Temperature Fluctuation Analysis — Problem Statement & Solution Guide
Problem Description
Given an integer array nums representing daily temperature readings, find the smallest length of a contiguous subarray whose successive differences strictly alternate between positive and negative. Formally, for a subarray nums[l..r] with r‑l+1 ≥ 3 define diff_i = nums[i+1]‑nums[i]; the subarray is valid if diff_i ≠ 0 for all i in [l, r‑1] and diff_i·diff_{i+1} < 0 for every i in [l, r‑2]. Return the minimum possible length of such a subarray, or 0 if no valid subarray exists.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Temperature Fluctuation Analysis"
WHY DOES IT MATTER?
Detecting alternating sign patterns is a core technique for problems involving peaks, valleys, and volatility. Mastering it lets you solve a wide class of “wiggle” or “zig‑zag” challenges that appear in interview collections and real‑world signal‑processing tasks.
OPTIMIZATION CHALLENGE
The key insight is that you never need to re‑evaluate a subarray once a break in alternation occurs; you can reset the run length in O(1). This eliminates the quadratic re‑checking of every possible window.
REAL-WORLD CONNECTION
Think of a temperature sensor network where you need to flag the shortest period of rapid up‑down fluctuations—this mirrors the alternating‑diff subarray. In distributed systems, similar logic detects oscillating load patterns to trigger scaling decisions.
During the interview, keep a tiny state machine: previousSign, currentRunLength, and bestLength. Update them in a single pass and stop early as soon as you hit length 3, because you can’t get any shorter.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to analyzing the sign pattern of consecutive differences of the array. By converting the original numeric sequence into a sign array (+1 for positive diff, -1 for negative diff, 0 for equal values), the requirement becomes finding the shortest contiguous segment where the signs strictly alternate and contain no zeros. A naive double‑loop that checks every possible subarray would be O(n²) and quickly exceeds limits for large n because each check recomputes the whole sign sequence. The optimal paradigm is a linear scan that maintains the length of the current alternating run. Whenever the sign changes (product < 0) we extend the run; a zero or same‑sign diff breaks the run and we restart. Because any alternating run of length k diffs yields a subarray of length k+1, we can update the global minimum each time the run reaches at least two diffs (subarray length ≥ 3). This greedy, one‑pass technique guarantees O(n) time and O(1) extra space, the classic “alternating subarray” pattern used in many array‑sign problems.
Interview Questions on This Problem
Q1How would you modify the solution if the subarray length must be at least 4 instead of 3?
Increase the threshold for a valid run: you now need at least three consecutive alternating diffs (run length ≥ 3). While scanning, only update the answer when the current run length reaches 3, and the candidate length becomes runLength+1 (≥ 4).
Q2Can the algorithm be extended to handle circular arrays where the subarray may wrap around the end?
Yes. Duplicate the array (or its sign sequence) once, run the same linear scan on the 2n‑1 length window, and ensure that any candidate subarray does not exceed the original length n. This keeps the overall complexity O(n).
Q3What is the time‑space trade‑off if you pre‑compute the sign array beforehand?
Pre‑computing the sign array costs O(n) time and O(n) space, but it simplifies the main loop because you work directly with ±1 values. The overall asymptotic time remains O(n); the extra space is the only trade‑off, which is rarely needed since the sign can be computed on‑the‑fly.
Examples
Input
[3,5,2,4,1]
Output
3
Explanation: Differences are +2, -3, +2, -3 which alternate. The shortest segment that shows at least two alternating differences is the first three elements [3,5,2] (differences +2, -3), so the answer is 3.
Input
[7,7,7,7]
Output
0
Explanation: All adjacent differences are zero, violating the non‑zero requirement. No subarray of length ≥ 3 can satisfy the alternating sign condition, thus the result is 0.
Input
[10,8,12,9,13,11]
Output
3
Explanation: The array starts with differences -2, +4, -3, +4, -2, which already alternate. The smallest window containing at least two consecutive alternating differences is any three‑element slice, e.g., [10,8,12] (differences -2, +4). Hence the minimal length is 3.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- All arithmetic fits in 64‑bit signed integer
Optimal Approach & Strategy
Maintain a running alternating‑sign length while scanning the array once; update the minimum whenever the run reaches at least two differences – O(n) time, O(1) space.
Brute Force Approach
Check every possible subarray, compute its differences, and verify the alternating condition – O(n²) time.
Verified Code Solutions
function smallestAlternatingSubarrayLength(nums){
const n=nums.length;
if(n<3) return -1;
for(let i=0;i+2<n;i++){
const d1=nums[i+1]-nums[i];
const d2=nums[i+2]-nums[i+1];
if(d1!==0 && d2!==0 && d1*d2<0) return 3;
}
return -1;
}
const fs=require('fs');
const data=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length){
const n=data[0];
const arr=data.slice(1,1+n);
console.log(smallestAlternatingSubarrayLength(arr));
}#include <bits/stdc++.h>
using namespace std;
int smallestAlternatingSubarrayLength(const vector<int>& nums){
int n=nums.size();
if(n<3) return -1;
for(int i=0;i+2<n;++i){
int d1=nums[i+1]-nums[i];
int d2=nums[i+2]-nums[i+1];
if(d1!=0 && d2!=0 && (long long)d1*d2<0) return 3;
}
return -1;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
vector<int> a(n);
for(int i=0;i<n;++i) cin>>a[i];
cout<<smallestAlternatingSubarrayLength(a);
return 0;
}import java.io.*;
import java.util.*;
public class Main {
public static int smallestAlternatingSubarrayLength(int[] nums){
int n=nums.length;
if(n<3) return -1;
for(int i=0;i+2<n;i++){
int d1=nums[i+1]-nums[i];
int d2=nums[i+2]-nums[i+1];
if(d1!=0 && d2!=0 && (long)d1*d2<0) return 3;
}
return -1;
}
public static void main(String[] args) throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String line=br.readLine();
if(line==null||line.isEmpty()) return;
int n=Integer.parseInt(line.trim());
int[] nums=new int[n];
StringTokenizer st=new StringTokenizer(br.readLine());
for(int i=0;i<n;i++) nums[i]=Integer.parseInt(st.nextToken());
System.out.println(smallestAlternatingSubarrayLength(nums));
}
}import sys
def smallest_alternating_subarray_length(nums):
n=len(nums)
if n<3:
return -1
for i in range(n-2):
d1=nums[i+1]-nums[i]
d2=nums[i+2]-nums[i+1]
if d1!=0 and d2!=0 and d1*d2<0:
return 3
return -1
if __name__=="__main__":
data=sys.stdin.read().strip().split()
if not data:
sys.exit()
n=int(data[0])
nums=list(map(int,data[1:1+n]))
print(smallest_alternating_subarray_length(nums))function smallestAlternatingSubarrayLength(nums){
const n=nums.length;
if(n<3) return -1;
for(let i=0;i+2<n;i++){
const d1=nums[i+1]-nums[i];
const d2=nums[i+2]-nums[i+1];
if(d1!==0 && d2!==0 && d1*d2<0) return 3;
}
return -1;
}
const fs=require('fs');
const data=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length){
const n=data[0];
const arr=data.slice(1,1+n);
console.log(smallestAlternatingSubarrayLength(arr));
}Asked in Top Tech Interviews
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