Interstellar Route Planner — Problem Statement & Solution Guide

RecursionMediumMixed

TimeO(n)

SpaceO(1)

Problem Description

Given a non‑negative integer n representing the index of a planetary system, a traveler may move forward by exactly 3 or exactly 5 systems in a single jump. Compute the total number of distinct ordered sequences of jumps whose total distance equals n. Two sequences are different if the order of jumps differs. The answer must be returned as a 64‑bit integer. Implement the solution using a recursive formulation (memoization or bottom‑up DP is acceptable).

DSA Pattern Breakdown

"Interstellar Route Planner"

medium

WHY DOES IT MATTER?

This pattern is essential because it teaches the transition from exponential recursive thinking to linear dynamic programming. It highlights the importance of memoization and state definition in counting problems, which are frequent in system design and algorithmic interviews. Mastering this pattern allows engineers to solve a wide range of combinatorial problems efficiently.

OPTIMIZATION CHALLENGE

The key insight is recognizing that the naive recursive approach recalculates the same subproblems exponentially. By using memoization (top-down) or tabulation (bottom-up), we reduce the time complexity from O(2^n) to O(n). The space optimization further reduces O(n) to O(1) by leveraging the fact that only the last few states are needed.

REAL-WORLD CONNECTION

This is analogous to packet routing in distributed systems, where a message can be forwarded via multiple intermediate nodes (jumps of 3 or 5 hops). Counting distinct ordered paths helps in network redundancy analysis and load balancing strategies, ensuring that traffic can be distributed across multiple valid routes without duplication.

In an interview, always start by defining the state clearly: 'dp[i] represents the number of ways to reach distance i.' Then, derive the recurrence relation explicitly. Mentioning the base cases (e.g., dp[0]=1, dp[i<0]=0) shows rigor. Finally, proactively discuss the space optimization to demonstrate senior-level optimization skills.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem 'Interstellar Route Planner' is a classic instance of the 'Counting Paths' or 'Coin Change' variation, specifically focusing on ordered sequences (compositions) rather than unordered sets. The core theoretical foundation lies in the principle of optimal substructure: the number of ways to reach distance n is the sum of the number of ways to reach n-3 and n-5. This recursive relationship, f(n) = f(n-3) + f(n-5), mirrors the Fibonacci sequence but with different step sizes. Understanding this recurrence is critical because it reveals that the problem is not about finding a single path, but about aggregating all possible permutations of jumps that sum to the target.

Interview Questions on This Problem

Q1At a fintech platform like Stripe, you need to calculate the number of ways a user can break down a transaction amount into specific fee tiers (e.g., $3 and $5 fees) for audit logging. How would you handle large transaction amounts where naive recursion times out?

I would implement a bottom-up dynamic programming approach using a 1D array dp where dp[i] stores the number of ways to reach amount i. By iterating from 0 to n, I avoid the exponential time complexity of naive recursion. This ensures O(n) time and O(n) space, which is efficient enough for large transaction amounts while maintaining 64-bit integer precision for the count.

Q2In a high-growth engineering startup, a logistics system needs to count distinct delivery route sequences using fixed distance increments. If the distance `n` is extremely large (e.g., 10^6), how do you optimize memory usage?

Since the recurrence f(n) = f(n-3) + f(n-5) only depends on the previous few states, I can optimize space to O(1) by using a sliding window of variables instead of a full array. I would maintain variables for the last 5 computed values and update them iteratively. This reduces memory overhead significantly while preserving the O(n) time complexity, which is crucial for memory-constrained edge devices or high-concurrency servers.

Q3At a global product company like Google, you are asked to generalize this problem to allow jumps of size 3, 5, and 7. How does the state transition change, and how do you ensure the solution remains efficient?

The state transition becomes f(n) = f(n-3) + f(n-5) + f(n-7). The algorithmic paradigm remains the same: bottom-up DP. The time complexity stays O(n) because for each state i, we perform a constant number of additions (3 in this case). The space complexity can be optimized to O(1) using a circular buffer of size 7. This generalization demonstrates scalability and the ability to adapt the core recursive logic to multi-step problems.

