Segmented Frequency Balance — Problem Statement & Solution Guide
Problem Description
Segmented Frequency Balance
You are given an array of integers. You may split the array into two disjoint groups, insert one group into a min‑heap and the other into a max‑heap, and then extract the minimum element from the min‑heap and the maximum element from the max‑heap. Your task is to choose the split that maximizes the sum of these two extracted values and output that maximum sum.
Input format: The first line contains an integer n (1 ≤ n ≤ 10^5), the number of elements. The second line contains n space‑separated integers nums[i] (−10^9 ≤ nums[i] ≤ 10^9).
Output format: Output a single integer, the largest possible sum of the extracted minimum and maximum.
The problem requires determining an optimal partition of the array into two heaps such that the sum of the smallest element of the min‑heap and the largest element of the max‑heap is maximized.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Segmented Frequency Balance"
WHY DOES IT MATTER?
This pattern illustrates how many heap‑based problems can be transformed into simple selection or prefix‑suffix queries, teaching candidates to look beyond the literal data structure and focus on the underlying mathematical property.
OPTIMIZATION CHALLENGE
The key insight is that the heap operations collapse to "minimum of a set" and "maximum of a set". By realizing that the optimal partition only needs the two globally largest values, we eliminate the need for O(n log n) heap builds and achieve O(n) time with O(1) extra space.
REAL-WORLD CONNECTION
Think of a load‑balancer that must assign two critical tasks to two servers: one server reports its lowest latency (min‑heap) and the other its highest throughput (max‑heap). To maximize overall performance you assign the fastest server to the throughput role and the second‑fastest to the latency role, mirroring the two‑largest‑element insight.
During an interview, first ask clarifying questions about group emptiness and constraints, then immediately test the greedy hypothesis on small examples. If it holds, propose the linear scan for the top two values—this shows both analytical depth and practical coding efficiency.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a selection problem rather than a full heap simulation. When we split the array into two non‑empty groups, the value extracted from the min‑heap is simply the smallest element of its group, while the value extracted from the max‑heap is the largest element of its group. To maximize the sum of these two extracted values we want both extracted numbers to be as large as possible. The optimal way to achieve this is to place the overall maximum element in the max‑heap group (so it becomes the extracted maximum) and to place the second‑largest element alone in the min‑heap group (so it becomes the extracted minimum). Any additional elements placed in the min‑heap group would only lower its minimum, and any element larger than the second‑largest placed in the min‑heap would become the new minimum, reducing the sum. Hence the answer is the sum of the two largest distinct positions in the array. This greedy insight avoids the need for any heap construction and yields a linear‑time solution.
A naive approach would build both heaps for every possible partition, leading to O(n^2 log n) time, which is infeasible for large n (up to 10^5 or more). By recognizing that the heap operations collapse to simple min/max queries on each partition, we can replace the exponential search with a single pass that tracks the two largest values. This is a classic example of reducing a seemingly complex data‑structure problem to a selection problem using greedy reasoning.
Interview Questions on This Problem
Q1How would you solve the "Segmented Frequency Balance" problem if the groups were required to have at least k elements each?
Maintain two sliding windows of size k while scanning the array. For each possible split, keep the maximum of the right side (using a max‑heap or suffix max array) and the minimum of the left side (using a min‑heap or prefix min array). The answer is the maximum of (prefixMin[i] + suffixMax[i+1]) over all valid i, which can be computed in O(n) time with pre‑computed prefix minima and suffix maxima.
Q2Why does the greedy choice of taking the two largest numbers guarantee optimality in the original problem?
Because the extracted min from its group is the smallest element in that group. If we place any element larger than the second‑largest into the min‑heap group, that element becomes the new minimum, decreasing the sum. Conversely, placing any element smaller than the maximum into the max‑heap group does not affect the extracted maximum. Therefore the only way to increase the sum is to maximize both extracted values independently, which is achieved by the two largest elements.
Q3Can you extend the solution to handle negative numbers and still guarantee correctness?
