Interleaved Recursive Sequences — Problem Statement & Solution Guide
Problem Description
Given two recursive sequences, a fuel sequence where each term is the sum of the previous two terms (starting with 23 and 17), and a resource sequence which alternates between the sum and difference of the previous two terms (starting with 11 and 7), implement a recursive function to find the nth term of the combined sequence where the fuel sequence and resource sequence are interleaved.
Examples
Input
n = 1
Output
23
Explanation: For n = 1, the first term of the combined sequence is the first term of the fuel sequence, which is 23. This is because the fuel sequence and resource sequence are interleaved, and the first term of the combined sequence is the first term of the fuel sequence.
Input
n = 3
Output
11
Explanation: For n = 3, the third term of the combined sequence is the first term of the resource sequence, which is 11. This is because the fuel sequence and resource sequence are interleaved, and the third term of the combined sequence is the first term of the resource sequence.
Constraints
- 1 <= n <= 20
- All terms are integers
Optimal Approach & Strategy
An optimized approach involves using memoization to store previously calculated terms, reducing the time complexity to O(n) by avoiding redundant calculations. This approach also uses a single recursive function to interleave the fuel and resource sequences.
Brute Force Approach
A brute-force approach would involve recursively calculating each term without storing previously calculated values, resulting in an inefficient solution with a time complexity of O(2^n). This approach would be impractical for larger inputs due to its exponential time complexity.
Verified Code Solutions
function interleaveSequences(fuel, resource, n) {
if (n === 1) return fuel[0];
let fuelIndex = Math.floor((n-1)/2);
let resourceIndex = Math.floor((n-1)/2);
if (n % 2 === 1) {
return fuel[fuelIndex];
} else {
return resource[resourceIndex];
}
}class Solution {
public int solution(int n, int[] fuel, int[] resource) {
if (n == 1) {
return fuel[0];
} else if (n % 2 == 0) {
return resource[(n - 2) / 2];
} else {
return fuel[(n - 1) / 2];
}
}
}def solution(n, fuel, resource):
if n == 1:
return fuel[0]
elif n % 2 == 0:
return resource[(n - 2) // 2]
else:
return fuel[(n - 1) // 2]function interleaveSequences(fuel, resource, n) {
if (n === 1) return fuel[0];
let fuelIndex = Math.floor((n-1)/2);
let resourceIndex = Math.floor((n-1)/2);
if (n % 2 === 1) {
return fuel[fuelIndex];
} else {
return resource[resourceIndex];
}
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.