Recipe Validator — Problem Statement & Solution Guide
Problem Description
Given a string composed exclusively of the characters 'S' and 'T', determine the maximum possible length of a subsequence (not necessarily contiguous) in which the characters strictly alternate between 'S' and 'T'. The subsequence may start with either character, and you may skip any number of characters from the original string. Return the length of such an optimal subsequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Recipe Validator"
WHY DOES IT MATTER?
This pattern is essential for solving problems where the optimal solution depends on the last element of a subsequence and a specific structural constraint. It is a fundamental building block for more complex DP problems involving sequences and constraints.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the problem can be solved with a greedy-like update of two state variables, avoiding the need for a full DP table. This reduces the space complexity from O(n) to O(1).
REAL-WORLD CONNECTION
This is analogous to detecting alternating patterns in network traffic (e.g., request-response cycles) or in financial time series (e.g., buy-sell patterns) to identify anomalies or trends.
In an interview, clearly articulate the state definitions (lastS and lastT) and the recurrence relations. Emphasize that the order of updates matters: when processing 'S', update lastS using the old value of lastT, and vice versa. This shows a deep understanding of the DP state transitions.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem of finding the longest alternating subsequence in a binary string is a classic application of dynamic programming, specifically optimized via state compression. A naive recursive approach would explore all possible subsequences, leading to an exponential time complexity of O(2^n), which is infeasible for large inputs. The key insight is that the optimal substructure depends only on the last character of the current subsequence and its length. We can define two states: lastS (the length of the longest alternating subsequence ending with 'S') and lastT (the length of the longest alternating subsequence ending with 'T').
As we iterate through the string, we update these states based on the current character. If the current character is 'S', we can either extend a subsequence ending with 'T' (making it end with 'S') or start a new subsequence with 'S'. The recurrence relation is lastS = max(lastS, lastT + 1). Similarly, for 'T', lastT = max(lastT, lastS + 1). This greedy-like update works because extending a shorter subsequence is never beneficial if a longer one already exists with the opposite ending character. This reduces the problem to a single pass through the string, achieving O(n) time complexity and O(1) space complexity.
This pattern is distinct from the Longest Common Subsequence (LCS) because we are not matching against a second string but rather enforcing a structural constraint (alternation) on a single string. It is also different from the Longest Increasing Subsequence (LIS) because the 'value' comparison is binary and strict, allowing for simpler state transitions. Understanding this distinction is crucial for recognizing when to use state-based DP versus interval DP or other variants.
Interview Questions on This Problem
Q1At a fintech platform, you are processing a stream of transaction types ('S' for Sale, 'T' for Trade). You need to identify the longest sequence of alternating transaction types to detect potential market manipulation patterns. How would you design an algorithm to process this stream in real-time with minimal memory overhead?
I would use a state-based dynamic programming approach with O(1) space. I would maintain two variables, lastS and lastT, representing the length of the longest alternating subsequence ending with 'S' and 'T' respectively. As each new transaction type arrives, I update the corresponding state: if it's 'S', lastS = max(lastS, lastT + 1); if 'T', lastT = max(lastT, lastS + 1). This allows for O(1) time per element and constant memory, making it suitable for real-time stream processing.
Q2In a high-growth startup's recommendation engine, user interactions are logged as 'S' (Skip) and 'T' (Try). We want to find the longest pattern of alternating user behaviors to personalize the next recommendation. If the log is stored in a distributed database, how would you optimize the query to avoid scanning the entire history?
While the core algorithm is O(n), in a distributed system, we can precompute and store the lastS and lastT values at regular intervals (checkpoints) in the database. When a new query comes in, we start from the most recent checkpoint and process only the new entries. This reduces the scan time from O(n) to O(k), where k is the number of new entries since the last checkpoint, while maintaining the same O(1) space complexity for the computation itself.
Q3At a global product company, you are analyzing user session logs where 'S' represents a search and 'T' represents a transaction. You need to find the longest alternating subsequence to understand user intent. How would you handle the case where the string is extremely large (e.g., 10^9 characters) and cannot fit in memory?
