Alternate Price Variations — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Alternate Price Variations problem optimally.
O(n)O(1)Problem Description
Given an integer array prices representing the closing price of a stock on successive days, determine whether the sequence strictly alternates between a rise and a fall. In other words, for every adjacent pair the relation must be either "<" then ">" then "<" … or ">" then "<" then ">" … throughout the entire array. Equality is not allowed. The first comparison may be either an increase or a decrease. Return true if the whole array follows this alternating pattern; otherwise return false.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Alternate Price Variations"
WHY DOES IT MATTER?
Detecting strict alternation is a fundamental pattern for signal processing, trend analysis, and validation of oscillatory behavior, which appears in finance, sensor data, and UI animations. Recognizing this pattern quickly prevents costly downstream processing on malformed data.
OPTIMIZATION CHALLENGE
The insight that only the previous comparison’s sign matters reduces the state to a single boolean, collapsing an O(n²) verification into O(n) time and O(1) space.
REAL-WORLD CONNECTION
Think of a heartbeat monitor that must see a regular up‑down pulse; any flat segment (equality) or double‑up violates the rhythm, similar to how distributed logs must alternate between write and read phases to avoid contention.
During an interview, scan once, flip an expectedDirection flag after each successful comparison, and exit early on mismatch – this shows both correctness and early‑exit optimization.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The alternating‑price problem is essentially a check for a strict zig‑zag pattern in a one‑dimensional array. A naive solution would compare every adjacent pair and keep a full history of the direction, but the key is that only the sign of the previous comparison matters – you do not need any additional state or look‑ahead beyond the immediate neighbor. This observation lets us reduce the problem to a single linear scan, updating a boolean flag that represents the expected direction for the next pair. Naive approaches that attempt to rebuild subsequences or use nested loops explode to O(n²) time on large inputs, which is unnecessary because the condition is local and deterministic. The optimal paradigm falls under greedy verification: at each step we greedily enforce the required alternating relation, and any violation immediately disproves the whole sequence, allowing early termination.
Because the pattern is strictly local, the algorithm runs in O(n) time with O(1) auxiliary space, which is optimal for any algorithm that must read each element at least once. This aligns with classic array‑traversal techniques used in problems like "wiggle subsequence" and "alternating sign" checks, where the core insight is that the global property can be validated by maintaining only the last comparison result.
Interview Questions on This Problem
Q1How would you modify the solution to also return the length of the longest alternating subarray instead of just a boolean?
Maintain two counters, one for a subarray ending with a '<' and another ending with a '>'. Update them based on the current comparison, resetting the opposite counter to 1 when the direction flips, and track the maximum of both counters throughout the scan.
Q2Can this problem be solved using a stack or any other data structure? If so, why is it unnecessary?
A stack could store the direction history, but because only the most recent direction matters, the stack adds O(n) space without any benefit; a simple variable suffices, making the stack approach over‑engineered.
Q3In a distributed system where price updates arrive out‑of‑order, how would you ensure the alternating property is still validated efficiently?
Buffer out‑of‑order updates until you can order them by timestamp, then apply the same linear scan on the ordered sequence; alternatively, maintain a sliding window with the last two ordered prices and validate each new arrival in O(1) as it becomes ordered.
Examples
Input
[3,5,2,8,1]
Output
true
Explanation: 3<5 (rise), 5>2 (fall), 2<8 (rise), 8>1 (fall). The direction changes at each step, so the array is alternating.
Input
[4,4,2,6]
Output
false
Explanation: The first two elements are equal (4==4), which violates the strict "rise" or "fall" requirement, therefore the sequence cannot be alternating.
Input
[10,5,7,3,9]
Output
true
Explanation: 10>5 (fall), 5<7 (rise), 7>3 (fall), 3<9 (rise). The comparisons flip direction at every step, satisfying the alternating condition.
Constraints
- 1 <= prices.length <= 100000
- -1000000000 <= prices[i] <= 1000000000
- All comparisons must be strict; equal adjacent values invalidate the pattern
Optimal Approach & Strategy
Perform a single pass, tracking only the last comparison’s direction and flipping the expected direction after each successful check, achieving O(n) time and O(1) space.
Brute Force Approach
Check every possible sub‑array or use nested loops to compare each pair with all previous directions, leading to O(n²) time.
Code Solutions
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim();
if(data.length===0) process.exit(0);
function isAlternating(prices){
const n=prices.length;
if(n<2) return true;
let prevDiff=prices[1]-prices[0];
if(prevDiff===0) return false;
let prevSign=prevDiff>0?1:-1;
for(let i=2;i<n;i++){
const diff=prices[i]-prices[i-1];
if(diff===0) return false;
const sign=diff>0?1:-1;
if(sign===prevSign) return false;
prevSign=sign;
}
return true;
}
const tokens=data.split(/\s+/).map(Number);
const n=tokens[0];
const prices=tokens.slice(1,1+n);
console.log(isAlternating(prices));#include <bits/stdc++.h>
using namespace std;
bool isAlternating(const vector<int>& prices){
int n=prices.size();
if(n<2) return true;
int prevDiff=prices[1]-prices[0];
if(prevDiff==0) return false;
int prevSign=(prevDiff>0)?1:-1;
for(int i=2;i<n;++i){
int diff=prices[i]-prices[i-1];
if(diff==0) return false;
int sign=(diff>0)?1:-1;
if(sign==prevSign) return false;
prevSign=sign;
}
return true;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
vector<int> prices(n);
for(int i=0;i<n;++i)cin>>prices[i];
cout<<(isAlternating(prices)?"true":"false")<<'\n';
return 0;
}import java.util.*;
public class Main{
static boolean isAlternating(int[] prices){
int n=prices.length;
if(n<2) return true;
int prevDiff=prices[1]-prices[0];
if(prevDiff==0) return false;
int prevSign=prevDiff>0?1:-1;
for(int i=2;i<n;i++){
int diff=prices[i]-prices[i-1];
if(diff==0) return false;
int sign=diff>0?1:-1;
if(sign==prevSign) return false;
prevSign=sign;
}
return true;
}
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
if(!sc.hasNextInt()) return;
int n=sc.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++) arr[i]=sc.nextInt();
System.out.println(isAlternating(arr));
}
}import sys
def is_alternating(prices):
n=len(prices)
if n<2:
return True
prev_diff=prices[1]-prices[0]
if prev_diff==0:
return False
prev_sign=1 if prev_diff>0 else -1
for i in range(2,n):
diff=prices[i]-prices[i-1]
if diff==0:
return False
sign=1 if diff>0 else -1
if sign==prev_sign:
return False
prev_sign=sign
return True
if __name__=='__main__':
data=sys.stdin.read().strip()
if not data:
sys.exit()
tokens=list(map(int,data.split()))
n=tokens[0]
prices=tokens[1:1+n]
print(str(is_alternating(prices)).lower())const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim();
if(data.length===0) process.exit(0);
function isAlternating(prices){
const n=prices.length;
if(n<2) return true;
let prevDiff=prices[1]-prices[0];
if(prevDiff===0) return false;
let prevSign=prevDiff>0?1:-1;
for(let i=2;i<n;i++){
const diff=prices[i]-prices[i-1];
if(diff===0) return false;
const sign=diff>0?1:-1;
if(sign===prevSign) return false;
prevSign=sign;
}
return true;
}
const tokens=data.split(/\s+/).map(Number);
const n=tokens[0];
const prices=tokens.slice(1,1+n);
console.log(isAlternating(prices));Asked in Top Tech Interviews
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