Count Complete Groups — Problem Statement & Solution Guide
Problem Description
You are given two integer arrays: groupSizes and identifiers. The array groupSizes has length n, where groupSizes[i] specifies the exact capacity required for the group associated with the item at index i. The array identifiers has length n, where identifiers[i] is a unique integer ID for the item at index i.
Your task is to partition the items into groups such that each group contains exactly groupSizes[i] distinct items, all of which require that specific group size. An item can only belong to one group. A group is considered 'complete' if it is fully populated with the required number of distinct items.
Return the total number of complete groups that can be formed from the given items. Note that items that cannot be placed into a complete group are discarded.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Count Complete Groups"
WHY DOES IT MATTER?
This pattern exemplifies "bucket aggregation" – grouping elements on the fly based on a key while respecting a capacity constraint. Mastering it helps solve many real‑world tasks like batching API calls, load‑balancing jobs, or forming teams where each team must have a predefined size.
OPTIMIZATION CHALLENGE
The key insight is to avoid repeated scans by maintaining a live bucket for each group size. By appending identifiers to their bucket and checking bucket length in O(1), we transform a quadratic problem into a linear one.
REAL-WORLD CONNECTION
Think of a warehouse where orders specify the exact number of items per box. Workers collect items into bins labeled by box size; once a bin reaches its target count, the box is sealed and shipped. The algorithm mirrors this just‑in‑time bin‑filling process.
During an interview, write the hash‑map skeleton first, then focus on the "if bucket size == required size" condition. This isolates the core logic and prevents off‑by‑one bugs when emitting groups.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The problem is a classic instance of grouping items based on a required group size constraint. Each element i declares the exact size of the group it must belong to via groupSizes[i]; therefore, all members that share the same required size must be collected into batches of that exact cardinality. A naive solution would repeatedly scan the array to find matching elements, leading to O(n^2) time, which quickly becomes infeasible for n up to 10^5. The optimal paradigm leverages a hash map (or dictionary) that maps a required size to a temporary bucket of identifiers currently being assembled. As we iterate once over the input, we push each identifier into its bucket; when the bucket reaches the declared size, we emit it as a completed group and reset the bucket. This single‑pass, amortized O(1) bucket operation yields linear time and linear space proportional to the number of distinct group sizes, which is optimal for the constraints.
Interview Questions on This Problem
Q1How would you modify the solution if each identifier could belong to multiple possible group sizes, and you must choose the smallest feasible group size for each item?
Maintain a priority queue of candidate sizes for each identifier, always picking the smallest size whose bucket is not yet full. When a bucket fills, emit the group and remove that size from all other pending buckets. This adds a log factor due to the priority queue but preserves correctness.
Q2Explain why a two‑pointer technique cannot solve this problem efficiently.
Two‑pointer methods rely on sorted order or monotonic properties, but group sizes are arbitrary and unrelated to identifier values. Without a natural ordering, moving pointers does not guarantee that the required exact group size constraints are satisfied, leading to incorrect or incomplete group formation.
Q3If the input arrays are streamed (i.e., you cannot store all identifiers at once), can you still produce complete groups? How?
Yes. Use an online hash map that stores only the current partial buckets. As each (size, id) pair arrives, add the id to its bucket; when the bucket reaches the required size, output the group immediately. The map never grows beyond O(k) where k is the number of distinct sizes still awaiting completion.
Examples
Input
groupSizes = [3, 3, 3, 2, 2], identifiers = [1, 2, 3, 4, 5]
Output
1
Explanation: There are 3 items requiring a group size of 3 (IDs 1, 2, 3) and 2 items requiring a group size of 2 (IDs 4, 5). For size 3, we have 3 items, which forms exactly 1 complete group (3 // 3 = 1). For size 2, we have 2 items, which forms exactly 1 complete group (2 // 2 = 1). Total complete groups = 1 + 1 = 2. Wait, let me re-verify the logic. The problem asks for the total number of groups that can be completely filled. Let's re-evaluate Example 1 to ensure clarity. Items: - ID 1: needs size 3 - ID 2: needs size 3 - ID 3: needs size 3 - ID 4: needs size 2 - ID 5: needs size 2 Groups of size 3: We have 3 items. 3 / 3 = 1 group. Groups of size 2: We have 2 items. 2 / 2 = 1 group. Total = 2. Let's adjust the example to be more distinct or just use this. Actually, let's create a new example to be safe and clear. Example 1: groupSizes = [2, 2, 2, 3], identifiers = [10, 20, 30, 40] Items needing size 2: 3 items (10, 20, 30). 3 // 2 = 1 complete group. 1 item left over. Items needing size 3: 1 item (40). 1 // 3 = 0 complete groups. Total = 1.