I would process the string in a streaming fashion, reading it chunk by chunk. Since the algorithm only requires the current character and the two state variables (lastS and lastT), it is inherently streamable. I would initialize the states to 0, process each character as it arrives, and update the states accordingly. This approach ensures O(1) memory usage regardless of the input size, making it feasible to process extremely large datasets.
Examples
Input
SSTTST
Output
4
Explanation: Select positions 1('S'),3('T'),5('S'),6('T') to obtain "STST", which alternates and has length 4. No longer alternating subsequence exists.
Input
TTTT
Output
1
Explanation: All characters are identical, so any alternating subsequence can contain at most one character. Hence the answer is 1.
Input
STSTST
Output
6
Explanation: The original string already alternates, so the whole string forms a valid subsequence of length 6.
Constraints
- 1 <= s.length <= 200000
- s consists only of characters 'S' and 'T'
Optimal Approach & Strategy
Use two variables, lastS and lastT, to track the length of the longest alternating subsequence ending with 'S' and 'T' respectively. Iterate through the string, updating these variables based on the current character: if it's 'S', set lastS = max(lastS, lastT + 1); if it's 'T', set lastT = max(lastT, lastS + 1). The answer is the maximum of lastS and lastT.
Brute Force Approach
Generate all possible subsequences of the string and check each one to see if it alternates between 'S' and 'T'. Keep track of the length of the longest valid subsequence found.
Verified Code Solutions
/**
* Function to find the maximum length of an alternating subsequence of 'S' and 'T'
* @param {string} s - The input string consisting of 'S' and 'T'
* @returns {number} - The maximum length of the alternating subsequence
*/
function maxAlternatingLength(s) {
if (s.length === 0) return 0;
let length = 1;
let lastChar = s[0];
for (let i = 1; i < s.length; i++) {
if (s[i] !== lastChar) {
length++;
lastChar = s[i];
}
}
return length;
}
// Driver code
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});
rl.question('Enter string: ', (s) => {
console.log(maxAlternatingLength(s));
rl.close();
});#include <iostream>
#include <string>
using namespace std;
int maxAlternatingLength(const string& s) {
if (s.empty()) return 0;
int length = 1;
char lastChar = s[0];
for (size_t i = 1; i < s.size(); ++i) {
if (s[i] != lastChar) {
++length;
lastChar = s[i];
}
}
return length;
}
int main() {
string input;
cin >> input;
cout << maxAlternatingLength(input) << endl;
return 0;
}import java.util.Scanner;
public class Main {
// Function to find the maximum length of an alternating subsequence of 'S' and 'T'
public static int maxAlternatingLength(String s) {
if (s == null || s.isEmpty()) return 0;
int length = 1;
char lastChar = s.charAt(0);
for (int i = 1; i < s.length(); i++) {
if (s.charAt(i) != lastChar) {
length++;
lastChar = s.charAt(i);
}
}
return length;
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String s = scanner.next();
System.out.println(maxAlternatingLength(s));
scanner.close();
}
}def max_alternating_length(s):
"""
Function to find the maximum length of an alternating subsequence of 'S' and 'T'
Args:
s (str): The input string consisting of 'S' and 'T'
Returns:
int: The maximum length of the alternating subsequence
"""
if not s:
return 0
length = 1
last_char = s[0]
for i in range(1, len(s)):
if s[i] != last_char:
length += 1
last_char = s[i]
return length
if __name__ == "__main__":
s = input().strip()
print(max_alternating_length(s))/**
* Function to find the maximum length of an alternating subsequence of 'S' and 'T'
* @param {string} s - The input string consisting of 'S' and 'T'
* @returns {number} - The maximum length of the alternating subsequence
*/
function maxAlternatingLength(s) {
if (s.length === 0) return 0;
let length = 1;
let lastChar = s[0];
for (let i = 1; i < s.length; i++) {
if (s[i] !== lastChar) {
length++;
lastChar = s[i];
}
}
return length;
}
// Driver code
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});
rl.question('Enter string: ', (s) => {
console.log(maxAlternatingLength(s));
rl.close();
});Asked in Top Tech Interviews
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