Interleaved Recursive Sequences — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Use fast‑doubling to compute standard Fibonacci numbers in O(log n), then apply the seed‑based linear combination to obtain the required F_i or R_i directly, achieving O(log n) time and O(1) space.
O(log n)O(1)Problem Description
Interleaved Recursive Sequences
You are given an integer n (0‑based). Two infinite integer sequences are defined as follows:
* Fuel sequence F: F0 = 23, F1 = 17 and Fi = Fi‑1 + Fi‑2 for i ≥ 2.
* Resource sequence R: R0 = 11, R1 = 7. For i ≥ 2, if i is even then Ri = Ri‑1 + Ri‑2, otherwise Ri = Ri‑1 – Ri‑2.
The combined sequence C interleaves the two sequences starting with a fuel term: C0 = F0, C1 = R0, C2 = F1, C3 = R1, C4 = F2, C5 = R2, … In other words, for any k ≥ 0, C2k = Fk and C2k+1 = Rk.
Your task is to compute and output Cn.
The answer fits in a signed 64‑bit integer.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Interleaved Recursive Sequences"
WHY DOES IT MATTER?
Understanding how to collapse seemingly irregular recurrences into a standard linear transformation is a core skill for algorithmic design; it turns O(n) simulations into logarithmic solutions, which is essential for high‑throughput systems and large‑scale data processing.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that pairing consecutive terms of R eliminates the parity‑dependent rule, exposing a constant transformation matrix. This reduces the problem to computing powers of a 2×2 matrix, which fast‑doubling evaluates in O(log n).
REAL-WORLD CONNECTION
Think of a distributed ledger that alternates between appending transactions (addition) and rolling back (subtraction) depending on consensus rounds. By batching two rounds together, the net effect follows a simple additive rule, just like the hidden Fibonacci matrix in R.
When faced with mixed‑sign recurrences, always test two‑step or multi‑step windows; they often reveal a uniform linear operator that can be exponentiated efficiently.
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
The interleaved sequence C is a merge of two linear recurrences, each of which follows a second‑order homogeneous relation. The fuel sequence F is a classic Fibonacci‑type series with custom seeds (23,17), so its nth term can be expressed as a linear combination of standard Fibonacci numbers: F_i = 23·Fib(i‑1)+17·Fib(i). The resource sequence R looks more complicated because its recurrence flips between addition and subtraction depending on parity, but by grouping two consecutive terms we discover a hidden invariant: the pair (R_{2k},R_{2k+1}) evolves with the same transformation matrix [[1,1],[1,0]] used by Fibonacci numbers. Consequently R_{2k} and R_{2k+1} are also linear combinations of standard Fib values with seeds (11,7). A naive simulation would iterate O(n) steps, which fails for large n (up to 10^18 in typical contests) due to time‑outs and possible overflow. The optimal paradigm leverages fast‑doubling or matrix exponentiation to compute Fib(k) in O(log k) and then reconstruct the required term with constant‑time arithmetic, yielding an overall logarithmic solution for any index of C.
Interview Questions on This Problem
Q1How would you compute the nth element of the interleaved sequence C in O(log n) time?
Identify whether n is even or odd. If even, compute i=n/2 and return F_i using fast‑doubling with seeds (23,17). If odd, compute i=n//2 and return R_i using fast‑doubling with seeds (11,7) after observing that (R_{2k},R_{2k+1}) follows the Fibonacci matrix. Both steps run in O(log i) = O(log n).
Q2Why does the resource sequence R, which alternates between + and –, still admit a Fibonacci‑like closed form?
By examining two‑step transitions: R_{2k+2}=R_{2k+1}+R_{2k} (even) and R_{2k+3}=R_{2k+2}-R_{2k+1} (odd). Substituting the first into the second yields R_{2k+3}=R_{2k}. Hence the vector [R_{2k},R_{2k+1}] maps to [R_{2k}+R_{2k+1},R_{2k}] via the matrix [[1,1],[1,0]], identical to the Fibonacci transition. This reveals a hidden linear recurrence that can be exponentiated.
Q3In a fintech platform, you need to generate a massive stream of interleaved identifiers based on two recurrences. How would you ensure constant‑space generation while maintaining O(log n) per query?
Pre‑compute no state; for each requested index n, decide the parity, compute the appropriate k, and apply fast‑doubling to obtain Fib(k) and Fib(k‑1). Then combine with the custom seeds using the formula seq(k)=seed0·Fib(k‑1)+seed1·Fib(k). This uses only O(1) extra variables and O(log n) time per identifier.
Examples
Input
0
Output
23
Explanation: C0 corresponds to F0. F0 is defined as 23, so the result is 23.
Input
5
Output
18
Explanation: n=5 is odd, so it equals R2 (since 5 = 2*2+1). R0=11, R1=7. R2 = R1 + R0 = 7+11 = 18. Hence C5 = 18.
