Renovating Homes with Limited Materials — Problem Statement & Solution Guide

ArraysMediumNEW OR EXISTING ID

TimeO(n)

SpaceO(k)

Problem Description

You are given an array rooms where rooms[i] represents the material type used in the i-th room. Each room must be assigned a material ID, which is a positive integer. The assignment must satisfy the rule that no two rooms that use the same material type receive the same ID. You may reuse IDs across different material types. Your task is to determine the smallest possible number of distinct IDs that must be used to satisfy this rule.

Input format: The first line contains an integer n (1 ≤ n ≤ 10^5), the number of rooms. The second line contains n integers rooms[0], rooms[1], …, rooms[n-1] (1 ≤ rooms[i] ≤ 10^9), the material type of each room.

Output format: Output a single integer – the minimum number of distinct material IDs required.

The optimal number of IDs equals the maximum frequency of any material type in the array, because each room of the same type must receive a unique ID while IDs can be shared across different types.

DSA Pattern Breakdown

"Renovating Homes with Limited Materials"

medium

WHY DOES IT MATTER?

Identifying the maximum frequency pattern is essential because it transforms a seemingly complex assignment problem into a simple counting problem, enabling linear-time solutions. It also illustrates how constraints can often be reduced to a single statistic that dictates the answer.

OPTIMIZATION CHALLENGE

The key insight is that inter-type ID reuse eliminates the need for a combinatorial search; only the intra-type uniqueness matters, so the bottleneck is the largest group size.

REAL-WORLD CONNECTION

In distributed systems, this is akin to assigning unique session tokens per user group while reusing token pools across groups. The maximum group size dictates the pool size needed, just as the maximum material frequency dictates the ID pool size.

When explaining this to an interviewer, emphasize the reduction to a frequency count and the reuse property, as it showcases both algorithmic insight and practical efficiency.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(k)

Core Theory — Why This Approach?

The problem reduces to finding the maximum number of rooms that share the same material type. Each material type must have distinct IDs for its rooms, but IDs can be reused across different types. Therefore, the minimal number of distinct IDs required is exactly the highest frequency of any material type in the array. A naive approach would try to assign IDs to each room individually and check all combinations, leading to exponential time. The optimal paradigm is a single pass frequency count using a hash map, followed by a linear scan to find the maximum value, achieving linear time and linear space in the number of distinct types.

Interview Questions on This Problem

Q1How would you determine the minimum number of distinct IDs needed for a list of rooms with various material types?

Count the occurrences of each material type using a hash map, then return the maximum count among all types. This maximum represents the minimum number of distinct IDs required because each type needs unique IDs for its rooms, and IDs can be reused across types.

Q2What is the time and space complexity of your solution, and why is it efficient for large inputs?

Time complexity is O(n) where n is the number of rooms, because we traverse the array once to build frequencies and once to find the maximum. Space complexity is O(k), where k is the number of distinct material types, for storing the frequency map. This linear performance scales well even for millions of rooms.

Q3Can you explain why a greedy assignment of IDs would work in this scenario?

A greedy assignment works because we can assign IDs 1..maxFreq to the most frequent material type, then reuse the same set of IDs for all other types. Since each type only requires unique IDs within itself, reusing IDs across types never violates the constraint, ensuring the minimal number of distinct IDs equals maxFreq.

Examples

Example 1

Input

5
1 2 1 3 2

Output

Explanation: Material 1 appears twice, material 2 appears twice, material 3 appears once. The maximum frequency is 2, so at least two IDs are needed. Assign IDs 1 and 2 to the two rooms of material 1, IDs 1 and 2 to the two rooms of material 2, and ID 1 to the room of material 3. Only two distinct IDs are used, which is minimal.

Example 2

Input

4
5 5 5 5

Output

Explanation: All four rooms use material 5, so each must receive a different ID. The maximum frequency is 4, thus four IDs are required.

Example 3

Input

6
1 2 3 4 5 6

Output

Explanation: All rooms use distinct materials, so a single ID can be assigned to every room without conflict.

Constraints

1 ≤ rooms.length ≤ 10^4
1 ≤ materialIds.length < 10^5
Each room has at least one material
Each material can be assigned to any number of rooms
Material IDs start at 1 and increase by 1

Optimal Approach & Strategy

Count frequencies of each material type in one pass, then return the maximum count; this runs in linear time and uses linear space for the frequency map.

Brute Force Approach

A naive method would try every possible ID assignment for each room and check all constraints, leading to factorial or exponential time complexity.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * @param {number[]} rooms
 * @return {number}
 */
function renovateHomes(rooms) {
    const freq = {};
    for (const room of rooms) {
        freq[room] = (freq[room] || 0) + 1;
    }
    let maxFreq = 0;
    for (const key in freq) {
        if (freq[key] > maxFreq) {
            maxFreq = freq[key];
        }
    }
    return maxFreq;
}

// Example usage
const n = 5;
const rooms = [1, 2, 1, 3, 2];
console.log(renovateHomes(rooms));

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

class Solution {
public:
    int renovateHomes(vector<int>& rooms) {
        unordered_map<int, int> freq;
        for (int room : rooms) {
            freq[room]++;
        }
        int maxFreq = 0;
        for (auto& p : freq) {
            if (p.second > maxFreq) {
                maxFreq = p.second;
            }
        }
        return maxFreq;
    }
};

int main() {
    int n;
    cin >> n;
    vector<int> rooms(n);
    for (int i = 0; i < n; i++) {
        cin >> rooms[i];
    }
    Solution sol;
    cout << sol.renovateHomes(rooms) << endl;
    return 0;
}

import java.util.*;

class Solution {
    public int renovateHomes(int[] rooms) {
        Map<Integer, Integer> freq = new HashMap<>();
        for (int room : rooms) {
            freq.put(room, freq.getOrDefault(room, 0) + 1);
        }
        int maxFreq = 0;
        for (int count : freq.values()) {
            if (count > maxFreq) {
                maxFreq = count;
            }
        }
        return maxFreq;
    }
}

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] rooms = new int[n];
        for (int i = 0; i < n; i++) {
            rooms[i] = sc.nextInt();
        }
        Solution sol = new Solution();
        System.out.println(sol.renovateHomes(rooms));
    }
}

def renovate_homes(rooms):
    freq = {}
    for room in rooms:
        freq[room] = freq.get(room, 0) + 1
    return max(freq.values()) if freq else 0

if __name__ == "__main__":
    n = int(input())
    rooms = list(map(int, input().split()))
    print(renovate_homes(rooms))

/**
 * @param {number[]} rooms
 * @return {number}
 */
function renovateHomes(rooms) {
    const freq = {};
    for (const room of rooms) {
        freq[room] = (freq[room] || 0) + 1;
    }
    let maxFreq = 0;
    for (const key in freq) {
        if (freq[key] > maxFreq) {
            maxFreq = freq[key];
        }
    }
    return maxFreq;
}

// Example usage
const n = 5;
const rooms = [1, 2, 1, 3, 2];
console.log(renovateHomes(rooms));

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Renovating Homes with Limited Materials Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Arrays

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Renovating Homes with Limited Materials — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor