1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance

Description

There are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [from_i, to_i, weight_i] represents a bidirectional and weighted edge between cities from_i and to_i, and given the integer distanceThreshold.

Return the city with the smallest number of cities that are reachable through some path and whose distance is at most distanceThreshold, If there are multiple such cities, return the city with the greatest number.

Notice that the distance of a path connecting cities i and j is equal to the sum of the edges' weights along that path.

Example 1:

Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
Output: 3
Explanation: The figure above describes the graph. 
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -> [City 1, City 2] 
City 1 -> [City 0, City 2, City 3] 
City 2 -> [City 0, City 1, City 3] 
City 3 -> [City 1, City 2] 
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.

Example 2:

Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
Output: 0
Explanation: The figure above describes the graph. 
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -> [City 1] 
City 1 -> [City 0, City 4] 
City 2 -> [City 3, City 4] 
City 3 -> [City 2, City 4]
City 4 -> [City 1, City 2, City 3] 
The city 0 has 1 neighboring city at a distanceThreshold = 2.

Constraints:

2 <= n <= 100
1 <= edges.length <= n * (n - 1) / 2
edges[i].length == 3
0 <= from_i < to_i < n
1 <= weight_i, distanceThreshold <= 10^4
All pairs (from_i, to_i) are distinct.

Solutions

class Solution {
    private int n;
    private int[][] g;
    private int[] dist;
    private boolean[] vis;
    private final int inf = 1 << 30;
    private int distanceThreshold;

    public int findTheCity(int n, int[][] edges, int distanceThreshold) {
        this.n = n;
        this.distanceThreshold = distanceThreshold;
        g = new int[n][n];
        dist = new int[n];
        vis = new boolean[n];
        for (var e : g) {
            Arrays.fill(e, inf);
        }
        for (var e : edges) {
            int f = e[0], t = e[1], w = e[2];
            g[f][t] = w;
            g[t][f] = w;
        }
        int ans = n, cnt = inf;
        for (int i = n - 1; i >= 0; --i) {
            int t = dijkstra(i);
            if (t < cnt) {
                cnt = t;
                ans = i;
            }
        }
        return ans;
    }

    private int dijkstra(int u) {
        Arrays.fill(dist, inf);
        Arrays.fill(vis, false);
        dist[u] = 0;
        for (int i = 0; i < n; ++i) {
            int k = -1;
            for (int j = 0; j < n; ++j) {
                if (!vis[j] && (k == -1 || dist[k] > dist[j])) {
                    k = j;
                }
            }
            vis[k] = true;
            for (int j = 0; j < n; ++j) {
                dist[j] = Math.min(dist[j], dist[k] + g[k][j]);
            }
        }
        int cnt = 0;
        for (int d : dist) {
            if (d <= distanceThreshold) {
                ++cnt;
            }
        }
        return cnt;
    }
}

class Solution {
public:
    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
        int g[n][n];
        int dist[n];
        bool vis[n];
        memset(g, 0x3f, sizeof(g));
        for (auto& e : edges) {
            int f = e[0], t = e[1], w = e[2];
            g[f][t] = g[t][f] = w;
        }
        auto dijkstra = [&](int u) {
            memset(dist, 0x3f, sizeof(dist));
            memset(vis, 0, sizeof(vis));
            dist[u] = 0;
            for (int i = 0; i < n; ++i) {
                int k = -1;
                for (int j = 0; j < n; ++j) {
                    if (!vis[j] && (k == -1 || dist[j] < dist[k])) {
                        k = j;
                    }
                }
                vis[k] = true;
                for (int j = 0; j < n; ++j) {
                    dist[j] = min(dist[j], dist[k] + g[k][j]);
                }
            }
            return count_if(dist, dist + n, [&](int d) { return d <= distanceThreshold; });
        };
        int ans = n, cnt = n + 1;
        for (int i = n - 1; ~i; --i) {
            int t = dijkstra(i);
            if (t < cnt) {
                cnt = t;
                ans = i;
            }
        }
        return ans;
    }
};