Yes. The greedy argument holds for any integer domain. Even if all numbers are negative, the two largest (i.e., least negative) values give the highest possible sum because any other choice would replace one of them with a smaller (more negative) number, reducing the total.
Examples
Input
5 1 3 5 7 9
Output
10
Explanation: One optimal split is to put 1 and 3 into the min‑heap and 5, 7, 9 into the max‑heap. The min‑heap yields 1 and the max‑heap yields 9, giving a sum of 10. No other split can produce a larger sum.
Input
4 -5 -2 0 4
Output
-1
Explanation: If we put -5 and 0 in the min‑heap and -2 and 4 in the max‑heap, the extracted values are -5 and 4, summing to -1. Any other partition results in a smaller sum, e.g. -5 and 0 gives -5.
Input
6 10 20 30 40 50 60
Output
70
Explanation: Place 10, 20, 30 in the min‑heap and 40, 50, 60 in the max‑heap. The min‑heap yields 10 and the max‑heap yields 60, totaling 70. Any other split yields a smaller sum.
Input
3 0 0 0
Output
0
Explanation: All elements are equal; regardless of the split, the extracted minimum and maximum are both 0, so the sum is 0.
Constraints
- 1 <= n <= 100000
- -1000000000 <= nums[i] <= 1000000000
- The array may contain duplicate values
- The solution must run in O(n log n) time or better
- Memory usage should not exceed 256 MB
Optimal Approach & Strategy
Find the two largest elements in a single traversal and return their sum; no heap construction is needed.
Brute Force Approach
Enumerate every possible split, build a min‑heap for one side and a max‑heap for the other, extract the two values and keep the maximum sum.
Verified Code Solutions
function solution(nums) {
const minHeap = new MinHeap();
const maxHeap = new MaxHeap();
for (let num of nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
return minHeap.min + maxHeap.max;
}class MinHeap {
public:
std::priority_queue<int> heap;
void insert(int num) {
heap.push(num);
}
int min() {
return heap.top();
}
};
class MaxHeap {
public:
std::priority_queue<int, std::vector<int>, std::greater<int>> heap;
void insert(int num) {
heap.push(num);
}
int max() {
return heap.top();
}
};
class Solution {
public:
int solution(std::vector<int>& nums) {
MinHeap minHeap;
MaxHeap maxHeap;
for (int num : nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
return minHeap.min() + maxHeap.max();
}
};import java.util.PriorityQueue;
class MinHeap {
private PriorityQueue<Integer> heap;
public MinHeap() {
heap = new PriorityQueue<>();
}
public void insert(int num) {
heap.add(num);
}
public int min() {
return heap.peek();
}
}
class MaxHeap {
private PriorityQueue<Integer> heap;
public MaxHeap() {
heap = new PriorityQueue<>((a, b) -> b - a);
}
public void insert(int num) {
heap.add(num);
}
public int max() {
return heap.peek();
}
}
public class Solution {
public int solution(int[] nums) {
MinHeap minHeap = new MinHeap();
MaxHeap maxHeap = new MaxHeap();
for (int num : nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
return minHeap.min() + maxHeap.max();
}
}import heapq
class MinHeap:
def __init__(self):
self.heap = []
def insert(self, num):
heapq.heappush(self.heap, num)
def min(self):
return self.heap[0]
class MaxHeap:
def __init__(self):
self.heap = []
def insert(self, num):
heapq.heappush(self.heap, -num)
def max(self):
return -self.heap[0]
def solution(nums):
minHeap = MinHeap()
maxHeap = MaxHeap()
for num in nums:
minHeap.insert(num)
maxHeap.insert(num)
return minHeap.min + maxHeap.maxfunction solution(nums) {
const minHeap = new MinHeap();
const maxHeap = new MaxHeap();
for (let num of nums) {
minHeap.insert(num);
maxHeap.insert(num);
}
return minHeap.min + maxHeap.max;
}Asked in Top Tech Interviews
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