Galactic Resource Allocation 3 — Problem Statement & Solution Guide
Problem Description
Given an integer array nums of length n where nums[i] denotes the amount of a particular resource available on day i, and an integer k representing the exact number of consecutive days a spaceship must stay docked, compute the maximum total resource that can be collected over any contiguous segment of length k. The input consists of n, the array nums, and k. Output a single integer – the greatest possible sum of k consecutive elements in nums.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Resource Allocation 3"
WHY DOES IT MATTER?
The Sliding Window pattern is essential for optimizing problems involving contiguous subarrays or substrings of fixed length. It demonstrates the power of reusing previous computations to avoid redundant work, a core principle in algorithmic efficiency.
OPTIMIZATION CHALLENGE
The key insight is recognizing the overlap between consecutive windows. By maintaining a running sum and updating it incrementally, we reduce the time complexity from quadratic to linear.
REAL-WORLD CONNECTION
This pattern is analogous to a moving average in financial time-series analysis, where the average of the last N days is updated daily by removing the oldest day's value and adding the newest, rather than recalculating the average from scratch.
During interviews, explicitly state the time complexity improvement (O(n*k) to O(n)) and verify the boundary conditions, such as when k equals the array length or when the array is empty.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem of finding the maximum sum of a contiguous subarray of fixed length k is a classic application of the Sliding Window technique. The naive approach involves iterating through every possible starting index and summing the next k elements, resulting in a time complexity of O(n*k). For large inputs where n can be up to 10^5 or 10^6, this quadratic behavior leads to Time Limit Exceeded (TLE) errors, making it unsuitable for production-grade systems or high-frequency trading environments where latency is critical.
The optimal paradigm relies on the observation that consecutive windows of size k overlap by k-1 elements. When the window slides from index i to i+1, the element at index i leaves the window, and the element at index i+k enters. Therefore, the sum of the new window can be computed in O(1) time by subtracting the outgoing element and adding the incoming element to the previous sum. This transforms the problem from O(n*k) to O(n), as each element is added and subtracted exactly once.
This technique is foundational in array and string processing, extending to problems involving maximum/minimum subarray sums, counting anagrams, and finding the first unique character in a sliding window. Understanding the invariant that the window size remains constant is crucial for correctly implementing the boundary conditions, particularly when k equals n or k is 1.
Interview Questions on This Problem
Q1At a fintech platform, we need to calculate the maximum 5-day trading volume for a stock to identify volatility spikes. How would you design an efficient algorithm to process a stream of daily volumes?
I would use a sliding window approach. By maintaining a running sum of the last 5 days, I can update the sum in O(1) time for each new day by subtracting the volume from 5 days ago and adding the current day's volume. This ensures the solution scales linearly with the number of days, O(n), which is essential for real-time data processing.
Q2In a distributed system, we need to find the maximum load on any 10-minute interval from a log of request counts per minute. How do you optimize the computation if the log is very large?
I would treat the log as an array and apply a sliding window of size 10. Instead of recalculating the sum for every 10-minute interval, I will maintain a cumulative sum that updates incrementally. This reduces the computational complexity from O(n*10) to O(n), ensuring the system can handle high-throughput logs without latency issues.
Q3A high-growth startup needs to find the best 3-day period for a marketing campaign based on daily user engagement scores. What is the most efficient way to compute this, and how would you handle edge cases where the array length is less than 3?
The most efficient way is to use a sliding window of size 3. I would first check if the array length is less than k; if so, I would return the sum of the entire array or handle it as an invalid input per requirements. Otherwise, I initialize the sum of the first k elements, then slide the window by adding the next element and subtracting the one leaving the window, tracking the maximum sum encountered.
Examples
Input
7 4 2 1 7 5 3 6 3
Output
15
Explanation: All windows of size 3 are: [4,2,1]→7, [2,1,7]→10, [1,7,5]→13, [7,5,3]→15, [5,3,6]→14. The largest sum is 15.
Input
5 -2 -3 -1 -4 -6 2
Output
-4
Explanation: Windows of size 2 are: [-2,-3]→-5, [-3,-1]→-4, [-1,-4]→-5, [-4,-6]→-10. The maximum among them is -4.
Input
8 10 -2 3 5 -1 2 8 -3 4
Output
16
Explanation: Sliding windows of length 4 produce sums: [10,-2,3,5]→16, [-2,3,5,-1]→5, [3,5,-1,2]→9, [5,-1,2,8]→14, [-1,2,8,-3]→6. The highest sum is 16.
