
Problem Statement
In practical programming scenarios, such as analyzing trends in a time series dataset, one common task might be to identify the longest sequence within an array where each element successively increases. Think of situations like tracking consecutive days of rising stock prices to evaluate momentum. This problem specifically addresses a variation of this challenge. Here, we need to determine the maximum length of a strictly increasing continuous subsequence from an unsorted array of integers, nums
.
A continuous increasing subsequence is clearly defined by identifying any two indices, l
and r
(l < r
), that derive a sequence starting at nums[l]
and ending at nums[r]
, such that each integer in this subarray follows an increasing trend directly from its predecessor. That is, for any index i
which lies between l
and r
inclusive, the value at nums[i]
must be strictly less than the value at nums[i + 1]
.
Examples
Example 1
Input:
nums = [1,3,5,4,7]
Output:
3
Explanation:
The longest continuous increasing subsequence is [1,3,5] with length 3. Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element 4.
Example 2
Input:
nums = [2,2,2,2,2]
Output:
1
Explanation:
The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly increasing.
Constraints
1 <= nums.length <= 104
-109 <= nums[i] <= 109
Approach and Intuition
Analyzing the problem and examples provided, detecting the longest continuous increasing subsequence within an unsorted list implies a traversal and comparison operation to dynamically determine lengths of increasing sequences as they appear. Here are the intuitive steps:
- Initialize
max_length
to remember the longest length found. - Start with
current_length
at 1 as each individual element is itself a trivial increasing sequence. - Iterate through the list
nums
from the second element:- Compare the current element with the previous one:
- If the current element is greater than the previous, it's part of an increasing sequence, thus increment
current_length
. - Else, update
max_length
withcurrent_length
if the former is smaller, then resetcurrent_length
to 1 since we start a new sequence.
- If the current element is greater than the previous, it's part of an increasing sequence, thus increment
- Compare the current element with the previous one:
- Ensure to update
max_length
once more after the loop if the longest sequence appears at the end of the list.
The approach's efficiency hinges on a single traversal of the list, leading to an O(n) time complexity, where n is the number of elements in the array. This single pass efficiently updates both lengths and compares them to find the required maximal sequence length, aligning well with the given constraints.
Solutions
- Java
class Solution {
public int longestContinuousIncreasingSubsequence(int[] arr) {
int maxLength = 0, start = 0;
for (int index = 0; index < arr.length; ++index) {
if (index > 0 && arr[index-1] >= arr[index]) start = index;
maxLength = Math.max(maxLength, index - start + 1);
}
return maxLength;
}
}
The provided Java solution efficiently calculates the length of the longest continuous increasing subsequence in an array. Here’s a concise explanation of how the solution functions:
- Initialize
maxLength
to 0 to keep track of the maximum length of any increasing subsequence found. - Initialize
start
to 0 to mark the beginning of a new potential increasing subsequence. - Iterate through the array using a loop. For each element:
- Check if the current element is smaller than or equal to the previous element. If true, set
start
to the current index to mark a new beginning for an increasing subsequence. - Update
maxLength
using theMath.max
function by comparing it with the difference between the current index and thestart
index plus one, representing the length of the current increasing subsequence.
- Check if the current element is smaller than or equal to the previous element. If true, set
- After the loop completes, return
maxLength
as it now holds the length of the longest continuous increasing subsequence found in the array.
This approach ensures a linear time complexity, making it efficient for handling large arrays.
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