3350. Adjacent Increasing Subarrays Detection II

Problem 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.

Intuition

The task is to determine the largest possible k for which two adjacent strictly increasing subarrays of length k can be formed. To tackle this, consider the following:

Traverse the Array: Iterate through nums to examine potential increasing sequences.
Track Lengths: Utilize two variables, cur and pre, to keep track of current and previous strictly increasing subarray lengths. Initialize both to zero at the start.
Check Strict Increase: For each element in the array, check if the current element forms a continuation of an increasing sequence compared to the next element. If an element is not part of a strictly increasing sequence, evaluate the possible k value using the smaller of pre and cur.
Calculate Maximum k: For any completed increasing sequence, attempt to update the maximum k by determining the maximum between the current known k and possible values based on completed sequences.
Move Sequence Forward: Once a strictly increasing sequence breaks, update pre to the value of cur and reset cur to start a new sequence check.

By iteratively updating these values, you can efficiently determine the largest k that allows for two adjacent strictly increasing subarrays.

Learn more about Binary Search patterns.

Quick Interview Experience

Help others by sharing your interview experience

Have you seen this problem before?

Solution Approach

The solution uses a single pass through the array nums to efficiently determine the maximum length k for which there exist two adjacent strictly increasing subarrays.

Initialize Variables:
- ans: To store the maximum value of k.
- pre: To store the length of the previous strictly increasing sequence.
- cur: To track the length of the current strictly increasing sequence.
Iterate Through the Array:
- Use a loop to traverse through each element x in nums using its index i.
Track Current Increasing Sequence:
- Increment cur for each element, as it is part of the current sequence.
- Check if the current element x is greater than or equal to the next one nums[i + 1]. This indicates the end of a strictly increasing sequence.
Update Maximum k:
- When the sequence ends, calculate potential k values:
  - cur // 2: Half of the current length can determine the k when both sequences are identical.
  - min(pre, cur): Minimum of the previous and current sequence lengths determines the largest possible adjacent subarray length.
- Update ans with the largest possible k from these calculations.
Prepare for Next Sequence:
- Assign cur value to pre and reset cur to zero, indicating a new sequence check.

This approach ensures that the algorithm runs in O(n) time complexity since it makes a single pass through the array, keeping track of necessary lengths and updating the maximum k efficiently using basic arithmetic and comparisons.

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Start Evaluator

Example Walkthrough

Consider the array nums = [1, 2, 3, 1, 2, 3, 4]. We want to find the maximum value of k, where there are two adjacent strictly increasing subarrays of length k.

Initialize Variables:
- Set ans = 0 to store the maximum k.
- Set pre = 0 for the previous increasing sequence length.
- Set cur = 0 for the current increasing sequence length.
Iterate Through the Array:
- Start at index 0 (nums[i] = 1):
  - Increment cur to 1 since the sequence is starting.
  - nums[0] < nums[1], so continue the sequence (current sequence: [1, 2, 3]).
- At index 2 (nums[2] = 3):
  - nums[2] > nums[3] indicates the sequence is ending.
  - Calculate maximum k:
    - cur = 3, update ans using cur // 2 = 1 and min(pre, cur) = 0.
    - Update ans = 1.
  - Set pre = 3, reset cur = 0.
- Resume at index 3 (nums[3] = 1):
  - Increment cur to 1. Start new sequence.
  - Continue as nums[3] < nums[4] < nums[5] < nums[6] (current sequence: [1, 2, 3, 4]).
- At index 6 (nums[6] = 4):
  - End of array, complete sequence check.
  - Calculate maximum k:
    - cur = 4, update ans using cur // 2 = 2 and min(pre, cur) = 3.
    - Update ans to max(1, 2) = 2.

The largest possible k for two adjacent strictly increasing subarrays is 2. The subarrays are [1, 2] and [2, 3] starting at indices 2 and 3 respectively.

