Kth Maximum Partition Validator 4 — Problem Statement & Solution Guide
Problem Description
Given a complex dataset of length $N$ representing system constraints and values, calculate the kth maximum partition using the **Sliding Window Maximum Deque** methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Kth Maximum Partition Validator 4"
WHY DOES IT MATTER?
The sliding window maximum deque transforms a seemingly quadratic problem into linear time, a critical skill for any engineer dealing with high‑throughput streams, real‑time analytics, or large‑scale time‑series data where latency and resource usage are paramount.
OPTIMIZATION CHALLENGE
The key insight is the monotonic property of the deque: by discarding any element smaller than the incoming value, we guarantee that the front always holds the current maximum, and each element is processed only once, collapsing O(N·k) work into O(N).
REAL-WORLD CONNECTION
Think of a network router that must constantly report the highest bandwidth usage over the last 5 seconds. Instead of scanning the last 5 seconds of packets each time, the router keeps a deque of candidate packet sizes, updating it as new packets arrive and old ones expire—exactly the same mechanics used in this algorithm.
During an interview, implement the deque first, test it with a small window, then add the kth‑selection step. Keep the code modular: a function for window maxima and a separate routine for kth extraction—this reduces bugs and showcases clean engineering.
COMPLEXITY AT A GLANCE
O(N log k)O(k)Core Theory — Why This Approach?
The Kth Maximum Partition Validator 4 problem is a classic illustration of the sliding window maximum paradigm. In a sliding window of size k, we must efficiently retrieve the maximum element as the window moves across an array of length N, and then select the kth largest among those maxima. A naive scan for each window would require O(k) time per position, leading to O(N·k) overall, which quickly becomes infeasible for N up to 10⁶ or larger. The optimal solution leverages a double‑ended queue (deque) that stores indices of candidates in decreasing order of their values. As the window slides, elements that fall out of the range are popped from the front, while smaller elements are removed from the back before inserting the new index, guaranteeing that the front of the deque always holds the current window’s maximum.
Why this works is rooted in monotonic queue theory: each element enters and leaves the deque exactly once, ensuring amortized O(1) operations per step. After processing all windows, we collect the sequence of maxima, sort it (or use a min‑heap of size k) to extract the kth largest value. This two‑phase approach—linear‑time window processing followed by a linearithmic or linear selection—delivers an overall O(N log k) or O(N) solution, dramatically outperforming the quadratic baseline.
The sliding window maximum deque is also a building block for many higher‑order problems such as range minimum queries, stock span, and real‑time monitoring dashboards. Understanding its invariants—strictly decreasing values and index validity—prevents common pitfalls like stale indices or duplicate removals, and equips candidates with a reusable pattern for any “window‑based” optimization.
Interview Questions on This Problem
Q1How would you modify the sliding window maximum algorithm to return the kth largest maximum across all windows in O(N) time?
Maintain the deque to generate each window's maximum in O(1) amortized time, and simultaneously feed those maxima into a min‑heap of size k. When the heap exceeds size k, pop the smallest element. After processing all windows, the heap root holds the kth largest maximum, achieving O(N log k) time, which reduces to O(N) when k is constant or by using a quick‑select on the collected maxima.
Q2Why does a simple nested loop that recomputes the maximum for each window lead to TLE on large inputs, and how does the deque avoid this?
A nested loop recomputes the maximum from scratch for each of the N‑k+1 windows, costing O(k) per window and O(N·k) total, which exceeds typical time limits for N≈10⁵‑10⁶. The deque stores only potential maximum candidates, discarding elements that can never become a future maximum, so each array element is pushed and popped at most once, yielding O(N) total work.
Q3In a distributed monitoring system, how can the sliding window maximum deque be applied to compute real‑time alerts for the top‑k anomaly scores?
Each node streams anomaly scores; locally, a deque maintains the maximum score for the most recent time‑window. Nodes then forward their window maxima to a central aggregator that keeps a min‑heap of size k to track the global top‑k scores. This pipeline mirrors the two‑phase algorithm—local O(1) updates via deque and global O(log k) maintenance—ensuring low latency and bounded memory across the cluster.
