Find the City With the Smallest Number of Neighbors at a Threshold Distance

Posted: 18 Mar, 2021
Difficulty: Moderate

PROBLEM STATEMENT

Note:

1. If multiple such cities exist, you have to find the city with the greatest number.

2. The distance of a path connecting two cities, ‘U’ and ‘V’, is the sum of the weight of the edges along that path.

3. The distance between two cities is the minimum of all possible path distances.
Input Format:
The first line contains an integer ‘T’, which denotes the number of test cases to be run. Then, the T test cases follow.

The first line of each test case contains three positive integers, ‘N’,  ‘M’, and ‘distanceThreshold’, as described in the problem statement.

The next ‘M’ lines of each test case contain three integers, ‘U’, ‘V’, and ‘W’ each, representing each edge of the graph.

The edge U V W represents an edge between vertices ‘U’ and ‘V’, having weight ‘W’.
Note:
The ‘edges’ will be passed to the function as an array of arrays. Each array will contain three integers, ‘U’, ‘V’, and ‘W’ in that order.
Output Format:
For each test case, print a single line containing a single integer denoting the required ‘city’ number, as described in the problem statement.

The output for each test case will be printed in a separate line.
Note:
You do not need to print anything. It has already been taken care of. Just implement the given function.
Constraints:
1 <= T <= 10
2 <= N <= 100
1 <= M <= (N * (N - 1)) / 2
0 <= U, V < N
1 <= W, distanceThreshold <= 100

Where ‘T’ denotes the number of test cases, ‘N’ represents the number of cities, and ‘M’ denotes the number of edges.
‘U’, ‘V’, and ‘W’ denote the edge between city ‘U’ and ‘W’ having weight ‘W’.

Time limit: 1 sec. Approach 1

The idea here is to use Dijkstra’s algorithm to compute the distance between cities as the edge weights are non-negative. This algorithm fixed one node and treated it as a source and compute the distance of other nodes to the source. First, we need to make the adjacency list for the graph which contains for each city the city to which it is connected, and the edge weight of that edge. Now, we have to run Dijkstra’s algorithm for each city to find the distance of all other cities to it. Please read more about Dijkstra’s algorithm from Wikipedia.

Now, for each city, we have to calculate the reachable cities within the threshold. We can use the vector of pairs for the same, where the 1st element denotes the number of reachable cities to a particular city and the 2nd element represents the number of that city (that is used to break the tie). Sort the vector of pairs in a way that the 1st element of the vector will contain the desired output, and the second of the 1st element is the required city number.

Steps:

• Create a vector of pair of size N named adj for the adjacency list.
• Run a loop from i = 0 to edges.size(), in each iteration:
• Create a vector of pair named ans, where 1st element denotes the number of reachable cities to a particular city and 2nd element represents the number of that city.
• Run a loop with variable i = 0 to N, in each iteration:
• Call dijkstra(i, N, adj, distanceThreshold, ans), where i is the current city, N is the number of cities, adj is the adjacency list, distanceThreshold is the given value, and ans is the vector of pair for storing the answer.
• This function will update ans with the number of cities reachable to city i.
• Sort the vector ans with the help of the compare.
• Finally, return ans.second.

bool compare(pair x, pair y):

• This function is used to sort the vector of pairs.
• If x.first != y.first, then return x.first < y.first.
• Return x.second > y.second, because to break the tie we need the city with the greatest number first.

void dijkstra(int src, int n, vector of pair adj[], int distanceThreshold, vector of pair ans):

• This function is computing distance between src and all other cities.
• Create a HashSet of type pair named findDist, where the 1st element denotes the distance between src and the city whose id is in the 2nd element.
• Create a distance array of size N, and initialize it with INT_MAX.
• Set distance[src] = 0, and insert {0, src} into the set.
• Run a while loop until findDist becomes empty, in each iteration:
• Declare a variable v, and initialize it with the second element of the top.
• Remove the top element.
• Run a loop to find all the direct connections of v, let u be its direct connections, and w be the edge weight in each iteration:
• If distance[u] > distance[v] + w, then update distance[u] = distance[v] + w, and insert {distance[u], u} into the set findDist.
• Declare a variable cnt = 0.
• Run a loop from i = 0 to N, in each iteration:
• If i != src and distance[i] <= distanceThreshold, then increase cnt by 1.
• Add {cnt, i} into the vector of pairs ans.