Constraints
- 1 <= n <= 100000
- 1 <= k <= n
- -1000000000 <= nums[i] <= 1000000000
- Result fits in a 64‑bit signed integer
Optimal Approach & Strategy
Initialize the sum of the first k elements, then slide the window by subtracting the outgoing element and adding the incoming element in each step. Track the maximum sum encountered during the iteration, achieving O(n) time complexity.
Brute Force Approach
Iterate through each starting index from 0 to n-k, and for each index, sum the next k elements to find the maximum sum. This approach has a time complexity of O(n*k) and is inefficient for large arrays.
Verified Code Solutions
/**
* @param {number[]} nums - Array of integers representing daily resources.
* @param {number} k - The exact number of consecutive days.
* @return {number} The maximum total resource collected over any contiguous segment of length k.
*/
function maxResource(nums, k) {
const n = nums.length;
if (n === 0 || k <= 0 || k > n) {
return 0;
}
let currentSum = 0;
for (let i = 0; i < k; i++) {
currentSum += nums[i];
}
let maxSum = currentSum;
for (let i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
if (currentSum > maxSum) {
maxSum = currentSum;
}
}
return maxSum;
}
// Driver code
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
terminal: false
});
let lines = [];
rl.on('line', line => {
lines.push(line);
});
rl.on('close', () => {
const input = lines.join(' ').split(' ').map(Number);
const n = input[0];
const nums = input.slice(1, 1 + n);
const k = input[1 + n];
console.log(maxResource(nums, k));
});#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;
int maxResource(const vector<int>& nums, int k) {
int n = nums.size();
if (n == 0 || k <= 0 || k > n) {
return 0;
}
long long currentSum = 0;
for (int i = 0; i < k; ++i) {
currentSum += nums[i];
}
long long maxSum = currentSum;
for (int i = k; i < n; ++i) {
currentSum += nums[i] - nums[i - k];
maxSum = max(maxSum, currentSum);
}
return static_cast<int>(maxSum);
}
int main() {
int n;
if (!(cin >> n)) return 0;
vector<int> nums(n);
for (int i = 0; i < n; ++i) {
cin >> nums[i];
}
int k;
cin >> k;
cout << maxResource(nums, k) << endl;
return 0;
}import java.util.*;
import java.io.*;
public class Main {
public static int maxResource(int[] nums, int k) {
int n = nums.length;
if (n == 0 || k <= 0 || k > n) {
return 0;
}
long currentSum = 0;
for (int i = 0; i < k; i++) {
currentSum += nums[i];
}
long maxSum = currentSum;
for (int i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
if (currentSum > maxSum) {
maxSum = currentSum;
}
}
return (int) maxSum;
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine().trim());
int[] nums = new int[n];
String[] parts = br.readLine().trim().split("\\s+");
for (int i = 0; i < n; i++) {
nums[i] = Integer.parseInt(parts[i]);
}
int k = Integer.parseInt(br.readLine().trim());
System.out.println(maxResource(nums, k));
}
}def max_resource(nums, k):
"""
Compute the maximum total resource collected over any contiguous segment of length k.
Args:
nums (list[int]): List of integers representing daily resources.
k (int): The exact number of consecutive days.
Returns:
int: The maximum total resource collected.
"""
n = len(nums)
if n == 0 or k <= 0 or k > n:
return 0
current_sum = sum(nums[:k])
max_sum = current_sum
for i in range(k, n):
current_sum += nums[i] - nums[i - k]
if current_sum > max_sum:
max_sum = current_sum
return max_sum
if __name__ == "__main__":
import sys
data = sys.stdin.read().split()
if not data:
sys.exit(0)
n = int(data[0])
nums = list(map(int, data[1:1+n]))
k = int(data[1+n])
print(max_resource(nums, k))/**
* @param {number[]} nums - Array of integers representing daily resources.
* @param {number} k - The exact number of consecutive days.
* @return {number} The maximum total resource collected over any contiguous segment of length k.
*/
function maxResource(nums, k) {
const n = nums.length;
if (n === 0 || k <= 0 || k > n) {
return 0;
}
let currentSum = 0;
for (let i = 0; i < k; i++) {
currentSum += nums[i];
}
let maxSum = currentSum;
for (let i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
if (currentSum > maxSum) {
maxSum = currentSum;
}
}
return maxSum;
}
// Driver code
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
terminal: false
});
let lines = [];
rl.on('line', line => {
lines.push(line);
});
rl.on('close', () => {
const input = lines.join(' ').split(' ').map(Number);
const n = input[0];
const nums = input.slice(1, 1 + n);
const k = input[1 + n];
console.log(maxResource(nums, k));
});Asked in Top Tech Interviews
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