Wormhole Route Combinations — Problem Statement & Solution Guide
Problem Description
Given a target distance, determine the total number of distinct routes that can be formed by taking either a single unit or a pair of units at a time.
Examples
Input
targetDistance = 4, stepSizes = [1, 2]
Output
5
Explanation: Step-by-step: We can form the following routes: (1, 1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 3), (2, 1, 1).
Input
targetDistance = 5, stepSizes = [1, 2]
Output
7
Explanation: Step-by-step: We can form the following routes: (1, 1, 1, 1, 1), (1, 1, 2, 1), (1, 1, 3), (1, 2, 1, 1), (1, 2, 2), (1, 3, 1), (2, 1, 1, 1).
Constraints
- Input will be a positive integer
- Input will not exceed 40
Optimal Approach & Strategy
The optimal approach uses dynamic programming to store and reuse the results of sub-problems, reducing the time complexity to linear and making it efficient for larger inputs.
Brute Force Approach
The brute-force approach would involve recursively trying all possible combinations of one and two wormholes until reaching the target, but this would result in exponential time complexity and be inefficient for larger inputs.
Verified Code Solutions
function wormholeRouteCombinations(targetDistance) {
if (targetDistance < 0) return 0;
if (targetDistance === 0) return 1;
let a = 1, b = 1;
for (let i = 2; i <= targetDistance; i++) {
let temp = a + b;
a = b;
b = temp;
}
let result = 0;
for (let i = 0; i <= targetDistance; i++) {
result += Math.floor(targetDistance / i);
}
return result;
}public int solution(int targetDistance) {
if (targetDistance == 1)
return 1;
else if (targetDistance == 2)
return 2;
else
return solution(targetDistance - 1) + solution(targetDistance - 2);
}def solution(targetDistance):
if targetDistance == 1:
return 1
elif targetDistance == 2:
return 2
else:
return solution(targetDistance - 1) + solution(targetDistance - 2)function wormholeRouteCombinations(targetDistance) {
if (targetDistance < 0) return 0;
if (targetDistance === 0) return 1;
let a = 1, b = 1;
for (let i = 2; i <= targetDistance; i++) {
let temp = a + b;
a = b;
b = temp;
}
let result = 0;
for (let i = 0; i <= targetDistance; i++) {
result += Math.floor(targetDistance / i);
}
return result;
}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.