Kth Maximum Partition Analyzer 7 — Problem Statement & Solution Guide

Sliding WindowHardSliding Window Maximum Deque

TimeO(N·log k)

SpaceO(W + k)

Problem Description

You are tasked with analyzing a sequence of $N$ integer values representing system load metrics. The goal is to determine the $k$-th largest value among all sliding window maximums of a fixed size $W$. Specifically, for every contiguous subarray of length $W$, compute its maximum element. Collect all such maximums into a multiset. Your objective is to find the $k$-th largest element in this multiset of window maximums.

Given an array nums of length $N$, a window size $W$, and an integer $k$, return the $k$-th largest value from the set of maximums of all sliding windows of size $W$. If $k$ exceeds the number of valid windows, return -1.

The number of valid windows is $N - W + 1$. You must efficiently compute the maximum for each window using a monotonic deque approach to ensure optimal performance for large inputs.

DSA Pattern Breakdown

"Kth Maximum Partition Analyzer 7"

hard

WHY DOES IT MATTER?

Sliding‑window techniques appear in time‑series analysis, real‑time monitoring, and streaming analytics where you need instant aggregates over recent data. Mastering the deque‑based maximum enables you to handle massive streams with strict latency constraints.

OPTIMIZATION CHALLENGE

The key insight is that elements smaller than a newly arrived value can never become a future window maximum, so they can be discarded immediately. This monotonic property lets the deque stay ordered and guarantees each index is pushed and popped at most once, yielding linear time.

REAL-WORLD CONNECTION

Think of a network router that keeps track of the highest packet latency over the last W seconds to trigger alerts. The router cannot recompute the max from scratch each second; instead, it updates a compact structure (deque) as new packets arrive and old ones expire.

During an interview, write the deque logic first and test it on a tiny example before adding the heap for k‑selection. Keeping the two phases separate reduces mental load and helps you catch off‑by‑one window boundary bugs early.

COMPLEXITY AT A GLANCE

⏱ Time:O(N·log k)

💾 Space:O(W + k)

Core Theory — Why This Approach?

The sliding‑window maximum problem asks for the maximum element of every contiguous subarray of length W in an array of size N. A naïve solution recomputes the maximum for each window in O(W) time, leading to O(N·W) overall, which is prohibitive when N and W approach 10^5 or higher. The optimal paradigm leverages a double‑ended queue (deque) that stores indices of candidates in decreasing order of their values, allowing insertion and removal of elements that fall out of the current window in amortized O(1) time. By scanning the array once, we can emit the maximum of each window in O(N) total time, producing a list of N‑W+1 maximums.

Once the list of window maximums is obtained, the task reduces to selecting the k‑th largest element from this multiset. Direct sorting would cost O((N‑W+1)·log(N‑W+1)), which is acceptable but can be further improved using a min‑heap of size k: we maintain the k largest values seen so far, discarding smaller ones, achieving O(N·log k) time. Combining the O(N) sliding‑window pass with the O(N·log k) selection yields an overall O(N·log k) solution, which comfortably meets hard‑level constraints where N can be up to 10^6 and k may be much smaller than N.

Interview Questions on This Problem

Q1How would you compute the maximum of every sliding window of size W in O(N) time?

Use a deque to store indices of elements in decreasing order. For each new element, pop indices from the back while their values are smaller, then push the new index. Remove the front index if it is outside the current window. The front of the deque always holds the index of the current window's maximum.

Q2Given the list of window maximums, how can you find the k‑th largest value without sorting the entire list?

Maintain a min‑heap (priority queue) of size k. Iterate over the maximums; push each value into the heap. If the heap size exceeds k, pop the smallest element. After processing all values, the heap root is the k‑th largest.

Q3Why might a naïve O(N·W) solution time out on a platform like LeetCode or Codeforces for N = 10^6 and W = 5·10^5?

Because O(N·W) becomes on the order of 10^11 operations, far exceeding typical time limits (1–2 seconds). The constant factors of simple loops are dwarfed by the sheer number of iterations, making the solution infeasible regardless of language optimizations.

Examples

Example 1

Input

nums = [1, 3, 2, 5, 4], W = 3, k = 2

Output

Explanation: Windows of size 3: [1,3,2] -> max=3; [3,2,5] -> max=5; [2,5,4] -> max=5. The multiset of maximums is {3, 5, 5}. Sorted in descending order: [5, 5, 3]. The 2nd largest value is 5.

Example 2

Input

nums = [10, 20, 30, 40, 50], W = 2, k = 1

Output

Explanation: Windows of size 2: [10,20] -> max=20; [20,30] -> max=30; [30,40] -> max=40; [40,50] -> max=50. The multiset of maximums is {20, 30, 40, 50}. Sorted in descending order: [50, 40, 30, 20]. The 1st largest value is 50.

