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3350. Adjacent Increasing Subarrays Detection II

Description

Given an array nums of n integers, your task is to find the maximum value of k for which there exist two adjacent subarrays of length k each, such that both subarrays are strictly increasing. Specifically, check if there are two subarrays of length k 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 the maximum possible value of k.

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]

Output: 3

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, and 3 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

Example 2:

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

Output: 2

Explanation:

  • The subarray starting at index 0 is [1, 2], which is strictly increasing.
  • The subarray starting at index 2 is [3, 4], which is also strictly increasing.
  • These two subarrays are adjacent, and 2 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

 

Constraints:

  • 2 <= nums.length <= 2 * 105
  • -109 <= nums[i] <= 109

Solutions

Solution 1

  • class Solution {
    public int maxIncreasingSubarrays(List<Integer> nums) {
        int ans = 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)) {
                ans = Math.max(ans, Math.max(cur / 2, Math.min(pre, cur)));
                pre = cur;
                cur = 0;
            }
        }
        return ans;
    }
}


  • class Solution {
public:
    int maxIncreasingSubarrays(vector<int>& nums) {
        int ans = 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]) {
                ans = max({ans, cur / 2, min(pre, cur)});
                pre = cur;
                cur = 0;
            }
        }
        return ans;
    }
};


  • class Solution:
    def maxIncreasingSubarrays(self, nums: List[int]) -> int:
        ans = pre = cur = 0
        for i, x in enumerate(nums):
            cur += 1
            if i == len(nums) - 1 or x >= nums[i + 1]:
                ans = max(ans, cur // 2, min(pre, cur))
                pre, cur = cur, 0
        return ans


  • func maxIncreasingSubarrays(nums []int) (ans int) {
	pre, cur := 0, 0
	for i, x := range nums {
		cur++
		if i == len(nums)-1 || x >= nums[i+1] {
			ans = max(ans, max(cur/2, min(pre, cur)))
			pre, cur = cur, 0
		}
	}
	return
}


  • function maxIncreasingSubarrays(nums: number[]): number {
    let [ans, 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]) {
            ans = Math.max(ans, (cur / 2) | 0, Math.min(pre, cur));
            [pre, cur] = [cur, 0];
        }
    }
    return ans;
}


  • /**
 * @param {number[]} nums
 * @return {number}
 */
var maxIncreasingSubarrays = function (nums) {
    let [ans, 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]) {
            ans = Math.max(ans, cur >> 1, Math.min(pre, cur));
            [pre, cur] = [cur, 0];
        }
    }
    return ans;
};


  • impl Solution {
    pub fn max_increasing_subarrays(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let (mut ans, mut pre, mut cur) = (0, 0, 0);

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

        ans
    }
}

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