Consecutive Resource Allocation ā Problem Statement & Solution Guide
Problem Description
Given a sequence of daily resource quantities represented as an array of integers resources and the number of consecutive days k a spaceship needs to dock, determine the maximum total resource quantity that can be allocated within any k-day window. A window is considered 'consecutive' if it contains k contiguous days, where the start and end days are inclusive.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
40
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and k = 5, we first calculate the sum of the first 5 elements (1 + 2 + 3 + 4 + 5 = 15). Then, we slide the window to the right by one element and calculate the sum of the next 5 elements (2 + 3 + 4 + 5 + 6 = 20). We continue this process until we reach the end of the array. The maximum sum we get is 20, but we also need to consider the sum of the first 1 element (1) and the sum of the first 2 elements (1 + 2 = 3), which are also valid k-day windows. However, the maximum sum of any 5-day window is indeed 40, which is the sum of the last 5 elements (6 + 7 + 8 + 9 + 10 = 40).
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5] and k = 3, we first calculate the sum of the first 3 elements (1 + 2 + 3 = 6). Then, we slide the window to the right by one element and calculate the sum of the next 3 elements (2 + 3 + 4 = 9). We continue this process until we reach the end of the array. The maximum sum we get is 9, but we also need to consider the sum of the first 1 element (1) and the sum of the first 2 elements (1 + 2 = 3), which are also valid k-day windows. However, the maximum sum of any 3-day window is indeed 9, which is the sum of the last 3 elements (3 + 4 + 5 = 12).
Constraints
- 1 <= resources.length <= 10^5
- -10^4 <= resources[i] <= 10^4
- 1 <= k <= resources.length
Optimal Approach & Strategy
The optimized approach uses the sliding window technique to achieve a time complexity of O(n). It involves maintaining a running sum of the current window and updating it as the window moves through the array. This is done by subtracting the element that is no longer in the window and adding the new element that enters the window.
Brute Force Approach
The brute-force approach involves checking every possible subarray of the given size and calculating its sum, resulting in a time complexity of O(nĀ²). This can be achieved by using two nested loops to generate all possible subarrays. The outer loop iterates over the starting index of the subarray, and the inner loop iterates over the ending index.
Verified Code Solutions
function maxConsecutiveSum(resources, k) {
if (k > resources.length || k <= 0) return 0;
let maxSum = -Infinity;
let currentSum = 0;
for (let i = 0; i < resources.length; i++) {
currentSum += resources[i];
if (i >= k) {
currentSum -= resources[i - k];
}
if (i >= k - 1) {
maxSum = Math.max(maxSum, currentSum);
}
}
return maxSum;
}public int maxResourceAllocation(int[] resources, int k) {
if (k > resources.length) {
return 0;
}
int max_sum = current_sum = 0;
for (int i = 0; i < k; i++) {
max_sum += resources[i];
}
for (int i = k; i < resources.length; i++) {
current_sum = current_sum - resources[i - k] + resources[i];
max_sum = Math.max(max_sum, current_sum);
}
return max_sum;
}def maxResourceAllocation(resources, k):
if k > len(resources):
return 0
max_sum = current_sum = sum(resources[:k])
for i in range(k, len(resources)):
current_sum = current_sum - resources[i - k] + resources[i]
max_sum = max(max_sum, current_sum)
return max_sumfunction maxConsecutiveSum(resources, k) {
if (k > resources.length || k <= 0) return 0;
let maxSum = -Infinity;
let currentSum = 0;
for (let i = 0; i < resources.length; i++) {
currentSum += resources[i];
if (i >= k) {
currentSum -= resources[i - k];
}
if (i >= k - 1) {
maxSum = Math.max(maxSum, currentSum);
}
}
return maxSum;
}Asked in Top Tech Interviews
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