Network Delay Time

Leetcode Link

You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target.

We will send a signal from a given node k. Return the minimum time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.

Example 1:

Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
Output: 2

Example 2:
Input: times = [[1,2,1]], n = 2, k = 1
Output: 1

Example 3:
Input: times = [[1,2,1]], n = 2, k = 2
Output: -1

Constraints:

1 <= k <= n <= 100
1 <= times.length <= 6000
times[i].length == 3
1 <= ui, vi <= n
ui != vi
0 <= wi <= 100
All the pairs (ui, vi) are unique. (i.e., no multiple edges.)

Solution

We wish to find the shortest path from k to all nodes in the graph. Dijkstra's algorithm is the best approach in finding shortest path from a single source. We will keep track of the delay time from k to each node, and find the maximum in the end.

Implementation

1def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
2    graph = defaultdict(list)
3    for src, dst, time in times:
4        graph[src].append((time, dst)) 
5    
6    queue = [(0, k)]
7    delay_time = [inf for _ in range(n+1)]
8    while queue:
9        cur_time, cur_node = heapq.heappop(queue) # pop min from min-heap
10        if cur_time > delay_time[cur_node]:       # a shorter path exists
11            continue
12        for edge in graph[cur_node]:
13            new_time, new_node = edge
14            if delay_time[new_node] > cur_time + new_time:
15                delay_time[new_node] = cur_time + new_time
16                heapq.heappush(queue, (delay_time[new_node], new_node))
17    
18    max_delay = max(delay_time)
19    return max_delay if max_delay != inf else -1

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Question 1 out of 10

What does the following code do?

1def f(arr1, arr2):
2  i, j = 0, 0
3  new_arr = []
4  while i < len(arr1) and j < len(arr2):
5      if arr1[i] < arr2[j]:
6          new_arr.append(arr1[i])
7          i += 1
8      else:
9          new_arr.append(arr2[j])
10          j += 1
11  new_arr.extend(arr1[i:])
12  new_arr.extend(arr2[j:])
13  return new_arr
14

1public static List<Integer> f(int[] arr1, int[] arr2) {
2  int i = 0, j = 0;
3  List<Integer> newArr = new ArrayList<>();
4
5  while (i < arr1.length && j < arr2.length) {
6      if (arr1[i] < arr2[j]) {
7          newArr.add(arr1[i]);
8          i++;
9      } else {
10          newArr.add(arr2[j]);
11          j++;
12      }
13  }
14
15  while (i < arr1.length) {
16      newArr.add(arr1[i]);
17      i++;
18  }
19
20  while (j < arr2.length) {
21      newArr.add(arr2[j]);
22      j++;
23  }
24
25  return newArr;
26}
27

1function f(arr1, arr2) {
2  let i = 0, j = 0;
3  let newArr = [];
4  
5  while (i < arr1.length && j < arr2.length) {
6      if (arr1[i] < arr2[j]) {
7          newArr.push(arr1[i]);
8          i++;
9      } else {
10          newArr.push(arr2[j]);
11          j++;
12      }
13  }
14  
15  while (i < arr1.length) {
16      newArr.push(arr1[i]);
17      i++;
18  }
19  
20  while (j < arr2.length) {
21      newArr.push(arr2[j]);
22      j++;
23  }
24  
25  return newArr;
26}
27

Find the intersection of two arrays.

Check if one array is a subsequence of the other.

Merge two sorted arrays.

Finding median values of 2 arrays.

I don’t know

Network Delay Time

Solution

Implementation

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