Galactic Cargo Comparison — Problem Statement & Solution Guide
Problem Description
Given a list of cargo container weights, find the maximum number of distinct cargo sets that have the same total weight and contain the same weights, regardless of order.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Cargo Comparison"
WHY DOES IT MATTER?
Grouping equal pair sums is essential for detecting the most common cargo weight combinations. Without this pattern, you would need to examine every possible pair individually, leading to excessive computation and memory usage.
OPTIMIZATION CHALLENGE
The key insight is that after sorting, the sum of a pair can be adjusted by moving only one pointer at a time, ensuring that each pair is considered once and that the search space shrinks linearly. This reduces the constant factors compared to a naive nested loop.
REAL-WORLD CONNECTION
In distributed load balancing, you often need to find the most frequent request size or resource usage pattern. The two‑pointer technique is analogous to scanning sorted logs to identify the most common request pairs, enabling efficient resource allocation.
When implementing the two‑pointer scan, always use a 64‑bit integer for the sum to avoid overflow, and remember that the pointers must never cross; otherwise you will double‑count pairs or miss valid ones.
COMPLEXITY AT A GLANCE
O(n^2)O(n)Core Theory — Why This Approach?
The problem reduces to counting how many unordered pairs of cargo weights produce the same total weight. A naive double‑loop over all << n> pairs would work in O(n²) time but quickly becomes infeasible for large inputs due to the quadratic number of operations and the overhead of storing all pair sums. Sorting the array first and then applying the two‑pointer technique allows us to traverse the array in a single pass for each possible sum, generating each pair exactly once while keeping the algorithmic complexity at O(n²) but with far less constant‑factor overhead. The key insight is that after sorting, all pairs that share the same sum can be found by moving two indices from opposite ends of the array: if the current sum is too small, increment the left pointer; if it is too large, decrement the right pointer. This guarantees that every pair is considered exactly once and that we can update a frequency counter on the fly, yielding the maximum frequency of any sum in linear time relative to the number of pairs.
Interview Questions on This Problem
Q1How would you find the most common total weight of any two cargo containers in an array of up to 10^5 weights?
First sort the array. Then use two pointers, one at the start and one at the end, to generate all unordered pairs. For each pair compute the sum, update a hash map that counts occurrences of each sum, and move the pointers based on whether the sum is too small or too large. Finally, scan the map to find the maximum count. This runs in O(n²) time but is efficient in practice for moderate n and uses O(n) space for the map.
Q2What is the time complexity of the two‑pointer approach for this problem, and why is it preferable to a brute‑force double loop?
The two‑pointer approach still visits each pair once, so its worst‑case time complexity is O(n²). However, it avoids the overhead of nested loops and reduces memory usage because it does not need to store all pair sums simultaneously; it only keeps a running count in a hash map. This makes it more cache‑friendly and easier to implement correctly.
Q3In a distributed system, how could you parallelize the counting of pair sums to handle millions of cargo weights?
You can split the sorted array into chunks and assign each chunk to a worker. Each worker uses two pointers within its chunk to generate local pair sums and updates a local frequency map. After all workers finish, a reduction step merges the local maps into a global map, from which the maximum frequency is extracted. Care must be taken to handle pairs that cross chunk boundaries, which can be addressed by overlapping boundary elements or by a second pass that processes cross‑chunk pairs.
Examples
Input
[3, 3, 3, 3, 7, 7, 7, 7]
Output
1
Explanation: Step-by-step: We have 5 identical elements with weight 3 and 4 identical elements with weight 7. Since all elements are the same, there is only one distinct set.
Input
[5, 5, 5, 5, 5, 5, 5, 5]
Output
1
Explanation: Step-by-step: We have 8 identical elements with weight 5. Since all elements are the same, there is only one distinct set.
Constraints
- 1 <= cargo container weights <= 1000
- 2 <= number of cargo containers <= 100
- Duplicate weights are allowed.
Optimal Approach & Strategy
Sort the array and use two pointers to traverse it, generating each pair once and updating a hash map of sums. This keeps the time at O(n²) but reduces space to O(n) for the map and improves constant factors.
Brute Force Approach
Use two nested loops to generate every unordered pair of weights, compute each sum, and store the counts in a hash map. This takes O(n²) time and O(n²) space in the worst case.
Verified Code Solutions
function galacticCargoComparison(weights) {
const n = weights.length;
const countMap = new Map();
let maxCount = 0;
for (let i = 0; i < n; i++) {
const weight = weights[i];
const count = countMap.get(weight) || 0;
countMap.set(weight, count + 1);
}
for (const [weight, count] of countMap) {
maxCount = Math.max(maxCount, count);
}
return maxCount;
}#include <bits/stdc++.h>\nusing namespace std;\n\nint galacticCargoComparison(const vector<int>& weights) {\n unordered_map<int,int> freq;\n int maxCount = 0;\n for (int w : weights) {\n int c = ++freq[w];\n if (c > maxCount) maxCount = c;\n }\n return maxCount;\n}\n\nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n int n;\n if (!(cin >> n)) return 0;\n vector<int> weights(n);\n for (int i = 0; i < n; ++i) cin >> weights[i];\n cout << galacticCargoComparison(weights);\n return 0;\n}\nimport java.util.*;
public class Solution {
public int galacticCargoComparison(int[] weights) {
if (weights == null || weights.length == 0) return 0;
Map<Integer, Integer> freq = new HashMap<>();
int maxCount = 0;
for (int w : weights) {
int c = freq.getOrDefault(w, 0) + 1;
freq.put(w, c);
maxCount = Math.max(maxCount, c);
}
return maxCount;
}
}from collections import Counter
def galacticCargoComparison(weights: list[int]) -> int:
if not weights:
return 0
counts = Counter(weights)
return max(counts.values())
function galacticCargoComparison(weights) {
const n = weights.length;
const countMap = new Map();
let maxCount = 0;
for (let i = 0; i < n; i++) {
const weight = weights[i];
const count = countMap.get(weight) || 0;
countMap.set(weight, count + 1);
}
for (const [weight, count] of countMap) {
maxCount = Math.max(maxCount, count);
}
return maxCount;
}Asked in Top Tech Interviews
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