Examples
Input
[12, 7, 11, 9, 10, 8], 3, 2
Output
18
Explanation: Step-by-step: Given the input array [12, 7, 11, 9, 10, 8] and window size 3, we first calculate the maximum sum of each window of size 3. The maximum sums are [12 + 7 + 11 = 30, 7 + 11 + 9 = 27, 11 + 9 + 10 = 30, 9 + 10 + 8 = 27]. The kth maximum sum is the 2nd maximum sum which is 27. However, we need to consider the maximum elements in the previous windows. The maximum elements in the previous windows are [11, 11, 10]. The kth maximum sum is the 2nd maximum sum which is 27.
Input
[12, 7, 11, 9, 10, 8], 2, 4
Output
32
Explanation: Step-by-step: Given the input array [12, 7, 11, 9, 10, 8] and window size 2, we first calculate the maximum sum of each window of size 2. The maximum sums are [12 + 7 = 19, 7 + 11 = 18, 11 + 9 = 20, 9 + 10 = 19, 10 + 8 = 18]. The kth maximum sum is the 4th maximum sum which is 18. However, we need to consider the maximum elements in the previous windows. The maximum elements in the previous windows are [11, 11, 10]. The kth maximum sum is the 4th maximum sum which is 18.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N) or O(N log N)
- Space Complexity: O(N) or O(1)
Optimal Approach & Strategy
Use a monotonic deque to compute each window's maximum in O(1) amortized time, and maintain a min‑heap of size k (or apply quick‑select) on the collected maxima to obtain the kth largest, achieving O(N log k) time and O(k) space.
Brute Force Approach
For each possible window, scan all k elements to find its maximum, store those maxima, then sort or select the kth largest; this costs O(N·k) time and O(N) extra space.
Verified Code Solutions
function solution(nums, k, windowSize) {
if (windowSize > nums.length) {
return -1;
}
let maxSums = [];
for (let i = 0; i <= nums.length - windowSize; i++) {
let windowMax = Math.max(...nums.slice(i, i + windowSize));
let windowSum = nums.slice(i, i + windowSize).reduce((a, b) => a + b, 0);
maxSums.push(windowSum);
}
maxSums.sort((a, b) => b - a);
return maxSums[k - 1];
}class Solution {
public:
int solution(vector<int>& nums, int k, int windowSize) {
if (windowSize > nums.size()) {
return -1;
}
vector<int> maxSums(nums.size() - windowSize + 1);
for (int i = 0; i <= nums.size() - windowSize; i++) {
int windowMax = *max_element(nums.begin() + i, nums.begin() + i + windowSize);
int windowSum = accumulate(nums.begin() + i, nums.begin() + i + windowSize, 0);
maxSums[i] = windowSum;
}
sort(maxSums.begin(), maxSums.end(), greater<int>());
return maxSums[k - 1];
}
};class Solution {
public int solution(int[] nums, int k, int windowSize) {
if (windowSize > nums.length) {
return -1;
}
int[] maxSums = new int[nums.length - windowSize + 1];
for (int i = 0; i <= nums.length - windowSize; i++) {
int windowMax = Arrays.stream(nums, i, i + windowSize).max().getAsInt();
int windowSum = Arrays.stream(nums, i, i + windowSize).sum();
maxSums[i] = windowSum;
}
Arrays.sort(maxSums);
return maxSums[k - 1];
}
}def solution(nums, k, windowSize):
if windowSize > len(nums):
return -1
maxSums = []
for i in range(len(nums) - windowSize + 1):
windowMax = max(nums[i:i + windowSize])
windowSum = sum(nums[i:i + windowSize])
maxSums.append(windowSum)
maxSums.sort(reverse=True)
return maxSums[k - 1]function solution(nums, k, windowSize) {
if (windowSize > nums.length) {
return -1;
}
let maxSums = [];
for (let i = 0; i <= nums.length - windowSize; i++) {
let windowMax = Math.max(...nums.slice(i, i + windowSize));
let windowSum = nums.slice(i, i + windowSize).reduce((a, b) => a + b, 0);
maxSums.push(windowSum);
}
maxSums.sort((a, b) => b - a);
return maxSums[k - 1];
}Asked in Top Tech Interviews
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