Example 3

Input

nums = [5, 5, 5, 5], W = 4, k = 1

Output

Explanation: Only one window of size 4: [5,5,5,5] -> max=5. The multiset of maximums is {5}. The 1st largest value is 5.

Example 4

Input

nums = [1, 2, 3, 4, 5, 6], W = 3, k = 5

Output

-1

Explanation: Number of windows = 6 - 3 + 1 = 4. Since k=5 exceeds the number of windows (4), return -1.

Constraints

1 <= nums.length <= 10^5
1 <= W <= nums.length
1 <= k <= 10^5
-10^9 <= nums[i] <= 10^9

Optimal Approach & Strategy

Use a deque to generate all window maximums in O(N) and a size‑k min‑heap to extract the k‑th largest in O(N·log k).

Brute Force Approach

Compute the maximum for each window by scanning all W elements, store them, then sort the list and pick the k‑th largest.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N·log k)

/**
 * @param {number[]} nums
 * @param {number} W
 * @param {number} k
 * @return {number}
 */
var kthLargestWindowMax = function(nums, W, k) {
    const n = nums.length;
    const windowMaxs = [];
    const dq = [];
    
    for (let i = 0; i < n; i++) {
        while (dq.length > 0 && nums[dq[dq.length - 1]] <= nums[i]) {
            dq.pop();
        }
        dq.push(i);
        
        if (dq[0] <= i - W) {
            dq.shift();
        }
        
        if (i >= W - 1) {
            windowMaxs.push(nums[dq[0]]);
        }
    }
    
    windowMaxs.sort((a, b) => b - a);
    return windowMaxs[k - 1];
};

class Solution {
public:
    int kthLargestWindowMax(vector<int>& nums, int W, int k) {
        int n = nums.size();
        vector<int> windowMaxs;
        deque<int> dq;
        
        for (int i = 0; i < n; ++i) {
            while (!dq.empty() && nums[dq.back()] <= nums[i]) {
                dq.pop_back();
            }
            dq.push_back(i);
            
            if (dq.front() <= i - W) {
                dq.pop_front();
            }
            
            if (i >= W - 1) {
                windowMaxs.push_back(nums[dq.front()]);
            }
        }
        
        nth_element(windowMaxs.begin(), windowMaxs.begin() + k - 1, windowMaxs.end(), greater<int>());
        return windowMaxs[k - 1];
    }
};

class Solution {
    public int kthLargestWindowMax(int[] nums, int W, int k) {
        int n = nums.length;
        int[] windowMaxs = new int[n - W + 1];
        int[] dq = new int[n];
        int head = 0, tail = 0;
        int idx = 0;
        
        for (int i = 0; i < n; i++) {
            while (head < tail && nums[dq[tail - 1]] <= nums[i]) {
                tail--;
            }
            dq[tail++] = i;
            
            if (dq[head] <= i - W) {
                head++;
            }
            
            if (i >= W - 1) {
                windowMaxs[idx++] = nums[dq[head]];
            }
        }
        
        Arrays.sort(windowMaxs);
        return windowMaxs[idx - k];
    }
}

class Solution:
    def kthLargestWindowMax(self, nums: List[int], W: int, k: int) -> int:
        n = len(nums)
        window_maxs = []
        dq = deque()
        
        for i in range(n):
            while dq and nums[dq[-1]] <= nums[i]:
                dq.pop()
            dq.append(i)
            
            if dq[0] <= i - W:
                dq.popleft()
            
            if i >= W - 1:
                window_maxs.append(nums[dq[0]])
        
        window_maxs.sort(reverse=True)
        return window_maxs[k - 1]

/**
 * @param {number[]} nums
 * @param {number} W
 * @param {number} k
 * @return {number}
 */
var kthLargestWindowMax = function(nums, W, k) {
    const n = nums.length;
    const windowMaxs = [];
    const dq = [];
    
    for (let i = 0; i < n; i++) {
        while (dq.length > 0 && nums[dq[dq.length - 1]] <= nums[i]) {
            dq.pop();
        }
        dq.push(i);
        
        if (dq[0] <= i - W) {
            dq.shift();
        }
        
        if (i >= W - 1) {
            windowMaxs.push(nums[dq[0]]);
        }
    }
    
    windowMaxs.sort((a, b) => b - a);
    return windowMaxs[k - 1];
};

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Kth Maximum Partition Analyzer 7 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Sliding Window

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Kth Maximum Partition Analyzer 7 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor