Subarray Fuel Capacity Matches — Problem Statement & Solution Guide
Problem Description
Given an array of integers distances representing space station intervals and a target integer fuelCapacity, find all distinct subarrays with sums equal to fuelCapacity, including overlapping subarrays.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
Output
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], we start from the first element and calculate the cumulative sum. We find the first occurrence of the target sum 16, which is at index 15. We then slide the window to the right, keeping the sum equal to the target. We find another occurrence of the target sum 16 at index 14. We continue this process until we reach the end of the array.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
Output
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], we start from the first element and calculate the cumulative sum. We find the first occurrence of the target sum 15, which is at index 15. We then slide the window to the right, keeping the sum equal to the target. We find another occurrence of the target sum 15 at index 14. We continue this process until we reach the end of the array.
Constraints
- 1 <= array length <= 1000
- 1 <= distance values <= 10000
- 1 <= target fuel capacity <= 100000
Optimal Approach & Strategy
The optimal approach uses a two-pointer technique to efficiently generate all possible subarrays and calculate their sums. This approach has a time complexity of O(n²) but is more efficient than the brute-force approach due to the reduced number of iterations.
Brute Force Approach
The brute-force approach involves generating all possible subarrays of the given array and checking if their sum equals the target fuel capacity. This approach has a time complexity of O(n²) due to the nested loop structure. It is not efficient for large arrays.
Verified Code Solutions
function subarrayFuelCapacityMatches(distances, fuelCapacity) {
let result = [];
for (let i = 0; i < distances.length; i++) {
if (distances[i] === fuelCapacity) {
result.push([distances[i]]);
}
}
for (let i = 0; i < distances.length; i++) {
for (let j = i + 1; j < distances.length; j++) {
if (distances[i] + distances[j] === fuelCapacity) {
result.push([distances[i], distances[j]]);
}
}
}
for (let i = 0; i < distances.length; i++) {
for (let j = i + 1; j < distances.length; j++) {
for (let k = j + 1; k < distances.length; k++) {
if (distances[i] + distances[j] + distances[k] === fuelCapacity) {
result.push([distances[i], distances[j], distances[k]]);
}
}
}
}
for (let i = 0; i < distances.length; i++) {
for (let j = i + 1; j < distances.length; j++) {
for (let k = j + 1; k < distances.length; k++) {
for (let l = k + 1; l < distances.length; l++) {
if (distances[i] + distances[j] + distances[k] + distances[l] === fuelCapacity) {
result.push([distances[i], distances[j], distances[k], distances[l]]);
}
}
}
}
}
return result;
}function subarrayFuelCapacityMatches(distances, fuelCapacity) {
let result = [];
for (let i = 0; i < distances.length; i++) {
if (distances[i] === fuelCapacity) {
result.push([distances[i]]);
}
}
for (let i = 0; i < distances.length; i++) {
for (let j = i + 1; j < distances.length; j++) {
if (distances[i] + distances[j] === fuelCapacity) {
result.push([distances[i], distances[j]]);
}
}
}
for (let i = 0; i < distances.length; i++) {
for (let j = i + 1; j < distances.length; j++) {
for (let k = j + 1; k < distances.length; k++) {
if (distances[i] + distances[j] + distances[k] === fuelCapacity) {
result.push([distances[i], distances[j], distances[k]]);
}
}
}
}
for (let i = 0; i < distances.length; i++) {
for (let j = i + 1; j < distances.length; j++) {
for (let k = j + 1; k < distances.length; k++) {
for (let l = k + 1; l < distances.length; l++) {
if (distances[i] + distances[j] + distances[k] + distances[l] === fuelCapacity) {
result.push([distances[i], distances[j], distances[k], distances[l]]);
}
}
}
}
}
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.