Optimizing Textbook Sequences — Problem Statement & Solution Guide
Problem Description
Given an array A of length n where A[i] denotes the author identifier of the i‑th textbook chapter, determine the maximum possible length of a subsequence S of A that satisfies two conditions: (1) any two consecutive elements of S have different author identifiers, and (2) the author of the first element of S is different from the author of the last element of S. The subsequence must preserve the original order of chapters. Output the length of the longest such subsequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimizing Textbook Sequences"
WHY DOES IT MATTER?
Understanding how to collapse redundant information (consecutive duplicates) is essential for many string‑processing and scheduling problems where adjacency constraints exist.
OPTIMIZATION CHALLENGE
The insight that the answer depends solely on the count of distinct runs and the relationship of the first and last runs eliminates the need for any DP or sliding‑window, cutting both time and space to linear and constant respectively.
REAL-WORLD CONNECTION
Think of a production line where identical consecutive tasks are merged into a single batch; the line’s throughput depends only on the number of distinct batches, not on how many identical items were in each batch.
During an interview, first state the adjacency rule, then immediately propose compressing runs – this shows you can simplify the problem before reaching for heavy machinery.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to analyzing the pattern of author identifiers along the textbook chapters. A naïve solution would try every subsequence, leading to exponential time, which is infeasible for n up to 2·10^5. The key observation is that any valid subsequence can be transformed into one that only keeps the first occurrence of each contiguous block of identical authors – because consecutive equal identifiers violate condition (1). After compressing the array into its run‑length representation, the subsequence length equals the number of runs, provided the first and last runs have different authors. If they are the same, we must drop one endpoint, decreasing the length by one. This yields a linear‑time algorithm that merely scans the array once, counting runs and checking the endpoints. The optimal paradigm is a greedy reduction to a simpler structure (run compression) followed by constant‑time post‑processing, a classic technique for problems that forbid adjacent duplicates.
Interview Questions on This Problem
Q1How would you compute the maximum length of a subsequence where adjacent elements differ and the first and last elements are also different?
Scan the array, count the number of times the author changes (runs). Let runs be this count plus one. If runs==1 return 0; else if first author != last author return runs; otherwise return runs‑1.
Q2Why does removing consecutive duplicate chapters not affect the optimal answer?
Consecutive duplicates can never appear together in a valid subsequence because condition (1) forbids equal adjacent identifiers. Dropping all but the first element of each block preserves any feasible subsequence and cannot reduce the maximum possible length.
Q3Can this problem be solved with DP? If so, why is DP overkill compared to the greedy run‑compression method?
A DP that tracks the last chosen author would run in O(n) time but uses O(n) space and extra logic. The greedy run‑compression directly yields the answer in O(1) extra space, making DP unnecessary.
Examples
Input
[1,2,1,3,2,2,4]
Output
6
Explanation: Select chapters at indices 0,1,2,3,4,6 → authors [1,2,1,3,2,4]. No two adjacent authors are equal and the first author (1) differs from the last (4). This subsequence has length 6, which is maximal because only one of the two consecutive 2's at indices 4 and 5 needs to be omitted.
Input
[5,5,5,5]
Output
1
Explanation: Any subsequence with two or more elements would contain two identical consecutive authors, violating condition 1, and would also have the same first and last author, violating condition 2. The best we can do is pick a single chapter, giving length 1.
Input
[7,8,9,10]
Output
4
Explanation: All authors are distinct, so the whole array forms a valid subsequence: [7,8,9,10]. The first (7) and last (10) authors differ, satisfying both conditions. Hence the maximum length is 4.
Constraints
- 1 <= n <= 200000
- -1000000000 <= A[i] <= 1000000000
- A[i] are integers
Optimal Approach & Strategy
Compress consecutive equal authors into runs, count runs, and adjust by one if the first and last runs share the same author.
Brute Force Approach
Try every possible subsequence, checking the two conditions, which is exponential in n.
Verified Code Solutions
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if (data.length===0) process.exit(0);
let pos=0;
const n = data[pos++];
const arr = data.slice(pos, pos+n);
function maxSubseqLength(arr){
if(arr.length===0) return 0;
let runs=0;
let firstVal=null, lastVal=null;
for(let i=0;i<arr.length;){
++runs;
const cur=arr[i];
if(runs===1) firstVal=cur;
while(i<arr.length && arr[i]===cur) i++;
lastVal=cur;
}
if(runs===1) return 0;
return (firstVal===lastVal)? runs-1 : runs;
}
console.log(maxSubseqLength(arr));#include <bits/stdc++.h>
using namespace std;
int maxSubseqLength(const vector<int>& A){
if(A.empty()) return 0;
int runs = 0;
int firstVal = 0, lastVal = 0;
for(size_t i=0;i<A.size();){
++runs;
int cur = A[i];
if(runs==1) firstVal = cur;
while(i<A.size() && A[i]==cur) ++i; // skip equal run
lastVal = cur;
}
if(runs==1) return 0; // first==last, cannot use any element
return (firstVal==lastVal) ? runs-1 : runs;
}
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<<maxSubseqLength(A);
return 0;
}import java.io.*;
import java.util.*;
public class Main {
public static int maxSubseqLength(int[] A){
if(A.length==0) return 0;
int runs=0;
int firstVal=0, lastVal=0;
int i=0;
while(i<A.length){
runs++;
int cur=A[i];
if(runs==1) firstVal=cur;
while(i<A.length && A[i]==cur) i++;
lastVal=cur;
}
if(runs==1) return 0;
return (firstVal==lastVal)? runs-1 : runs;
}
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[] A = new int[n];
StringTokenizer st = new StringTokenizer(br.readLine());
for(int i=0;i<n;i++) A[i]=Integer.parseInt(st.nextToken());
System.out.print(maxSubseqLength(A));
}
}import sys
def max_subseq_length(arr):
if not arr:
return 0
runs = 0
first_val = None
last_val = None
i = 0
n = len(arr)
while i < n:
runs += 1
cur = arr[i]
if runs == 1:
first_val = cur
while i < n and arr[i] == cur:
i += 1
last_val = cur
if runs == 1:
return 0
return runs - 1 if first_val == last_val else runs
def main():
data = sys.stdin.read().strip().split()
if not data:
return
n = int(data[0])
arr = list(map(int, data[1:1+n]))
print(max_subseq_length(arr))
if __name__ == "__main__":
main()const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if (data.length===0) process.exit(0);
let pos=0;
const n = data[pos++];
const arr = data.slice(pos, pos+n);
function maxSubseqLength(arr){
if(arr.length===0) return 0;
let runs=0;
let firstVal=null, lastVal=null;
for(let i=0;i<arr.length;){
++runs;
const cur=arr[i];
if(runs===1) firstVal=cur;
while(i<arr.length && arr[i]===cur) i++;
lastVal=cur;
}
if(runs===1) return 0;
return (firstVal===lastVal)? runs-1 : runs;
}
console.log(maxSubseqLength(arr));Asked in Top Tech Interviews
Solve in Interative Editor
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