Payload Sequence Resolver 9 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a stream of data packets, each represented by a unique integer identifier. The system requires you to determine the 'Resolver Value' based on the frequency distribution of these identifiers within the sequence. The Resolver Value is defined as the sum of the squares of the frequencies of all distinct identifiers present in the sequence. For example, if an identifier appears 3 times, it contributes 9 to the total sum. Your goal is to compute this value efficiently for a given array of packet identifiers.
Given an array packets of integers, where each integer represents a packet identifier, calculate the sum of the squares of the count of each unique identifier in the array. This metric helps in assessing the concentration of traffic or data redundancy in the system. The solution must handle large input sizes efficiently, leveraging frequency counting techniques to avoid excessive computational overhead.
The input will be a single array of integers. The output should be a single integer representing the computed Resolver Value. Ensure that your algorithm runs in linear time relative to the size of the input array to meet performance requirements for high-throughput systems.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Resolver 9"
WHY DOES IT MATTER?
Frequency‑square aggregation appears in entropy, variance, and collision‑probability calculations.
OPTIMIZATION CHALLENGE
Reducing O(n²) pairwise counting to O(n) frequency hashing cuts runtime by orders of magnitude.
REAL-WORLD CONNECTION
Network routers often need to compute the sum of squared packet counts to detect traffic spikes or DDoS patterns.
Initialize the hash map with reserve() for expected distinct IDs to avoid rehash overhead.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The Resolver Value is the sum of the squares of the occurrence counts of each distinct packet identifier. Computing it efficiently requires a single pass to tally frequencies, then a second linear pass to aggregate the squared counts. Naïve double‑loop methods enumerate every pair of positions, leading to O(n²) time and quickly exhausting time limits for streams of millions of packets. The optimal paradigm leverages a hash map (or array when identifiers are bounded) to achieve O(n) time by converting the problem into frequency counting, a classic example of the “count‑and‑aggregate” pattern in combinatorial graph‑like data.
Because the final sum can exceed 32‑bit limits, the algorithm must use 64‑bit arithmetic. Moreover, the approach scales to any identifier range, as the hash map grows only with the number of distinct values, not the total stream length, ensuring linear space proportional to unique identifiers. This makes the solution robust for both dense and sparse identifier distributions, a key requirement in high‑throughput networking or log‑analysis systems.
Interview Questions on This Problem
Q1Why does a nested loop solution time out for large input sizes?
A nested loop examines every pair of elements, resulting in O(n²) operations. For n in the millions, this exceeds typical time limits.
Q2How can you compute the Resolver Value in a single pass?
Maintain a hash map that increments the count for each identifier as you read it. After the stream ends, iterate over the map and sum count².
Q3What data type should store the final answer and why?
Use a 64‑bit integer (long long) because the sum of squares can grow beyond 32‑bit range. This prevents overflow on large frequency values.
Examples
Input
packets = [1, 2, 2, 3, 3, 3]
Output
14
Explanation: Count the frequency of each identifier: 1 appears 1 time, 2 appears 2 times, 3 appears 3 times. Square each frequency: 1^2 = 1, 2^2 = 4, 3^2 = 9. Sum the squares: 1 + 4 + 9 = 14.
Input
packets = [5, 5, 5, 5]
Output
16
Explanation: Count the frequency of each identifier: 5 appears 4 times. Square the frequency: 4^2 = 16. Sum the squares: 16.
Input
packets = [1, 2, 3, 4, 5]
Output
5
Explanation: Count the frequency of each identifier: 1, 2, 3, 4, and 5 each appear 1 time. Square each frequency: 1^2 = 1 for each. Sum the squares: 1 + 1 + 1 + 1 + 1 = 5.
Input
packets = [7, 7, 8, 8, 8, 9]
Output
14
Explanation: Count the frequency of each identifier: 7 appears 2 times, 8 appears 3 times, 9 appears 1 time. Square each frequency: 2^2 = 4, 3^2 = 9, 1^2 = 1. Sum the squares: 4 + 9 + 1 = 14.
Constraints
- 1 <= packets.length <= 10^5
- 1 <= packets[i] <= 10^9
- The sum of squares of frequencies will not exceed 10^18
- All elements in packets are positive integers
- Time complexity must be O(n) where n is the length of packets
Optimal Approach & Strategy
Use a hash map to count frequencies in one pass and compute the sum of squares in a second linear pass.
Brute Force Approach
Use two nested loops to count occurrences for every element, then sum the squares.
Verified Code Solutions
/**
* @param {number[]} packets
* @return {number}
*/
var resolverValue = function(packets) {
const freq = new Map();
for (let p of packets) {
freq.set(p, (freq.get(p) || 0) + 1);
}
let sum = 0;
for (let count of freq.values()) {
sum += count * count;
}
return sum;
};class Solution {
public:
int resolverValue(vector<int>& packets) {
unordered_map<int, int> freq;
for (int p : packets) {
freq[p]++;
}
long long sum = 0;
for (auto& pair : freq) {
sum += (long long)pair.second * pair.second;
}
return (int)sum;
}
};class Solution {
public int resolverValue(int[] packets) {
Map<Integer, Integer> freq = new HashMap<>();
for (int p : packets) {
freq.put(p, freq.getOrDefault(p, 0) + 1);
}
int sum = 0;
for (int count : freq.values()) {
sum += count * count;
}
return sum;
}
}class Solution:
def resolverValue(self, packets: List[int]) -> int:
freq = {}
for p in packets:
freq[p] = freq.get(p, 0) + 1
return sum(c * c for c in freq.values())/**
* @param {number[]} packets
* @return {number}
*/
var resolverValue = function(packets) {
const freq = new Map();
for (let p of packets) {
freq.set(p, (freq.get(p) || 0) + 1);
}
let sum = 0;
for (let count of freq.values()) {
sum += count * count;
}
return sum;
};Asked in Top Tech Interviews
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