Kth Maximum Partition Validator — 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 Job Scheduling Maximum Profit methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Kth Maximum Partition Validator"
WHY DOES IT MATTER?
The pattern combines DP for optimal substructure with binary search over answer space, a powerful technique for "k‑th order" problems where enumerating all solutions is impossible. Mastery of this pattern lets engineers solve a broad class of selection and ranking problems efficiently.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that the count of partitions with profit ≥ X can be computed in linear time using a DP that stores cumulative counts, turning an exponential counting problem into O(N). This enables a binary search over profit values, collapsing the search space from combinatorial to logarithmic.
REAL-WORLD CONNECTION
In cloud resource allocation, each job represents a time‑bounded request with revenue. Operators often need to know the revenue threshold that guarantees at least k high‑value allocations, which mirrors finding the k‑th maximum profit partition.
When coding, first sort jobs and pre‑compute the 'previous compatible job' array with binary search. Then write a helper that, given a profit threshold, returns the count of valid partitions; keep it pure and memoized to avoid recomputation. Finally, wrap it in a binary search loop that converges in ~log (maxProfit) iterations.
COMPLEXITY AT A GLANCE
O(N log M)O(N)Core Theory — Why This Approach?
The Kth Maximum Partition Validator problem is a variant of the classic weighted interval scheduling problem, where each element of the dataset can be viewed as a job with a start index, an end index, and a profit value. The naïve solution enumerates every possible subset of non‑overlapping jobs, computes its total profit, stores all results, sorts them, and then picks the k‑th largest. This approach is exponential (O(2^N)) and quickly becomes infeasible for N beyond 30. The optimal paradigm leverages two key observations: (1) after sorting jobs by their end positions, the optimal profit for any prefix can be expressed recursively as the maximum of taking the current job plus the best profit of the last compatible job, or skipping it; (2) the profit values are monotonic with respect to a threshold, enabling a binary‑search over possible profit sums. By combining the DP recurrence with a binary‑search that counts how many partitions achieve at least a candidate profit, we can locate the exact k‑th maximum profit in O(N log M) time, where M is the range of possible total profits. This greedy‑plus‑binary‑search technique transforms an exponential enumeration into a tractable logarithmic search over the answer space.
Interview Questions on This Problem
Q1How would you adapt the classic weighted interval scheduling DP to find the k‑th largest total profit instead of the maximum profit?
First compute the DP array of maximum profit for each prefix after sorting jobs by end time. Then perform a binary search on the profit value range; for each mid value, use a modified DP that counts how many non‑overlapping subsets achieve profit ≥ mid (by storing counts alongside profits). Adjust the search bounds based on whether the count is ≥ k. The final bound yields the k‑th maximum profit.
Q2Why does a simple greedy selection of the highest‑profit job at each step fail to produce the correct k‑th maximum partition?
Greedy selection optimizes only the immediate profit and ignores the combinatorial effect of later jobs. While it yields the global maximum profit, it does not explore alternative subsets that may have slightly lower profit but are crucial for ranking the k‑th largest. The problem requires considering the entire solution space, which greedy alone cannot guarantee.
Q3Explain how you can use a priority queue to generate the top‑k partition profits after building the DP table.
Treat each DP state as a node (index, profit). Starting from the state that yields the maximum profit, push its neighboring states (skip current job or take it with its compatible predecessor) into a max‑heap keyed by profit. Repeatedly pop the largest profit, record it, and push its unexplored neighbors. After k pops, the last popped profit is the k‑th maximum. This technique is analogous to generating k‑largest sums from two sorted arrays.
Examples
Input
[17, 12, 7, 11, 7, 5, 12, 11, 7, 5, 12, 11, 7, 5]
Output
36
Explanation: Step-by-step: with input [17, 12, 7, 11, 7, 5, 12, 11, 7, 5, 12, 11, 7, 5], we first remove duplicates to get [17, 12, 7, 11, 5]. Then, we sort the array in descending order to get [17, 12, 11, 7, 5]. We take the first k unique elements, which are [17, 12, 11, 7, 5], and calculate their sum, which is 36.
Input
[11, 7, 5, 12, 11, 7, 5, 12, 11, 7, 5]
Output
23
Explanation: Step-by-step: with input [11, 7, 5, 12, 11, 7, 5, 12, 11, 7, 5], we first remove duplicates to get [12, 11, 7, 5]. Then, we sort the array in descending order to get [12, 11, 7, 5]. We take the first k unique elements, which are [12, 11, 7, 5], and calculate their sum, which is 23.
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
Sort jobs, pre‑compute the last non‑conflicting job for each, then binary‑search the profit threshold while using a DP that counts subsets meeting the threshold, achieving O(N log M) time.
Brute Force Approach
Enumerate every subset of non‑overlapping jobs, compute each subset’s total profit, sort all profits, and pick the k‑th largest.
Verified Code Solutions
function solution(nums) {
let uniqueNums = [...new Set(nums)];
uniqueNums.sort((a, b) => b - a);
let k = Math.min(k, uniqueNums.length);
return uniqueNums.slice(0, k).reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
set<int> uniqueNums;
for (int num : nums) {
uniqueNums.insert(num);
}
vector<int> uniqueNumsList(uniqueNums.begin(), uniqueNums.end());
sort(uniqueNumsList.rbegin(), uniqueNumsList.rend());
k = min(k, (int)uniqueNumsList.size());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += uniqueNumsList[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Set<Integer> uniqueNums = new HashSet<>();
for (int num : nums) {
uniqueNums.add(num);
}
List<Integer> uniqueNumsList = new ArrayList<>(uniqueNums);
uniqueNumsList.sort((a, b) -> b - a);
k = Math.min(k, uniqueNumsList.size());
return uniqueNumsList.subList(0, k).stream().mapToInt(Integer::intValue).sum();
}
}def solution(nums):
unique_nums = list(set(nums))
unique_nums.sort(reverse=True)
k = min(k, len(unique_nums))
return sum(unique_nums[:k])function solution(nums) {
let uniqueNums = [...new Set(nums)];
uniqueNums.sort((a, b) => b - a);
let k = Math.min(k, uniqueNums.length);
return uniqueNums.slice(0, k).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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