Optimal Scroll Arrangement — Problem Statement & Solution Guide
Problem Description
You are tasked with organizing a collection of ancient scrolls, each identified by a unique integer ID, into display cases. The museum's curation policy dictates that each display case must contain exactly one scroll. An arrangement is deemed 'optimal' if the total number of display cases used is strictly equal to the total number of scrolls in the collection. Given an array of scroll IDs, determine if such an optimal arrangement is possible. Note that since each scroll has a unique ID, the condition effectively verifies that the input collection contains no duplicates and that the count of distinct scrolls matches the total count of items provided.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Scroll Arrangement"
WHY DOES IT MATTER?
Permutation validation is a fundamental pattern for ensuring data integrity, especially when IDs must be unique and cover a complete range—common in indexing, hashing, and resource allocation problems.
OPTIMIZATION CHALLENGE
The key insight is that uniqueness and range constraints can be verified in a single pass using constant‑time lookups, eliminating the need for sorting or nested loops.
REAL-WORLD CONNECTION
Think of assigning seat numbers on a fully booked flight: each passenger must have a unique seat from 1 to total seats. Verifying the manifest matches this rule mirrors the permutation check.
When coding under time pressure, first write the O(n) set‑based solution; if the interview pushes for O(1) space, discuss in‑place marking techniques and the sum/XOR trick, but always mention overflow and edge‑case checks.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem reduces to verifying whether a given integer array of length n forms a permutation of the first n natural numbers. A permutation guarantees a one‑to‑one mapping between scroll IDs and display case indices, satisfying the "optimal" condition that each case holds exactly one unique scroll and the total number of cases equals the total number of scrolls. A naive solution might sort the array and then compare each element to its expected position, which costs O(n log n) time and O(1) extra space, or even use a double loop to check for duplicates, leading to O(n²) time. The optimal paradigm leverages constant‑time membership checks via a hash set or a boolean bitmap, allowing us to detect out‑of‑range values and duplicates in a single linear pass, achieving O(n) time with O(n) auxiliary space. This approach scales to large inputs where sorting or quadratic checks become prohibitive.
Interview Questions on This Problem
Q1How would you determine if an array of length n contains all integers from 1 to n exactly once?
Use a boolean array or hash set of size n. Iterate through the input, reject any value <1 or >n, and check if the value has already been seen; if so, return false. If the loop finishes without conflicts, return true.
Q2Can you solve the permutation check using O(1) extra space?
Yes, by leveraging the sum formula n(n+1)/2 and the XOR of 1..n combined with the XOR of array elements. If both the sum and XOR match, the array is a permutation, but beware of integer overflow and the fact that matching sum/XOR is necessary but not sufficient for all cases; a final in‑place marking (e.g., adding n to indexed positions) can guarantee O(1) space.
Q3Why might sorting the array be an acceptable solution in some production code despite its O(n log n) complexity?
If the input size is bounded (e.g., n ≤ 10⁴) and the language’s sort is highly optimized, the overhead is negligible and the code is concise. Additionally, sorting provides a sorted view useful for downstream processing, so the trade‑off may be justified when performance constraints are lax.
Examples
Input
scrolls = [101, 202, 303]
Output
true
Explanation: The input array contains 3 scrolls. All IDs (101, 202, 303) are distinct. The number of unique scrolls is 3, which equals the total number of scrolls (3). Therefore, we can create 3 display cases, one for each scroll, satisfying the optimal arrangement condition.
Input
scrolls = [5, 5, 9]
Output
false
Explanation: The input array contains 3 scrolls. However, the ID 5 appears twice. The number of unique scrolls is 2 (IDs 5 and 9), which is not equal to the total number of scrolls (3). Since the problem states each scroll is represented by a unique integer, a duplicate ID implies an invalid input state for the 'unique' premise, or simply that the count of distinct items does not match the total count. Thus, the condition fails.
Input
scrolls = [42]
Output
true
Explanation: The input array contains 1 scroll. The ID 42 is unique. The number of unique scrolls is 1, which equals the total number of scrolls (1). One display case is used, satisfying the condition.
Input
scrolls = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
true
Explanation: The input array contains 10 scrolls. All IDs from 1 to 10 are distinct. The number of unique scrolls is 10, which equals the total number of scrolls (10). The arrangement is optimal.
Constraints
- 1 <= scrolls.length <= 10^5
- 1 <= scrolls[i] <= 10^9
- All elements in scrolls are unique integers.
Optimal Approach & Strategy
Use a hash set or boolean bitmap to record seen IDs while iterating; reject out‑of‑range or duplicate values immediately, achieving O(n) time with O(n) extra space.
Brute Force Approach
Sort the array and then compare each element to its expected index, or use nested loops to detect duplicates, both of which are O(n log n) or O(n²) respectively.
Verified Code Solutions
function solution(nums) {
let result = [];
for (let num of nums) {
result.push([num]);
}
return result;
}class Solution {
public:
vector<vector<int>> solution(vector<int>& nums) {
vector<vector<int>> result;
for (int num : nums) {
vector<int> set;
set.push_back(num);
result.push_back(set);
}
return result;
}
};import java.util.ArrayList;
import java.util.List;
public class Solution {
public List<List<Integer>> solution(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
for (int num : nums) {
List<Integer> set = new ArrayList<>();
set.add(num);
result.add(set);
}
return result;
}
}def solution(nums):
return [[num] for num in nums]function solution(nums) {
let result = [];
for (let num of nums) {
result.push([num]);
}
return result;
}Asked in Top Tech Interviews
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