Examples

Example 1

Input

Output

Explanation: The only ways to sum to 8 using 3 and 5 are (3,5) and (5,3). Hence 2 distinct routes.

Example 2

Input

Output

Explanation: Solutions of 3a+5b=18 are (a=6,b=0) giving the sequence (3,3,3,3,3,3) and (a=1,b=3) giving four permutations of one 3 and three 5s. The number of permutations for the mixed case is C(4,1)=4. Total routes =1+4=5.

Example 3

Input

Output

Explanation: No combination of 3 and 5 can sum to 2, so there are no valid routes.

Constraints

0 <= n <= 100000
Result fits in a signed 64‑bit integer
Time limit allows O(n) or O(n) memory solutions

Optimal Approach & Strategy

The optimal approach uses bottom-up dynamic programming with a 1D array or a sliding window to store the number of ways to reach each distance. By iterating from 0 to n and computing dp[i] = dp[i-3] + dp[i-5], we achieve O(n) time complexity and O(1) space complexity.

Brute Force Approach

The naive approach uses a recursive function that explores all possible jump sequences by branching into two choices (jump 3 or jump 5) until the distance reaches 0 or goes negative. This results in an exponential time complexity of O(2^n) due to redundant calculations of overlapping subproblems.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function countRoutes(n) {
    const memo = new Map();
    
    function helper(rem) {
        if (rem < 0) return 0;
        if (rem === 0) return 1;
        if (memo.has(rem)) return memo.get(rem);
        
        const result = helper(rem - 3) + helper(rem - 5);
        memo.set(rem, result);
        return result;
    }
    
    return helper(n);
}

const readline = require('readline');
const rl = readline.createInterface({
    input: process.stdin,
    terminal: false
});

rl.on('line', (line) => {
    const n = parseInt(line.trim());
    console.log(countRoutes(n));
    rl.close();
});

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

class Solution {
private:
    unordered_map<int, long long> memo;
    
    long long helper(int n) {
        if (n < 0) return 0;
        if (n == 0) return 1;
        if (memo.find(n) != memo.end()) return memo[n];
        
        long long result = helper(n - 3) + helper(n - 5);
        memo[n] = result;
        return result;
    }

public:
    long long countRoutes(int n) {
        memo.clear();
        return helper(n);
    }
};

int main() {
    int n;
    if (cin >> n) {
        Solution sol;
        cout << sol.countRoutes(n) << endl;
    }
    return 0;
}

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
    private static Map<Integer, Long> memo;
    
    private static long helper(int n) {
        if (n < 0) return 0;
        if (n == 0) return 1;
        if (memo.containsKey(n)) return memo.get(n);
        
        long result = helper(n - 3) + helper(n - 5);
        memo.put(n, result);
        return result;
    }
    
    public static long countRoutes(int n) {
        memo = new HashMap<>();
        return helper(n);
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        if (scanner.hasNextInt()) {
            int n = scanner.nextInt();
            System.out.println(countRoutes(n));
        }
        scanner.close();
    }
}

def count_routes(n: int) -> int:
    memo = {}
    
    def helper(rem):
        if rem < 0:
            return 0
        if rem == 0:
            return 1
        if rem in memo:
            return memo[rem]
        
        result = helper(rem - 3) + helper(rem - 5)
        memo[rem] = result
        return result
    
    return helper(n)

if __name__ == "__main__":
    import sys
    data = sys.stdin.read().strip().split()
    if data:
        n = int(data[0])
        print(count_routes(n))

function countRoutes(n) {
    const memo = new Map();
    
    function helper(rem) {
        if (rem < 0) return 0;
        if (rem === 0) return 1;
        if (memo.has(rem)) return memo.get(rem);
        
        const result = helper(rem - 3) + helper(rem - 5);
        memo.set(rem, result);
        return result;
    }
    
    return helper(n);
}

const readline = require('readline');
const rl = readline.createInterface({
    input: process.stdin,
    terminal: false
});

rl.on('line', (line) => {
    const n = parseInt(line.trim());
    console.log(countRoutes(n));
    rl.close();
});

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Interstellar Route Planner Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Recursion

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Interstellar Route Planner — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor