Maximized Network Stream Analyzer 7 — Problem Statement & Solution Guide
Problem Description
You are given an array throughput of integers, where each element represents the data capacity of a specific network node. Your task is to construct a valid transmission sequence using all elements from the array such that no two consecutive nodes in the sequence have the same throughput value. If such a sequence exists, return the maximum possible sum of the sequence (which is simply the sum of all elements, as all must be used). However, if it is impossible to arrange the nodes to satisfy the non-adjacency constraint, return -1.
The core challenge lies in determining whether a valid permutation exists. A valid arrangement is possible if and only if the frequency of the most common value does not exceed the sum of the frequencies of all other values plus one. If this condition is met, the total effective stream is the sum of all throughput values. If the condition is violated, no valid sequence can be formed.
Input: An array throughput of integers.
Output: Return the sum of all elements in throughput if a valid non-adjacent sequence can be constructed; otherwise, return -1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Maximized Network Stream Analyzer 7"
WHY DOES IT MATTER?
The "most frequent first" pattern guarantees that we never run out of a different value to place next, which is essential for constructing a valid sequence. Without this strategy, we could get stuck with a single value left that would violate the adjacency rule.
OPTIMIZATION CHALLENGE
The key insight is that feasibility depends solely on the maximum frequency. By checking this condition first, we avoid unnecessary rearrangement work and can immediately return the sum if possible.
REAL-WORLD CONNECTION
In load balancing, you might need to schedule jobs on servers such that no server handles two consecutive heavy jobs. The same heap‑based approach ensures that the most demanding jobs are spread out, preventing overload.
When explaining this to an interviewer, emphasize the feasibility check first, then describe the heap construction as a greedy pairing of the two most frequent values. Highlight that the sum is trivial once feasibility is confirmed.
COMPLEXITY AT A GLANCE
O(n log k)O(k)Core Theory — Why This Approach?
The problem reduces to determining whether a permutation of the array exists such that no two adjacent elements are equal. A naive approach would generate all permutations and check each one, which is factorial time and infeasible for large n. The optimal solution uses frequency counting and a max‑heap (priority queue). By repeatedly extracting the two most frequent values and placing them next to each other, we can always avoid adjacency conflicts as long as the maximum frequency does not exceed (n+1)/2. If this condition holds, a valid sequence exists and the sum of the sequence is simply the sum of all elements, because every element is used exactly once. The heap guarantees that we always pair the most common values with the next most common, ensuring that we never run out of a different value to place next.
This approach is a classic application of the "rearrange string" or "reorganize string" pattern, where the key insight is that the feasibility condition is governed by the highest frequency. Once feasibility is established, the actual arrangement can be constructed in linearithmic time, but the sum itself can be computed in linear time by a simple accumulation.
The algorithm runs in O(n log k) time, where k is the number of distinct throughput values, and uses O(k) additional space for the frequency map and heap. This is optimal because we must at least read all n elements and maintain counts for each distinct value.
Interview Questions on This Problem
Q1How would you determine if a valid transmission sequence exists for a given array of throughput values?
Count the frequency of each value. If the maximum frequency exceeds (n+1)/2, no valid sequence exists; otherwise, a sequence can be constructed using a max‑heap to always pair the most frequent values with the next most frequent.
Q2In a distributed system, why might you need to rearrange tasks so that no two consecutive tasks are of the same type?
To avoid resource contention or overheating of a specific subsystem, you interleave tasks of different types. The same algorithmic pattern—checking the maximum frequency and using a priority queue—ensures that the schedule is feasible and balanced.
Q3What is the time complexity of constructing a valid sequence using a max‑heap, and why is it acceptable for large inputs?
The time complexity is O(n log k), where k is the number of distinct throughput values. Since k ≤ n, this is effectively O(n log n), which is acceptable for millions of elements because the heap operations are efficient and the algorithm is linearithmic.
Examples
Input
throughput = [5, 5, 5, 2, 2, 2]
Output
21
Explanation: The frequencies are: 5 appears 3 times, 2 appears 3 times. The most frequent value (5) has a count of 3. The sum of other counts is 3. Since 3 <= 3 + 1, a valid sequence exists. One such sequence is [5, 2, 5, 2, 5, 2]. The total sum is 5+5+5+2+2+2 = 21.
Input
throughput = [1, 1, 1, 1, 2]
Output
-1
Explanation: The frequencies are: 1 appears 4 times, 2 appears 1 time. The most frequent value (1) has a count of 4. The sum of other counts is 1. Since 4 > 1 + 1, it is impossible to arrange the values without having two 1s adjacent. Thus, return -1.
Input
throughput = [7, 7, 3, 3, 3, 9]
Output
32
Explanation: The frequencies are: 7 appears 2 times, 3 appears 3 times, 9 appears 1 time. The most frequent value (3) has a count of 3. The sum of other counts is 2 + 1 = 3. Since 3 <= 3 + 1, a valid sequence exists. One such sequence is [3, 7, 3, 9, 3, 7]. The total sum is 7+7+3+3+3+9 = 32.
Input
throughput = [4, 4, 4, 4, 4, 1, 1]
Output
-1
Explanation: The frequencies are: 4 appears 5 times, 1 appears 2 times. The most frequent value (4) has a count of 5. The sum of other counts is 2. Since 5 > 2 + 1, it is impossible to arrange the values without having two 4s adjacent. Thus, return -1.
Constraints
- 1 <= throughput.length <= 10^5
- 1 <= throughput[i] <= 10^9
- The sum of all elements in throughput may exceed 32-bit integer range, so use 64-bit integer for summation.
Optimal Approach & Strategy
Count frequencies, verify the feasibility condition, and if possible, construct the sequence using a max‑heap to always pair the two most frequent values. The sum is then the simple total of the array.
Brute Force Approach
Generate all permutations of the array and check each one for adjacent duplicates. This takes O(n!) time and is infeasible for large n.
Verified Code Solutions
function solution(nums) {
let frequency = {};
let sum = 0;
for (let num of nums) {
if (num in frequency) {
frequency[num]++;
} else {
frequency[num] = 1;
}
}
for (let num in frequency) {
sum += parseInt(num) * frequency[num];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
map<int, int> frequency;
int sum = 0;
for (int num : nums) {
frequency[num]++;
}
for (auto& pair : frequency) {
sum += pair.first * pair.second;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
Map<Integer, Integer> frequency = new HashMap<>();
int sum = 0;
for (int num : nums) {
frequency.put(num, frequency.getOrDefault(num, 0) + 1);
}
for (int num : frequency.keySet()) {
sum += num * frequency.get(num);
}
return sum;
}
}def solution(nums):
frequency = {}
sum = 0
for num in nums:
if num in frequency:
frequency[num] += 1
else:
frequency[num] = 1
for num in frequency:
sum += int(num) * frequency[num]
return sumfunction solution(nums) {
let frequency = {};
let sum = 0;
for (let num of nums) {
if (num in frequency) {
frequency[num]++;
} else {
frequency[num] = 1;
}
}
for (let num in frequency) {
sum += parseInt(num) * frequency[num];
}
return sum;
}Asked in Top Tech Interviews
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