Kth Maximum Partition Analyzer 3 — Problem Statement & Solution Guide
Problem Description
You are given an array of integers nums and an integer k. First, count how many times each distinct value appears in nums. Sort the distinct values in descending order of their frequencies; if two values have the same frequency, place the larger value first. Rebuild a new array by writing each value as many times as its frequency, following the sorted order. Next, split this rebuilt array into exactly k contiguous parts as evenly as possible: the first r = n mod k parts contain ceil(n/k) elements, and the remaining parts contain floor(n/k) elements, where n is the length of the rebuilt array. For each part, determine its maximum element. Finally, sort these k maximums in descending order and return the k‑th element of that sorted list. If k is larger than the number of parts (which cannot happen under the construction), return -1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Kth Maximum Partition Analyzer 3"
WHY DOES IT MATTER?
This pattern tests the ability to decouple frequency counting from ordering logic. It is essential for problems involving 'top K' elements with complex ranking criteria, which are common in recommendation systems, log analysis, and inventory management.
OPTIMIZATION CHALLENGE
The key insight is to operate on distinct values ($m$) rather than the entire array ($n$). By using a Hash Map to count frequencies first, we reduce the problem size from $n$ to $m$. Then, using a Heap allows us to avoid a full sort if we only need the top $k$ partitions or elements.
REAL-WORLD CONNECTION
Imagine a CDN (Content Delivery Network) ranking cache hits. You want to prioritize caching content that is hit most frequently. If two pieces of content have the same hit rate, you might prioritize the larger file to save more bandwidth. This problem models that exact ranking logic.
Always clarify the definition of 'partition' and 'k-th maximum'. Does it mean the k-th element in the reconstructed array, or the k-th distinct value? In interviews, ask for clarification on the output format before coding. If it's the k-th element, you might need to simulate the reconstruction partially.
COMPLEXITY AT A GLANCE
O(n + m log m)O(m)Core Theory — Why This Approach?
The problem fundamentally revolves around frequency analysis and stable sorting with custom comparators. The naive approach involves counting frequencies, sorting the distinct keys based on a composite key (frequency descending, value descending), and then reconstructing the array. While sorting $m$ distinct elements takes $O(m \log m)$, the reconstruction phase naively requires $O(n)$ time to write out the elements. However, the core challenge often lies in efficiently managing the 'k-th maximum partition' aspect, which implies identifying a specific segment or element in the reconstructed order. If the problem implies finding the k-th element in this specific sorted-by-frequency order, a Max-Heap (Priority Queue) is the optimal paradigm. By pushing all distinct values into a Max-Heap keyed by (frequency, value), we can extract the top elements in $O(m \log m)$ time. This is superior to sorting when we only need the top $k$ elements, reducing the complexity to $O(m + k \log m)$ if $k \ll m$.
Interview Questions on This Problem
Q1How would you optimize this solution if the array `nums` is extremely large (e.g., 10^8 elements) but the number of distinct values `m` is small (e.g., 10^3)?
Since $m$ is small, the bottleneck is not the heap operations but the initial frequency counting. We should use a hash map for $O(n)$ counting. Then, instead of a full sort or heap, we can simply sort the $m$ distinct values, which is $O(m \log m)$. The reconstruction is $O(n)$. The key insight is that when $m \ll n$, sorting the distinct keys is often faster in practice than heap operations due to cache locality and lower constant factors, but theoretically, both are acceptable. The critical part is avoiding $O(n \log n)$ sorting of the entire array.
Q2In a distributed system, if `nums` is split across multiple nodes, how would you compute the global frequency distribution efficiently?
Use a MapReduce-style approach. Each node computes local frequencies and sends them to a reducer. The reducer aggregates the counts. Then, the reducer performs the sorting/heap operation on the aggregated distinct values. This reduces network overhead by sending only distinct keys and counts rather than raw data. The complexity becomes $O(n)$ for local counting and $O(m \log m)$ for global aggregation, where $m$ is the global distinct count.
