Protocol Tome Consolidator 12 — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums and a positive integer K. Your task is to determine the greatest possible sum of any contiguous sub‑array that contains exactly K elements. The input consists of the size of the array, the value K, and the sequence of integers. Output the maximum sum achievable. The solution must run efficiently for large inputs.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Consolidator 12"
WHY DOES IT MATTER?
Fixed‑size sliding windows are a cornerstone for many linear‑time array problems.
OPTIMIZATION CHALLENGE
Transforming an O(N·K) brute force into O(N) hinges on reusing overlapping computation.
REAL-WORLD CONNECTION
They model real‑time data streams like monitoring the last K sensor readings for anomaly detection.
Initialize the first window once, then reuse its sum; avoid recomputing from scratch inside the loop.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The naive solution enumerates every possible sub‑array of length K, summing each in O(K) time, which leads to O(N·K) overall and quickly becomes infeasible for N up to 10^5 or more. This approach also repeats work because adjacent windows share K‑1 elements, causing unnecessary recomputation.
The optimal paradigm is the sliding‑window technique: compute the sum of the first K elements once, then slide the window one position at a time, subtracting the element exiting the window and adding the new entrant. This yields a linear O(N) traversal with O(1) auxiliary space, guaranteeing scalability for large inputs while preserving exact‑K length constraints.
Interview Questions on This Problem
Q1How does the sliding‑window method avoid recomputing sums for overlapping sub‑arrays?
It updates the current sum by subtracting the element that leaves the window and adding the new element that enters. This constant‑time update eliminates the O(K) recomputation for each shift.
Q2What edge case must you handle when K is larger than the array length?
If K > N the problem is undefined; a robust solution should return an error value or zero. Validating this condition up front prevents out‑of‑bounds access.
Q3Can the algorithm handle negative numbers, and does it affect the max‑sum logic?
Yes, the window sum may become negative, but the algorithm still tracks the maximum encountered sum. No special case is needed beyond initializing the max with the first window sum.
Examples
Input
5 3 1 2 -1 4 5
Output
8
Explanation: All sub‑arrays of length 3 are: [1,2,-1] (sum=2), [2,-1,4] (sum=5), [-1,4,5] (sum=8). The largest sum among them is 8.
Input
6 2 -3 -2 -5 -1 -4 -6
Output
-5
Explanation: The sums of every 2‑element window are: -3+-2=-5, -2+-5=-7, -5+-1=-6, -1+-4=-5, -4+-6=-10. The maximum (least negative) sum is -5.
Input
8 4 10 -2 3 5 -1 6 2 4
Output
16
Explanation: Sliding a window of size 4 yields sums: 10-2+3+5=16, -2+3+5-1=5, 3+5-1+6=13, 5-1+6+2=12, -1+6+2+4=11. The highest sum is 16.
Constraints
- 1 <= nums.length <= 100000
- 1 <= K <= nums.length
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
Use a sliding window: compute the first K‑element sum, then slide, updating the sum in O(1) per step while tracking the max.
Brute Force Approach
Iterate over every possible start index, sum K elements for each window, and keep the maximum.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => b - a);
let max_sum = -Infinity;
for (let i = 0; i <= nums.length - K; i++) {
let current_sum = 0;
for (let j = i; j < i + K; j++) {
current_sum += nums[j];
}
max_sum = Math.max(max_sum, current_sum);
}
return max_sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.rbegin(), nums.rend());
int max_sum = INT_MIN;
for (int i = 0; i <= nums.size() - K; i++) {
int current_sum = 0;
for (int j = i; j < i + K; j++) {
current_sum += nums[j];
}
max_sum = max(max_sum, current_sum);
}
return max_sum;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int max_sum = Integer.MIN_VALUE;
for (int i = 0; i <= nums.length - K; i++) {
int current_sum = 0;
for (int j = i; j < i + K; j++) {
current_sum += nums[j];
}
max_sum = Math.max(max_sum, current_sum);
}
return max_sum;
}
}def solution(nums, K):
nums.sort(reverse=True)
max_sum = float('-inf')
for i in range(len(nums) - K + 1):
current_sum = sum(nums[i:i + K])
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums, K) {
nums.sort((a, b) => b - a);
let max_sum = -Infinity;
for (let i = 0; i <= nums.length - K; i++) {
let current_sum = 0;
for (let j = i; j < i + K; j++) {
current_sum += nums[j];
}
max_sum = Math.max(max_sum, current_sum);
}
return max_sum;
}Asked in Top Tech Interviews
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