3349. Adjacent Increasing Subarrays Detection I

Problem Description

Given an array nums of n integers and an integer k, the task is to determine whether there exist two adjacent subarrays of length k such that both subarrays are strictly increasing. Specifically, you need to check for 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.

Intuition

The problem requires identifying two adjacent subarrays of length k that are strictly increasing. The solution involves iterating through the array and keeping track of increasing sequences.

Tracking Increasing Sequences:
- As you iterate through the array, keep a count (cur) of the current increasing sequence length. This is incremented each time the current element x is less than the next element (nums[i + 1]).
End of Increasing Sequence:
- If you reach the end of the array or find an element that is not less than its successor, you've reached the end of a strictly increasing sequence.
Evaluate for Adjacent Subarrays:
- Compute possible lengths for adjacent subarrays by utilizing previous (pre) and current (cur) sequence lengths.
- Ensure the ability to form two consecutive, strictly increasing sequences of length k.
- Use a variable (mx) to track the maximum possible length that can be achieved using consecutive subarrays and adjust it with half of the current length as well as the minimum of pre and cur to account for overlapping possibilities.
Decision Making:
- If the maximum possible overlapping subarray length (mx) is at least k, return true.
- Otherwise, if no such pair of subarrays exists, return false.

This approach efficiently checks for the required conditions by processing the array in a single pass.

Solution Approach

The solution utilizes a linear scan through the array to assess whether the conditions for two adjacent strictly increasing subarrays of length k are met. Here's a step-by-step breakdown of the approach:

Initialize Counters:
- Start with mx, pre, and cur set to 0. These variables hold the maximum length of overlapping sequences, the length of the previous increasing sequence, and the length of the current increasing sequence, respectively.
Iterate Through the Array:
- For each element x in the array nums, iterate through its index i.
Count Increasing Sequence:
- Increment cur for each element that is part of an increasing sequence. This is done when x is less than the next element in the array.
Encounter Sequence End:
- If you reach an element that is not followed by a strictly greater element (or the end of the array), compute potential lengths for overlapping subarrays: update mx to the maximum of its current value, half of cur (to handle odd-length sequences), and the minimum of pre and cur (to manage possible overlaps).
Update Sequence Lengths:
- Set pre to the current sequence length, and reset cur to zero to handle potential new increasing sequences starting.
Check Length Condition:
- Once all elements have been processed, if mx is at least k, return true, indicating two valid adjacent increasing subarrays were found.
Return Result:
- If no such subarrays can be found by the end of the iteration, return false.

This approach is effective due to its linear complexity, relying on simple arithmetic operations and comparisons to track and evaluate the potential subarray formations.

Ready to land your dream job?

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

Start Evaluator

Example Walkthrough

Let's take a small example to illustrate the solution approach.

Consider the array nums = [1, 2, 3, 2, 3, 4, 1] and k = 3.

Initialize Counters:

Start with mx = 0, pre = 0, and cur = 0.
Iterate Through the Array:
- At index 0, nums[0] = 1. Check if nums[0] < nums[1] (1 < 2). Since the condition is true, increment cur to 1.
- At index 1, nums[1] = 2. Check if nums[1] < nums[2] (2 < 3). The condition holds, increment cur to 2.
- At index 2, nums[2] = 3. Check if nums[2] < nums[3] (3 < 2). The condition fails, marking the end of an increasing sequence. Compute mx as the maximum of mx, cur // 2, and min(pre, cur). Here, mx = max(0, 2 // 2, min(0, 2)) = 1. Update pre = cur = 2, and reset cur = 0.
Continue Iterating:
- At index 3, nums[3] = 2. Check if nums[3] < nums[4] (2 < 3). The condition is true, increment cur to 1.
- At index 4, nums[4] = 3. Check if nums[4] < nums[5] (3 < 4). The condition holds, increment cur to 2.
- At index 5, nums[5] = 4. Check if nums[5] < nums[6] (4 < 1). The condition fails, marking the end of an increasing sequence. Compute mx as max(1, 2 // 2, min(2, 2)) = 2. Update pre = cur = 2, and reset cur = 0.
Check Length Condition:
- After processing the entire array, check if mx is at least k. Here, mx = 2 which is less than k = 3.
Return Result:
- Since mx < k, there are no two adjacent strictly increasing subarrays of length k. Therefore, the function returns false.

This example demonstrates how to track increasing sequences and calculate the possibility of having adjacent subarrays of the required length using the outlined approach.