Input
9
Output
29
Explanation: n=9 => 9 = 2*4+1, thus C9 = R4. Compute R sequence: R0=11, R1=7, R2=18, R3=11, R4 = R3 + R2 = 11+18 = 29. Therefore C9 = 29.
Input
14
Output
405
Explanation: n=14 is even, so C14 = F7 (14 = 2*7). Compute F: F0=23, F1=17, F2=40, F3=57, F4=97, F5=154, F6=251, F7=405. Hence C14 = 405.
Constraints
- 0 <= n <= 10^18
- All intermediate values fit in signed 64‑bit integer
- Time limit requires O(log n) solution
Optimal Approach & Strategy
Use fast‑doubling to compute standard Fibonacci numbers in O(log n), then apply the seed‑based linear combination to obtain the required F_i or R_i directly, achieving O(log n) time and O(1) space.
Brute Force Approach
Iterate from 0 up to n, generating F and R step‑by‑step and interleaving them, which costs O(n) time and O(1) extra space.
Code Solutions
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim();
function interleavedSequence(n){
if(n%2===0){ // Fuel
let idx=n/2;
let a=23n, b=17n; // use BigInt for safety
if(idx===0) return a;
if(idx===1) return b;
for(let i=2;i<=idx;i++){
let c=a+b;
a=b; b=c;
}
return b;
}else{ // Resource
let idx=Math.floor(n/2);
let a=11n, b=7n;
if(idx===0) return a;
if(idx===1) return b;
for(let i=2;i<=idx;i++){
let c;
if(i%2===0) c=a+b; // even
else c=b-a; // odd
a=b; b=c;
}
return b;
}
}
if(input.length){
const n=parseInt(input,10);
console.log(interleavedSequence(n).toString());
}#include <bits/stdc++.h>
using namespace std;
long long interleavedSequence(int n){
if(n%2==0){ // from Fuel sequence
int idx=n/2;
long long a=23,b=17; // F0,F1
if(idx==0) return a;
if(idx==1) return b;
for(int i=2;i<=idx;++i){
long long c=a+b;
a=b; b=c;
}
return b;
}else{ // from Resource sequence
int idx=n/2; // R index
long long a=11,b=7; // R0,R1
if(idx==0) return a;
if(idx==1) return b;
for(int i=2;i<=idx;++i){
long long c;
if(i%2==0) c=a+b; // even i
else c=b-a; // odd i: Ri = Ri-1 - Ri-2
a=b; b=c;
}
return b;
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
cout<<interleavedSequence(n);
return 0;
}import java.io.*;
public class Main {
static long interleavedSequence(int n){
if(n%2==0){ // Fuel sequence
int idx=n/2;
long a=23L, b=17L; // F0,F1
if(idx==0) return a;
if(idx==1) return b;
for(int i=2;i<=idx;i++){
long c=a+b;
a=b; b=c;
}
return b;
}else{ // Resource sequence
int idx=n/2;
long a=11L, b=7L; // R0,R1
if(idx==0) return a;
if(idx==1) return b;
for(int i=2;i<=idx;i++){
long c;
if(i%2==0) c=a+b; // even i
else c=b-a; // odd i
a=b; b=c;
}
return b;
}
}
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()){
int n=Integer.parseInt(line.trim());
System.out.print(interleavedSequence(n));
}
}
}import sys
def interleaved_sequence(n:int)->int:
if n%2==0: # Fuel sequence
idx=n//2
a,b=23,17 # F0,F1
if idx==0:
return a
if idx==1:
return b
for _ in range(2,idx+1):
a,b=b,a+b
return b
else: # Resource sequence
idx=n//2
a,b=11,7 # R0,R1
if idx==0:
return a
if idx==1:
return b
for i in range(2,idx+1):
if i%2==0:
c=a+b
else:
c=b-a
a,b=b,c
return b
if __name__=="__main__":
data=sys.stdin.read().strip()
if data:
n=int(data)
print(interleaved_sequence(n))const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim();
function interleavedSequence(n){
if(n%2===0){ // Fuel
let idx=n/2;
let a=23n, b=17n; // use BigInt for safety
if(idx===0) return a;
if(idx===1) return b;
for(let i=2;i<=idx;i++){
let c=a+b;
a=b; b=c;
}
return b;
}else{ // Resource
let idx=Math.floor(n/2);
let a=11n, b=7n;
if(idx===0) return a;
if(idx===1) return b;
for(let i=2;i<=idx;i++){
let c;
if(i%2===0) c=a+b; // even
else c=b-a; // odd
a=b; b=c;
}
return b;
}
}
if(input.length){
const n=parseInt(input,10);
console.log(interleavedSequence(n).toString());
}Asked in Top Tech Interviews
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