Galactic Transmission Analyzer — Problem Statement & Solution Guide
Problem Description
Galactic Transmission Analyzer
In a deep‑space communication protocol, data packets are encoded as strings where consecutive identical characters form a signal block. A signal block is defined as a maximal sequence of one or more identical characters. The protocol treats any block of length exactly one as noise and requires it to be discarded.
Given a transmission string S, determine how many signal blocks of length greater than one remain after removing all noise blocks. The answer should be output as a single integer.
Input format: a single line containing the string S.
Output format: a single integer representing the number of non‑noise signal blocks.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Transmission Analyzer"
WHY DOES IT MATTER?
Recognizing the run-length pattern allows you to collapse a potentially quadratic problem into a linear one, which is critical for scalability in production systems dealing with massive logs or network packets.
OPTIMIZATION CHALLENGE
The key insight is that you only need to remember the current character and its run length; once the character changes, you can decide immediately whether the run contributes to the answer, eliminating the need for auxiliary data structures.
REAL-WORLD CONNECTION
In distributed systems, similar logic is used in log aggregation where consecutive identical log entries are compressed into a single record to save storage and bandwidth.
When explaining this to an interviewer, emphasize that the algorithm’s simplicity is its strength—no extra memory, no nested loops, just a single pass with constant state.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to identifying maximal contiguous runs of identical characters in a string and discarding those runs whose length is exactly one. A naive approach would iterate over every possible substring or use nested loops, leading to O(n^2) time and unnecessary memory usage. The optimal paradigm is a single linear scan that tracks the current run length; whenever the character changes, we evaluate the run and decide whether to count it. This approach guarantees O(n) time and O(1) auxiliary space, making it suitable for very long transmissions.
Interview Questions on This Problem
Q1How would you modify the algorithm if the protocol required discarding blocks of length less than or equal to k instead of exactly one?
You would simply change the condition that checks the run length from "==1" to "<=k". The rest of the single-pass logic remains identical, still running in O(n) time and O(1) space.
Q2A fintech platform needs to process a stream of transaction codes in real time. How can you adapt this algorithm to work on a streaming input where the entire string is not available upfront?
Maintain a sliding window of the current character and its run length. As each new character arrives, update the run length; when the character changes, evaluate the run and output the result. This streaming version uses O(1) space and processes each character once, achieving O(n) time over the stream.
Q3During a high-growth startup interview, you’re asked to explain why a two-pointer technique is unnecessary here. What would you say?
Because the problem only requires counting runs, not comparing or extracting substrings, a two-pointer approach would add complexity without benefit. A single pointer that tracks the current character and a counter suffices, leading to cleaner code and lower overhead.
Examples
Input
aaabbcddd
Output
3
Explanation: The string splits into blocks: "aaa" (length 3), "bb" (length 2), "c" (length 1, noise), "ddd" (length 3). After discarding the noise block "c", three blocks remain, so the output is 3.
Input
abc
Output
0
Explanation: Each character forms a block of length 1, all are noise. No blocks remain, so the output is 0.
Input
zzzzzz
Output
1
Explanation: The entire string is a single block "zzzzzz" of length 6, which is not noise. Thus one block remains, yielding an output of 1.
Constraints
- 1 <= |S| <= 100000
- S consists of lowercase English letters
Optimal Approach & Strategy
Traverse the string once, maintaining the current character and run length; when the character changes, decide whether to count the run, achieving O(n) time and O(1) space.
Brute Force Approach
A naive solution would examine every possible substring, checking if it forms a maximal block and if its length is greater than one, leading to O(n^2) time.
Verified Code Solutions
const fs=require('fs');const input=fs.readFileSync(0,'utf8').trim();let s=input;let count=0;for(let i=0;i<s.length;){let j=i;while(j<s.length && s[j]===s[i]) j++;let len=j-i;if(len>1) count++;i=j;}console.log(count);#include <bits/stdc++.h>
using namespace std;
int main(){ios::sync_with_stdio(false);cin.tie(nullptr);string s;if(!(cin>>s)) return 0;int count=0;int i=0;while(i<(int)s.size()){int j=i;while(j<(int)s.size() && s[j]==s[i]) j++;int len=j-i;if(len>1) count++;i=j;}cout<<count;return 0;}
import java.io.*;
public class Main{
public static void main(String[] args) throws Exception{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String s=br.readLine();
if(s==null) return;
int count=0;
for(int i=0;i<s.length();){
int j=i;
while(j<s.length() && s.charAt(j)==s.charAt(i)) j++;
if(j-i>1) count++;
i=j;
}
System.out.print(count);
}
}
import sys
s=sys.stdin.read().strip()
count=0
i=0
while i<len(s):
j=i
while j<len(s) and s[j]==s[i]:
j+=1
if j-i>1:
count+=1
i=j
print(count)
const fs=require('fs');const input=fs.readFileSync(0,'utf8').trim();let s=input;let count=0;for(let i=0;i<s.length;){let j=i;while(j<s.length && s[j]===s[i]) j++;let len=j-i;if(len>1) count++;i=j;}console.log(count);
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