Solution Implementation

1from typing import List
2
3class Solution:
4    def hasIncreasingSubarrays(self, nums: List[int], k: int) -> bool:
5        max_length = 0  # Variable to store the maximum length of valid subarrays
6        previous = 0    # Length of the previous increasing subarray
7        current = 0     # Length of the current increasing subarray
8
9        # Iterate through the list with both the index and the element
10        for i, element in enumerate(nums):
11            current += 1  # Increment the current subarray length
12
13            # Check if the end of the list is reached or no longer increasing
14            if i == len(nums) - 1 or element >= nums[i + 1]:
15                # Find the maximum subarray length that satisfies the conditions
16                max_length = max(max_length, current // 2, min(previous, current))
17              
18                # Update the previous subarray length and reset current length
19                previous = current
20                current = 0
21      
22        # Return true if max_length is at least k, meaning there's a valid subarray
23        return max_length >= k
24

1import java.util.List;
2
3class Solution {
4    public boolean hasIncreasingSubarrays(List<Integer> nums, int k) {
5        int maxSubarrayLength = 0;  // Keeps track of the maximum length of the found increasing subarrays
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();        // Length of the input list
9
10        // Iterate over the list to find increasing subarrays
11        for (int i = 0; i < n; ++i) {
12            ++currentLength;  // Increase the current length as we're counting the current element
13
14            // Check if the current element is the last one or not part of an increasing sequence
15            if (i == n - 1 || nums.get(i) >= nums.get(i + 1)) {
16                // Calculate the maximum length found so far
17                maxSubarrayLength = Math.max(maxSubarrayLength, Math.max(currentLength / 2, Math.min(previousLength, currentLength)));
18              
19                // Set 'previousLength' to 'currentLength' to use it in the next iteration
20                previousLength = currentLength;
21              
22                // Reset current length since the current sequence has ended
23                currentLength = 0;
24            }
25        }
26
27        // Return whether the maximum length found is at least 'k'
28        return maxSubarrayLength >= k;
29    }
30}
31

1#include <vector>
2#include <algorithm>
3using namespace std;
4
5class Solution {
6public:
7    bool hasIncreasingSubarrays(vector<int>& nums, int k) {
8        int maxLength = 0; // Variable to store the maximum length of any valid subarray found
9        int previousLength = 0; // Length of the previous increasing subarray
10        int currentLength = 0; // Length of the current increasing subarray
11        int n = nums.size(); // Size of the input vector
12      
13        // Iterate through the vector
14        for (int i = 0; i < n; ++i) {
15            ++currentLength; // Increment the current subarray length
16
17            // Check if it's the last element or the current element is not less than the next element
18            if (i == n - 1 || nums[i] >= nums[i + 1]) {
19                // Update the maximum increasing subarray length
20                maxLength = max({maxLength, currentLength / 2, min(previousLength, currentLength)});
21              
22                // Update the previous length and reset the current length to start a new subarray
23                previousLength = currentLength;
24                currentLength = 0;
25            }
26        }
27      
28        // Return true if there is a subarray of at least length k, false otherwise
29        return maxLength >= k;
30    }
31};
32

1// Determine whether there exists a subarray of length at least k 
2// where elements are increasing.
3
4function hasIncreasingSubarrays(nums: number[], k: number): boolean {
5    let maxLen = 0; // Maximum length of a valid subarray encountered.
6    let prevLen = 0; // Length of the previous increasing subarray.
7    let currentLen = 0; // Length of the current increasing subarray.
8    const n = nums.length; // Total number of elements in the array.
9
10    for (let i = 0; i < n; ++i) {
11        // Increase the count of the current subarray.
12        ++currentLen;
13
14        // Check if the current element is the last or if an increasing sequence ends.
15        if (i === n - 1 || nums[i] >= nums[i + 1]) {
16            // Calculate the maximum length based on different possible segments.
17            maxLen = Math.max(
18                maxLen,
19                Math.floor(currentLen / 2), // Maximum subarray length considering half of the current increasing segment.
20                Math.min(prevLen, currentLen) // Maximum possible overlap between consecutive segments.
21            );
22
23            // Update previous and reset current increasing subarray lengths.
24            prevLen = currentLen;
25            currentLen = 0;
26        }
27    }
28
29    // Return true if there's at least one subarray with length >= k, false otherwise.
30    return maxLen >= k;
31}
32

Time and Space Complexity

The given solution processes each element in the nums array exactly once in a single loop. Therefore, the time complexity is O(n), where n is the number of elements in nums. The algorithm performs a constant amount of work per element, resulting in a linear time complexity.

As for space complexity, no additional space proportional to the input size is used; only a constant amount of extra variables (mx, pre, cur) are maintained. Thus, the space complexity is O(1).

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:

Question 1 out of 10

Which of these properties could exist for a graph but not a tree?

Contain cycles

There are multiple ways to traverse

Connected

Has no cycle

I don’t know