Q3What if the tie-breaking rule changes to 'smaller value first' for same frequencies? How does this affect the heap comparator?
The comparator in the Max-Heap must be adjusted. Instead of comparing value in descending order, it should compare value in ascending order. In code, this means if freq1 == freq2, return value1 < value2 for a max-heap that prioritizes larger values, or adjust the sign logic accordingly. This is a common pitfall where candidates forget to update the secondary sort key logic when the problem constraints change.
Examples
Input
{"nums":[1,2,2,3,3,3],"k":2}Output
2
Explanation: Frequencies: 3→3, 2→2, 1→1. Sorted order: 3, 2, 1. Rebuilt array: [3,3,3,2,2,1]. Split into 2 parts: [3,3,3] and [2,2,1]. Maxima: 3 and 2. Sorted maxima descending: [3,2]. The 2nd largest is 2.
Input
{"nums":[4,4,4,4,4],"k":1}Output
4
Explanation: Only value 4 with frequency 5. Rebuilt array: [4,4,4,4,4]. One part containing all elements. Its maximum is 4, which is also the 1st largest.
Input
{"nums":[5,1,5,2,1,3,5,2],"k":3}Output
2
Explanation: Frequencies: 5→3, 2→2, 1→2, 3→1. Sorted by frequency then value: 5, 2, 1, 3. Rebuilt array: [5,5,5,2,2,1,1,3]. n=8, k=3 → r=2, first two parts size 3, last part size 2. Parts: [5,5,5], [2,2,1], [1,3]. Maxima: 5, 2, 3. Sorted descending: [5,3,2]. The 3rd largest is 2.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- 1 <= k <= 100000
- The sum of frequencies equals nums.length
- All operations must run in O(n log n) time or better
Optimal Approach & Strategy
Count frequencies in O(n) time using a hash map. Sort the distinct keys in O(m log m) time. Iterate through the sorted keys to locate the k-th element in the reconstructed array in O(m) time. Total time is O(n + m log m).
Brute Force Approach
Sort the entire array nums based on a custom comparator that compares frequencies (requiring a pre-pass to count) and then values. This results in O(n log n) time complexity, which is inefficient for large n.
Verified Code Solutions
function solution(nums, k) {
let frequency = {};
for (let num of nums) {
frequency[num] = (frequency[num] || 0) + 1;
}
let sortedFrequency = Object.keys(frequency).sort((a, b) => frequency[b] - frequency[a]);
return sortedFrequency[k - 1];
}class Solution {
public:
int solution(vector<int>& nums, int k) {
unordered_map<int, int> frequency;
for (int num : nums) {
frequency[num]++;
}
vector<int> sortedFrequency(frequency.begin(), frequency.end());
sort(sortedFrequency.begin(), sortedFrequency.end(), [](const auto& a, const auto& b) {
return frequency[b.first] > frequency[a.first];
});
return sortedFrequency[k - 1].first;
}
};class Solution {
public int solution(int[] nums, int k) {
Map<Integer, Integer> frequency = new HashMap<>();
for (int num : nums) {
frequency.put(num, frequency.getOrDefault(num, 0) + 1);
}
List<Integer> sortedFrequency = new ArrayList<>(frequency.keySet());
Collections.sort(sortedFrequency, (a, b) -> frequency.get(b) - frequency.get(a));
return sortedFrequency.get(k - 1);
}
}def solution(nums, k):
frequency = {}
for num in nums:
frequency[num] = frequency.get(num, 0) + 1
sorted_frequency = sorted(frequency, key=frequency.get, reverse=True)
return sorted_frequency[k - 1]function solution(nums, k) {
let frequency = {};
for (let num of nums) {
frequency[num] = (frequency[num] || 0) + 1;
}
let sortedFrequency = Object.keys(frequency).sort((a, b) => frequency[b] - frequency[a]);
return sortedFrequency[k - 1];
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.