Consecutive Resource Allocation — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Consecutive Resource Allocation problem optimally.
O(n)O(1)Problem Description
Given an integer array quantities and an integer windowSize, determine the greatest possible sum of any contiguous sub‑array whose length is exactly windowSize. The function should return this maximum sum. The input consists of the array length n, the n elements of quantities, and the value windowSize. The output is a single integer representing the maximum sum among all valid windows.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Consecutive Resource Allocation"
WHY DOES IT MATTER?
Fixed‑size sliding windows appear in performance monitoring, financial time‑series analysis, and real‑time signal processing where you constantly need the latest aggregate over a recent interval. Mastering this pattern lets you turn quadratic‑time naïve loops into linear‑time solutions, a frequent interview expectation.
OPTIMIZATION CHALLENGE
The key insight is recognizing the overlap between consecutive windows—only one element changes—so you can update the window sum in O(1) instead of recomputing from scratch.
REAL-WORLD CONNECTION
Think of a moving average sensor on a production line: as each new item passes, you drop the oldest measurement and add the newest to keep the average current, without re‑scanning the entire recent history each time.
During the interview, compute the sum of the first window explicitly, then maintain a single variable ‘currentSum’. When you slide, do currentSum += arr[i] - arr[i‑windowSize]; update max if needed. This one‑liner slide shows you understand the pattern.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem is a classic fixed‑size sliding window scenario. A naïve solution would recompute the sum of each window from scratch, leading to O(n·windowSize) time, which quickly becomes prohibitive when n and windowSize are large (e.g., n≈10^6). The optimal paradigm leverages the fact that consecutive windows overlap by all but one element: when the window slides one position to the right, we subtract the element exiting the window and add the new element entering it. This constant‑time update yields a linear‑time algorithm. The sliding‑window technique is a special case of the more general two‑pointer method, where one pointer marks the start of the current segment and the other marks its end, allowing us to maintain aggregate information (here, the sum) efficiently as the segment moves across the array.
Interview Questions on This Problem
Q1How would you modify the solution if the window size could vary for each query, and you need to answer multiple queries efficiently?
Pre‑compute prefix sums in O(n) time; each query for a window [l,r] is answered in O(1) as prefix[r+1]‑prefix[l]. This transforms variable‑size window queries into constant‑time lookups.
Q2What is the time‑space trade‑off when using a deque to solve the maximum‑sum sub‑array of size k versus the simple sliding‑window sum?
Both approaches run in O(n) time, but the deque method (often used for maximum/minimum in a window) uses O(k) extra space, whereas the simple sum method only needs O(1) extra space because it stores a single running total.
Q3In a distributed system where each node holds a chunk of the array, how can you compute the global maximum window sum with minimal communication?
Each node computes the maximum sum of windows fully contained within its chunk and also the best prefix and suffix sums of length k‑1. Nodes then exchange these border aggregates; a final reduction combines them to evaluate windows that cross chunk boundaries, achieving O(n) local work and O(number_of_nodes) communication.
Examples
Input
[4,2,1,7,3,5], windowSize = 3
Output
15
Explanation: All length‑3 windows are: [4,2,1] → sum 7, [2,1,7] → sum 10, [1,7,3] → sum 11, [7,3,5] → sum 15. The largest sum is 15.
Input
[-5,-2,-3,-4], windowSize = 2
Output
-5
Explanation: Length‑2 windows: [-5,-2] → -7, [-2,-3] → -5, [-3,-4] → -7. The maximum among them is -5.
Input
[10,-2,3,5,-1,2], windowSize = 4
Output
16
Explanation: Length‑4 windows: [10,-2,3,5] → 16, [-2,3,5,-1] → 5, [3,5,-1,2] → 9. The highest sum is 16.
Constraints
- 1 <= quantities.length <= 100000
- 1 <= windowSize <= quantities.length
- -10^9 <= quantities[i] <= 10^9
- Result fits in a 64‑bit signed integer
Optimal Approach & Strategy
Use a sliding window: keep a running sum, update it in O(1) when the window moves, and track the maximum.
Brute Force Approach
Compute the sum of every possible window of size k by iterating over all start indices and summing k elements each time.
Code Solutions
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = data[idx++];
const quantities = data.slice(idx, idx+n); idx+=n;
const windowSize = data[idx++] || 0;
function maxWindowSum(arr, k) {
if (k <= 0 || k > arr.length) return 0;
let cur = 0, best = -Infinity;
for (let i = 0; i < arr.length; ++i) {
cur += arr[i];
if (i >= k) cur -= arr[i - k];
if (i >= k - 1) best = Math.max(best, cur);
}
return best;
}
console.log(maxWindowSum(quantities, windowSize).toString());#include <bits/stdc++.h>
using namespace std;
long long maxWindowSum(const vector<int>& quantities, int windowSize) {
int n = quantities.size();
if (windowSize <= 0 || windowSize > n) return 0LL; // undefined, return 0
long long cur = 0, best = LLONG_MIN;
for (int i = 0; i < n; ++i) {
cur += quantities[i];
if (i >= windowSize) cur -= quantities[i - windowSize];
if (i >= windowSize - 1) best = max(best, cur);
}
return best;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
vector<int> quantities(n);
for(int i=0;i<n;++i) cin>>quantities[i];
int windowSize; cin>>windowSize;
cout<<maxWindowSum(quantities,windowSize)<<"\n";
return 0;
}
import java.io.*;
import java.util.*;
public class Main {
public static long maxWindowSum(int[] quantities, int windowSize) {
int n = quantities.length;
if (windowSize <= 0 || windowSize > n) return 0L;
long cur = 0, best = Long.MIN_VALUE;
for (int i = 0; i < n; i++) {
cur += quantities[i];
if (i >= windowSize) cur -= quantities[i - windowSize];
if (i >= windowSize - 1) best = Math.max(best, cur);
}
return best;
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int[] quantities = new int[n];
st = new StringTokenizer(br.readLine());
for (int i = 0; i < n; i++) {
if (!st.hasMoreTokens()) {
st = new StringTokenizer(br.readLine());
}
quantities[i] = Integer.parseInt(st.nextToken());
}
int windowSize = Integer.parseInt(br.readLine().trim());
System.out.println(maxWindowSum(quantities, windowSize));
}
}
import sys
def max_window_sum(arr, k):
n = len(arr)
if k <= 0 or k > n:
return 0
cur = sum(arr[:k])
best = cur
for i in range(k, n):
cur += arr[i] - arr[i - k]
if cur > best:
best = cur
return best
if __name__ == "__main__":
data = list(map(int, sys.stdin.read().strip().split()))
if not data:
sys.exit(0)
n = data[0]
quantities = data[1:1+n]
windowSize = data[1+n] if len(data) > 1+n else 0
print(max_window_sum(quantities, windowSize))
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = data[idx++];
const quantities = data.slice(idx, idx+n); idx+=n;
const windowSize = data[idx++] || 0;
function maxWindowSum(arr, k) {
if (k <= 0 || k > arr.length) return 0;
let cur = 0, best = -Infinity;
for (let i = 0; i < arr.length; ++i) {
cur += arr[i];
if (i >= k) cur -= arr[i - k];
if (i >= k - 1) best = Math.max(best, cur);
}
return best;
}
console.log(maxWindowSum(quantities, windowSize).toString());
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.