Solution Implementation

1from typing import List
2
3class Solution:
4    def maxIncreasingSubarrays(self, nums: List[int]) -> int:
5        # Initialize answer, previous segment length, and current segment length.
6        answer = previous_length = current_length = 0
7      
8        # Iterate over each element in the list with its index.
9        for index, num in enumerate(nums):
10            current_length += 1  # Increment the length of the current increasing subarray.
11          
12            # Check if the end of the list is reached or the current number is not less than the next number.
13            if index == len(nums) - 1 or num >= nums[index + 1]:
14                # Update the answer with the maximum of current answer, the half of current length,
15                # or the minimum of previous and current lengths.
16                answer = max(answer, current_length // 2, min(previous_length, current_length))
17              
18                # Update previous_length with current_length for the next iteration, reset current_length.
19                previous_length, current_length = current_length, 0
20      
21        return answer
22

1import java.util.List;
2
3class Solution {
4    public int maxIncreasingSubarrays(List<Integer> nums) {
5        int ans = 0; // Variable to store the maximum length of an increasing subarray
6        int previousLength = 0; // Length of the previous increasing subarray
7        int currentLength = 0; // Length of the current increasing subarray
8        int n = nums.size(); // Number of elements in the list
9      
10        // Iterate through the list of numbers
11        for (int i = 0; i < n; ++i) {
12            ++currentLength; // Increment the length of the current subarray
13          
14            // Check if the end of the list is reached or the current number is not smaller than the next one
15            if (i == n - 1 || nums.get(i) >= nums.get(i + 1)) {
16                // Update the maximum length found so far, comparing current, half of the current, and the previous lengths
17                ans = Math.max(ans, Math.max(currentLength / 2, Math.min(previousLength, currentLength)));
18              
19                // Update previousLength to be the currentLength for the next round
20                previousLength = currentLength;
21              
22                // Reset currentLength for a new potential subarray
23                currentLength = 0;
24            }
25        }
26      
27        return ans; // Return the maximum length of increasing subarrays found
28    }
29}
30

1class Solution {
2public:
3    int maxIncreasingSubarrays(vector<int>& nums) {
4        int ans = 0; // Initialize the variable to hold the maximum length 
5        int pre = 0; // Variable to store the length of the previous subarray
6        int cur = 0; // Variable to track the current subarray length
7        int n = nums.size(); // Get the size of the input vector
8
9        // Iterate through each element in the vector
10        for (int i = 0; i < n; ++i) {
11            ++cur; // Increment the current subarray length
12
13            // Check if it's the end of the vector or the next number is not greater
14            if (i == n - 1 || nums[i] >= nums[i + 1]) {
15                // Calculate maximum of ans, half of the current subarray, and minimum of previous and current subarrays
16                ans = max({ans, cur / 2, min(pre, cur)});
17                pre = cur; // Update pre to the current subarray length
18                cur = 0; // Reset cur for the next subarray
19            }
20        }
21      
22        return ans; // Return the maximum length found
23    }
24};
25

1function maxIncreasingSubarrays(nums: number[]): number {
2    // Initialize variables.
3    let [maxLength, previousLength, currentLength] = [0, 0, 0];
4    const n = nums.length;
5
6    // Iterate over the array to find increasing subarrays.
7    for (let i = 0; i < n; ++i) {
8        ++currentLength; // Increment current subarray length.
9      
10        // If the end of an increasing subarray is reached.
11        if (i === n - 1 || nums[i] >= nums[i + 1]) {
12            // Update the maximum length found so far.
13            maxLength = Math.max(maxLength, Math.floor(currentLength / 2), Math.min(previousLength, currentLength));
14            // Set previousLength to currentLength and reset currentLength to zero for new subarray.
15            [previousLength, currentLength] = [currentLength, 0];
16        }
17    }
18    return maxLength; // Return the maximum length of a valid increasing subarray.
19}
20

Time and Space Complexity

The code has a time complexity of O(n), where n is the length of the nums list. This is because each element in the nums list is processed exactly once in a single pass through the list.

The space complexity of the code is O(1) since only a constant amount of extra space is used, regardless of the input size. The variables ans, pre, cur, and loop index i consume constant space.

Learn more about how to find time and space complexity quickly.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Given a sorted array of integers and an integer called target, find the element that equals to the target and return its index. Select the correct code that fills the ___ in the given code snippet.

1def binary_search(arr, target):
2    left, right = 0, len(arr) - 1
3    while left ___ right:
4        mid = (left + right) // 2
5        if arr[mid] == target:
6            return mid
7        if arr[mid] < target:
8            ___ = mid + 1
9        else:
10            ___ = mid - 1
11    return -1
12

1public static int binarySearch(int[] arr, int target) {
2    int left = 0;
3    int right = arr.length - 1;
4
5    while (left ___ right) {
6        int mid = left + (right - left) / 2;
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

1function binarySearch(arr, target) {
2    let left = 0;
3    let right = arr.length - 1;
4
5    while (left ___ right) {
6        let mid = left + Math.trunc((right - left) / 2);
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

<=, left, right

<, left, right

<=, right, left

<, right, left

I don't know