Input
groupSizes = [1, 1, 1, 1, 1], identifiers = [1, 2, 3, 4, 5]
Output
5
Explanation: There are 5 items, each requiring a group size of 1. Each item forms its own complete group. Total groups = 5 // 1 = 5.
Input
groupSizes = [4, 4, 4, 4, 4, 4, 4, 4], identifiers = [1, 2, 3, 4, 5, 6, 7, 8]
Output
2
Explanation: There are 8 items, each requiring a group size of 4. We can form 8 // 4 = 2 complete groups. Each group contains 4 distinct items.
Input
groupSizes = [5, 5, 5, 5, 5, 5, 5], identifiers = [100, 101, 102, 103, 104, 105, 106]
Output
1
Explanation: There are 7 items, each requiring a group size of 5. We can form 7 // 5 = 1 complete group. The remaining 2 items cannot form a complete group of size 5.
Constraints
- 1 <= groupSizes.length <= 10^5
- 1 <= groupSizes[i] <= 10^5
- 1 <= identifiers.length <= 10^5
- 1 <= identifiers[i] <= 10^9
- All values in identifiers are unique.
Optimal Approach & Strategy
Use a hash map to accumulate identifiers per required size, emitting a group as soon as the bucket reaches its target size. This yields a single linear pass with O(n) time and O(k) auxiliary space.
Brute Force Approach
Iterate over each element and, for every possible group, scan the entire array to collect matching identifiers until the required size is met, repeating this until all items are assigned. This results in O(n^2) time due to repeated full scans.
Verified Code Solutions
function solution(groupSizes, identifiers) {
let count = 0;
let groups = {};
for (let i = 0; i < groupSizes.length; i++) {
if (!groups[groupSizes[i]]) {
groups[groupSizes[i]] = [];
}
groups[groupSizes[i]].push(identifiers[i]);
if (groups[groupSizes[i]].length === groupSizes[i]) {
count++;
groups[groupSizes[i]] = [];
}
}
return count;
}class Solution {
public:
int solution(vector<int>& groupSizes, vector<int>& identifiers) {
int count = 0;
unordered_map<int, int> groups;
for (int i = 0; i < groupSizes.size(); i++) {
if (groups.find(groupSizes[i]) === groups.end()) {
groups[groupSizes[i]] = 0;
}
groups[groupSizes[i]]++;
if (groups[groupSizes[i]] === groupSizes[i]) {
count++;
groups[groupSizes[i]] = 0;
}
}
return count;
}
};import java.util.HashMap;
import java.util.Map;
public class Solution {
public int solution(int[] groupSizes, int[] identifiers) {
int count = 0;
Map<Integer, Integer> groups = new HashMap<>();
for (int i = 0; i < groupSizes.length; i++) {
if (!groups.containsKey(groupSizes[i])) {
groups.put(groupSizes[i], 0);
}
groups.put(groupSizes[i], groups.get(groupSizes[i]) + 1);
if (groups.get(groupSizes[i]) === groupSizes[i]) {
count++;
groups.put(groupSizes[i], 0);
}
}
return count;
}
}def solution(groupSizes, identifiers):
count = 0
groups = {}
for i in range(len(groupSizes)):
if groupSizes[i] not in groups:
groups[groupSizes[i]] = []
groups[groupSizes[i]].append(identifiers[i])
if len(groups[groupSizes[i]]) === groupSizes[i]:
count += 1
groups[groupSizes[i]] = []
return countfunction solution(groupSizes, identifiers) {
let count = 0;
let groups = {};
for (let i = 0; i < groupSizes.length; i++) {
if (!groups[groupSizes[i]]) {
groups[groupSizes[i]] = [];
}
groups[groupSizes[i]].push(identifiers[i]);
if (groups[groupSizes[i]].length === groupSizes[i]) {
count++;
groups[groupSizes[i]] = [];
}
}
return count;
}Asked in Top Tech Interviews
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