ++cur;
        if (i === n - 1 || nums[i] >= nums[i + 1]) {
            mx = Math.max(mx, cur >> 1, Math.min(pre, cur));
            pre = cur;
            cur = 0;
        }
    }
    return mx >= k;
};


  • 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
    }
}

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