Count Distinct Vial Pairs — Problem Statement & Solution Guide
Problem Description
Given three types of vials with quantities moonpetal, dragonBreath, and starlightDew, find the total number of distinct pairs that can be formed, considering each pair must contain at least one vial of an unidentified substance. The input will be an object with the quantities of each vial.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Count Distinct Vial Pairs"
WHY DOES IT MATTER?
Counting with exclusion allows us to transform a potentially quadratic enumeration into a constant-time calculation, which is essential for large-scale data processing and real-time systems where performance constraints are tight.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the only pairs we must exclude are those where both elements belong to the same type, and that these can be counted directly with the combination formula, avoiding any pairwise comparison.
REAL-WORLD CONNECTION
Think of a social network where users are grouped by interests. If you want to find the number of friend pairs that span different interest groups, you would count all possible friendships and subtract those within the same group, just as we do with vial types.
Always validate that your formula handles edge cases such as zero or one vial per type, and use 64-bit arithmetic to prevent overflow before performing the division.
COMPLEXITY AT A GLANCE
O(1)O(1)Core Theory — Why This Approach?
The problem reduces to a classic combinatorial counting task: we have a multiset of vials split into three distinct types and we must count unordered pairs that are not both of the same type. A naive approach would iterate over every possible pair of vials, which would require O((m+n+p)^2) time where m, n, and p are the counts of moonpetal, dragonBreath, and starlightDew respectively. This quickly becomes infeasible for large inputs because the number of pairs grows quadratically. The optimal paradigm leverages the combinatorial identity C(k,2) = k*(k-1)/2 to compute the number of ways to choose two items from a group of size k in constant time. By first computing the total number of unordered pairs from all vials, C(total,2), and then subtracting the pairs that consist of two vials of the same type, we obtain the desired count in O(1) time and O(1) space. This approach eliminates the need for explicit enumeration and guarantees scalability.
Interview Questions on This Problem
Q1How would you compute the number of distinct unordered pairs of vials that include at least one vial of a different type, given counts of three types?
Compute the total number of unordered pairs using C(total,2). Then subtract the number of same-type pairs: C(moonpetal,2) + C(dragonBreath,2) + C(starlightDew,2). The result is the count of pairs that contain at least one vial of a different type.
Q2What potential pitfalls should you watch for when implementing this solution in a language with 32-bit integers?
The intermediate multiplication k*(k-1) can overflow a 32-bit integer if k is large. Use 64-bit integers (long long in C++/Java, long in Python) to safely store the intermediate results before division.
Q3Can you explain why this problem is a good example of the 'counting with exclusion' pattern?
We first count all possible pairs (the inclusive set) and then exclude the undesired pairs (both vials of the same type). This is a classic inclusion–exclusion approach that simplifies the counting by handling the complement set.
Examples
Input
{"moonpetal": 2, "dragonBreath": 3, "starlightDew": 1}Output
11
Explanation: Step-by-step: with input {"moonpetal": 2, "dragonBreath": 3, "starlightDew": 1}, we calculate the number of distinct pairs by considering each type of vial paired with every other type, including itself. For moonpetal and dragonBreath, we have 2 * 3 = 6 pairs. For moonpetal and starlightDew, we have 2 * 1 = 2 pairs. For dragonBreath and starlightDew, we have 3 * 1 = 3 pairs. Therefore, the total number of distinct pairs is 6 + 2 + 3 = 11.
Input
{"moonpetal": 1, "dragonBreath": 1, "starlightDew": 1}Output
3
Explanation: Step-by-step: with input {"moonpetal": 1, "dragonBreath": 1, "starlightDew": 1}, we calculate the number of distinct pairs by considering each type of vial paired with every other type. For moonpetal and dragonBreath, we have 1 * 1 = 1 pair. For moonpetal and starlightDew, we have 1 * 1 = 1 pair. For dragonBreath and starlightDew, we have 1 * 1 = 1 pair. Therefore, the total number of distinct pairs is 1 + 1 + 1 = 3.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
Use combinatorial formulas: totalPairs = C(total,2); sameTypePairs = C(moonpetal,2)+C(dragonBreath,2)+C(starlightDew,2); answer = totalPairs - sameTypePairs. This runs in O(1) time and O(1) space.
Brute Force Approach
Enumerate every pair of vials by nested loops and count those that contain at least one vial of a different type, which takes O((m+n+p)^2) time.
Verified Code Solutions
function solution(vials) { let totalPairs = 0; totalPairs += vials.moonpetal * vials.dragonBreath; totalPairs += vials.moonpetal * vials.starlightDew; totalPairs += vials.dragonBreath * vials.starlightDew; return totalPairs; }class Solution { public: int solution(int moonpetal, int dragonBreath, int starlightDew) { int totalPairs = 0; totalPairs += moonpetal * dragonBreath; totalPairs += moonpetal * starlightDew; totalPairs += dragonBreath * starlightDew; return totalPairs; } };class Solution { public int solution(int moonpetal, int dragonBreath, int starlightDew) { int totalPairs = 0; totalPairs += moonpetal * dragonBreath; totalPairs += moonpetal * starlightDew; totalPairs += dragonBreath * starlightDew; return totalPairs; } }def solution(vials): total_pairs = 0; total_pairs += vials['moonpetal'] * vials['dragonBreath']; total_pairs += vials['moonpetal'] * vials['starlightDew']; total_pairs += vials['dragonBreath'] * vials['starlightDew']; return total_pairsfunction solution(vials) { let totalPairs = 0; totalPairs += vials.moonpetal * vials.dragonBreath; totalPairs += vials.moonpetal * vials.starlightDew; totalPairs += vials.dragonBreath * vials.starlightDew; return totalPairs; }Asked in Top Tech Interviews
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