Temperature Fluctuation Analysis 2 — Problem Statement & Solution Guide
Problem Description
You are given an array of integers, nums, representing a sequence of temperature readings. A subsequence of nums is a sequence that can be derived by deleting zero or more elements without changing the order of the remaining elements. For a subsequence to be considered *alternating*, the signs of the differences between each pair of consecutive elements must strictly alternate: a positive difference must be followed by a negative difference, which must be followed by a positive difference, and so on, or vice versa. Your task is to determine the maximum possible length of such an alternating subsequence that can be extracted from nums.
Input format: The first line contains an integer n (1 ≤ n ≤ 10^5), the number of temperature readings. The second line contains n space‑separated integers, each in the range [−10^9, 10^9].
Output format: Output a single integer, the length of the longest alternating subsequence that can be formed from the given array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Temperature Fluctuation Analysis 2"
WHY DOES IT MATTER?
This pattern is essential for optimizing sequence problems where the order matters but not the contiguity. It demonstrates how to reduce a quadratic DP state to a constant state by recognizing that only the most recent 'trend' (up or down) matters for the next step, a technique applicable to stock trading problems and signal processing.
OPTIMIZATION CHALLENGE
The key insight is that you do not need to remember the entire history of the subsequence, only whether the last step was an increase or a decrease. This allows the state space to collapse from O(n) to O(1).
REAL-WORLD CONNECTION
This is analogous to analyzing stock price movements to find the maximum profit from multiple transactions where you must alternate between buying and selling. It is also used in signal processing to count the number of oscillations or frequency changes in a time-series dataset.
In interviews, start with the O(n^2) DP to show you understand the problem structure, then derive the O(n) greedy solution by explaining why intermediate states are redundant. This demonstrates both depth and optimization skills.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem of finding the longest alternating subsequence (LAS) is a classic application of dynamic programming that can be optimized to linear time using a greedy insight. A naive DP approach defines dp[i][0] as the length of the longest alternating subsequence ending at index i with a positive difference, and dp[i][1] for a negative difference. While this yields an O(n^2) solution, it fails to exploit the structural properties of the sequence. The key theoretical insight is that the length of the LAS depends only on the number of 'turning points' (local maxima and minima) in the sequence, not the specific values, provided we handle plateaus correctly.
Interview Questions on This Problem
Q1How would you modify this solution to handle a sequence where equal consecutive elements are allowed but do not count as a difference?
You must skip equal elements when determining the sign of the difference. In the greedy approach, if nums[i] == nums[i-1], you simply continue without updating the state. In the DP approach, the transition logic remains the same, but the condition for updating dp[i][0] or dp[i][1] should only trigger if nums[i] > nums[j] or nums[i] < nums[j] strictly, ignoring equality.
Q2Can this problem be solved in O(1) space? If so, how do you track the state?
Yes. You only need to track two variables: up (length of LAS ending with an increase) and down (length of LAS ending with a decrease). Iterate through the array; if nums[i] > nums[i-1], update up = down + 1; if nums[i] < nums[i-1], update down = up + 1. This works because the optimal substructure allows us to discard previous states once the current trend is established.
Q3What is the relationship between the Longest Alternating Subsequence and the number of local extrema in the array?
The length of the LAS is equal to the number of local extrema (peaks and valleys) plus one, assuming the first and last elements are included in the subsequence. If the array starts with a monotonic segment, the first element of the LAS is the first element of that segment. The greedy algorithm effectively counts these turning points.
Examples
Input
6 1 7 4 9 2 5
Output
6
Explanation: The entire array itself is an alternating subsequence: 1 → 7 (difference +6, positive) 7 → 4 (difference −3, negative) 4 → 9 (difference +5, positive) 9 → 2 (difference −7, negative) 2 → 5 (difference +3, positive) All differences alternate in sign, so the maximum length is 6.
Input
5 4 3 2 1 5
Output
3
Explanation: One optimal alternating subsequence is 4, 3, 5: 4 → 3 (difference −1, negative) 3 → 5 (difference +2, positive) The differences alternate, giving a length of 3. No longer alternating subsequence exists because any additional element would break the sign alternation.
Input
5 1 2 3 4 5
Output
2
Explanation: All differences are positive, so we cannot have two consecutive differences with opposite signs. The longest alternating subsequence can only contain two elements, e.g., 1 and 2, which yields a single positive difference. Thus the maximum length is 2.
Input
3 10 10 10
Output
1
Explanation: All elements are equal, so every difference is zero and does not have a sign. A subsequence of length 1 has no differences and is trivially alternating. Any longer subsequence would contain zero differences, which cannot alternate. Therefore the maximum length is 1.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The input array may contain duplicate values.
Optimal Approach & Strategy
Use a greedy approach with two variables, up and down, to track the length of the longest alternating subsequence ending with an increase or decrease. Iterate through the array, updating up or down based on the comparison between consecutive elements, achieving O(n) time and O(1) space.
Brute Force Approach
Generate all possible subsequences and check if each one is alternating, keeping track of the maximum length. This approach has exponential time complexity O(2^n) and is infeasible for large inputs.
Verified Code Solutions
/**
* @param {number[]} nums
* @return {number}
*/
var maxAlternatingSubsequence = function(nums) {
if (nums.length === 0) return 0;
let up = 1, down = 1;
for (let i = 1; i < nums.length; i++) {
if (nums[i] > nums[i - 1]) {
up = down + 1;
} else if (nums[i] < nums[i - 1]) {
down = up + 1;
}
}
return Math.max(up, down);
};class Solution {
public:
int maxAlternatingSubsequence(vector<int>& nums) {
if (nums.empty()) return 0;
int n = nums.size();
int up = 1, down = 1;
for (int i = 1; i < n; ++i) {
if (nums[i] > nums[i - 1]) {
up = down + 1;
} else if (nums[i] < nums[i - 1]) {
down = up + 1;
}
}
return max(up, down);
}
};class Solution {
public int maxAlternatingSubsequence(int[] nums) {
if (nums.length == 0) return 0;
int up = 1, down = 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] > nums[i - 1]) {
up = down + 1;
} else if (nums[i] < nums[i - 1]) {
down = up + 1;
}
}
return Math.max(up, down);
}
}class Solution:
def maxAlternatingSubsequence(self, nums: List[int]) -> int:
if not nums:
return 0
up, down = 1, 1
for i in range(1, len(nums)):
if nums[i] > nums[i - 1]:
up = down + 1
elif nums[i] < nums[i - 1]:
down = up + 1
return max(up, down)/**
* @param {number[]} nums
* @return {number}
*/
var maxAlternatingSubsequence = function(nums) {
if (nums.length === 0) return 0;
let up = 1, down = 1;
for (let i = 1; i < nums.length; i++) {
if (nums[i] > nums[i - 1]) {
up = down + 1;
} else if (nums[i] < nums[i - 1]) {
down = up + 1;
}
}
return Math.max(up, down);
};Asked in Top Tech Interviews
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