The other question asked in the same OA - https://leetcode.com/discuss/interview-question/5856836/Hackerearth-or-OA-or-The-Prom
Happy Neighbourhood
In a neighborhood, there are N empty houses numbered from 1 to N arranged in a line. Each day, starting from day 1, one house will be occupied by residents. The sequence of occupied houses is given as a permutation of length N. On the ith day, the house with the number given by the ith element of the permutation will be occupied.
The neighborhood will be considered happy if there is at least one set of consecutive occupied houses. On which day will the neighborhood become happy?
Note: A permutation of length N is an array of N integers where each element is between 1 and N, with no repetitions.
Function Description
Complete the function solve. This function takes the following 3 parameters and returns the required answer.
- N: Represents the number of houses
- M: Represents the number of consecutive houses needed
- house: Represents an array indicating the house that will be filled on each day
Input format for custom testing
Note: Use this input format if you are testing against custom input or writing code in a language where we don't provide boilerplate code.
- The first line contains N denoting the number of houses.
- The second line contains M denoting the number of consecutive houses needed.
- The third line contains an array house denoting the house that will be filled on each day.
Output format
Print a single integer representing the first day on which the neighbourhood becomes happy.
Constraints
1 ≤ N ≤ 10^5
1 ≤ M ≤ N
1 ≤ house[i] ≤ N
Sample Input Sample Output
3 1
1
3 2 1Explanation
Given
- N = 3
- M = 1
- house = [3, 2, 1]
Output
1
Approach
On day 1, house 3 is filled. Only 1 consecutive house is needed, so the neighbourhood becomes happy.
Note:
Your code must be able to print the sample output from the provided sample input. However, your code is run against multiple hidden test cases. Therefore, your code must pass these hidden test cases to solve the problem statement.
Limits
Time Limit: 1.0 sec(s) for each input file
Memory Limit: 256 MB
Source Limit: 1024 KB
Scoring
Score is assigned if any testcase passes
Test case 1
Input Expected Output
10 10
8
8 4 9 10 3 1 7 2 6 5Test case 2
Input Expected Output
20 5
2
16 12 14 7 11 13 15 4 1 20 9 3 10 8 6 2 17 19 18 5Test case 3
Input Expected Output
5 5
5
5 4 1 2 3Working Solution
private static class UnionFind {
int[] size;
int[] parent;
public UnionFind(int n) {
parent = new int[n + 1];
size = new int[n + 1];
for (int i = 1; i <= n; i++) {
parent[i] = i;
size[i] = 1;
}
}
public int find(int x) {
if (parent[x] != x) {
parent[x] = find(parent[x]); // Path compression
}
return parent[x];
}
public void union(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX != rootY) {
if (size[rootX] < size[rootY]) {
int temp = rootX;
rootX = rootY;
rootY = temp;
}
parent[rootY] = rootX;
size[rootX] += size[rootY];
}
}
public int componentSize(int x) {
return size[find(x)];
}
}
public class Main {
public static int solve(int N, int M, int[] house) {
UnionFind uf = new UnionFind(N);
boolean[] occupied = new boolean[N + 1]; // Track which houses are occupied
for (int day = 0; day < N; day++) {
int currentHouse = house[day];
occupied[currentHouse] = true;
// Connect current house with its left and right neighbors
if (currentHouse > 1 && occupied[currentHouse - 1]) {
uf.union(currentHouse, currentHouse - 1);
}
if (currentHouse < N && occupied[currentHouse + 1]) {
uf.union(currentHouse, currentHouse + 1);
}
// Check if there is a component with size at least M
if (uf.componentSize(currentHouse) >= M) {
return day + 1; // Return the day (1-based index)
}
}
return -1; // Should not be reached for valid inputs
}
public static void main(String[] args) {
int N = 20;
int M = 2;
int[] house = {16, 12, 14, 7, 11, 13, 15, 4, 1, 20, 9, 3, 10, 8, 6, 2, 17, 19, 18, 5};
System.out.println(solve(N, M, house));
}
}