class Solution:
    def findTheCity(
        self, n: int, edges: List[List[int]], distanceThreshold: int
    ) -> int:
        def dijkstra(u: int) -> int:
            dist = [inf] * n
            dist[u] = 0
            vis = [False] * n
            for _ in range(n):
                k = -1
                for j in range(n):
                    if not vis[j] and (k == -1 or dist[k] > dist[j]):
                        k = j
                vis[k] = True
                for j in range(n):
                    # dist[j] = min(dist[j], dist[k] + g[k][j])
                    if dist[k] + g[k][j] < dist[j]:
                        dist[j] = dist[k] + g[k][j]
            return sum(d <= distanceThreshold for d in dist)

        g = [[inf] * n for _ in range(n)]
        for f, t, w in edges:
            g[f][t] = g[t][f] = w
        ans, cnt = n, inf
        for i in range(n - 1, -1, -1):
            if (t := dijkstra(i)) < cnt:
                cnt, ans = t, i
        return ans

func findTheCity(n int, edges [][]int, distanceThreshold int) int {
	g := make([][]int, n)
	dist := make([]int, n)
	vis := make([]bool, n)
	const inf int = 1e7
	for i := range g {
		g[i] = make([]int, n)
		for j := range g[i] {
			g[i][j] = inf
		}
	}
	for _, e := range edges {
		f, t, w := e[0], e[1], e[2]
		g[f][t], g[t][f] = w, w
	}

	dijkstra := func(u int) (cnt int) {
		for i := range vis {
			vis[i] = false
			dist[i] = inf
		}
		dist[u] = 0
		for i := 0; i < n; i++ {
			k := -1
			for j := 0; j < n; j++ {
				if !vis[j] && (k == -1 || dist[j] < dist[k]) {
					k = j
				}
			}
			vis[k] = true
			for j := 0; j < n; j++ {
				dist[j] = min(dist[j], dist[k]+g[k][j])
			}
		}
		for _, d := range dist {
			if d <= distanceThreshold {
				cnt++
			}
		}
		return
	}

	ans, cnt := n, inf
	for i := n - 1; i >= 0; i-- {
		if t := dijkstra(i); t < cnt {
			cnt = t
			ans = i
		}
	}
	return ans
}

function findTheCity(n: number, edges: number[][], distanceThreshold: number): number {
    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(Infinity));
    const dist: number[] = Array(n).fill(Infinity);
    const vis: boolean[] = Array(n).fill(false);
    for (const [f, t, w] of edges) {
        g[f][t] = g[t][f] = w;
    }

    const dijkstra = (u: number): number => {
        dist.fill(Infinity);
        vis.fill(false);
        dist[u] = 0;
        for (let i = 0; i < n; ++i) {
            let k = -1;
            for (let j = 0; j < n; ++j) {
                if (!vis[j] && (k === -1 || dist[j] < dist[k])) {
                    k = j;
                }
            }
            vis[k] = true;
            for (let j = 0; j < n; ++j) {
                dist[j] = Math.min(dist[j], dist[k] + g[k][j]);
            }
        }
        return dist.filter(d => d <= distanceThreshold).length;
    };

    let ans = n;
    let cnt = Infinity;
    for (let i = n - 1; i >= 0; --i) {
        const t = dijkstra(i);
        if (t < cnt) {
            cnt = t;
            ans = i;
        }
    }

    return ans;
}

function findTheCity(n, edges, distanceThreshold) {
    const g = Array.from({ length: n }, () => Array(n).fill(Infinity));
    const dist = Array(n).fill(Infinity);
    const vis = Array(n).fill(false);
    for (const [f, t, w] of edges) {
        g[f][t] = g[t][f] = w;
    }

    const dijkstra = u => {
        dist.fill(Infinity);
        vis.fill(false);
        dist[u] = 0;
        for (let i = 0; i < n; ++i) {
            let k = -1;
            for (let j = 0; j < n; ++j) {
                if (!vis[j] && (k === -1 || dist[j] < dist[k])) {
                    k = j;
                }
            }
            vis[k] = true;
            for (let j = 0; j < n; ++j) {
                dist[j] = Math.min(dist[j], dist[k] + g[k][j]);
            }
        }
        return dist.filter(d => d <= distanceThreshold).length;
    };

    let ans = n;
    let cnt = Infinity;
    for (let i = n - 1; i >= 0; --i) {
        const t = dijkstra(i);
        if (t < cnt) {
            cnt = t;
            ans = i;
        }
    }

    return ans;
}

1334 - Find the City With the Smallest Number of Neighbors at a Threshold Distance

1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance

Description

Solutions

All Problems

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1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance

Description

Solutions

All Problems

All Solutions