For a given bidirectional weighted graph, find the node (city) with the least number of nodes reachable within a given threshold.

Note: If there are multiple such nodes, then return the node with the highest value among them.

Time Complexity: O(n^3)

```
const findtheCity = (n, edges, threshold) => {
const adjMatrix = Array(n);
for (let i = 0; i < n; i++) {
adjMatrix[i] = [];
for (let j = 0; j < n; j++) {
adjMatrix[i][j] = (i === j) ? 0 : Number.MAX_VALUE;
}
}
for (const val of edges) {
adjMatrix[val[0]][val[1]] = val[2];
adjMatrix[val[1]][val[0]] = val[2];
}
fw(adjMatrix);
let minNeighbours = Number.MAX_VALUE;
let neighbour;
for (let i = 0; i < n; i++) {
let currNeighbours = 0;
for (let j = 0; j < n; j++) {
if (adjMatrix[i][j] <= threshold) {
currNeighbours++;
}
}
if (currNeighbours <= minNeighbours) {
minNeighbours = currNeighbours;
neighbour = i;
}
}
return neighbour;
};
const fw = (adjMatrix) => {
const n = adjMatrix.length;
for (let k = 0; k < n; k++) {
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
adjMatrix[i][j] = Math.min(adjMatrix[i][j],
adjMatrix[i][k] + adjMatrix[k][j]);
}
}
}
};
console.log(findtheCity(4, [
[0, 1, 3],
[1, 2, 1],
[1, 3, 4],
[2, 3, 1]
], 4)); // 3
```