Cheapest Cost to All Cities - Google Top Interview Questions

Problem Statement :

```You are given two lists of integers costs_from and costs_to of the same length where each index i represents a city.

Building a one-way road from city i to j costs costs_from[i] + costs_to[j].

You are also given a two-dimensional list of integers edges where each list contains [x, y] meaning there's already a one-way road from city x to y.

Given that we want to be able to go to any city from city 0 (not necessarily in one path), return the minimum cost to build the necessary roads.

Constraints

n ≤ 1,000 where n is the length of costs_from and costs_to

m ≤ 100,000 where m is the length of edges

Example 1

Input

costs_from = [5, 1, 1, 10]

costs_to = [1, 1, 2, 1]

edges = [

[0, 3]

]

Output

9

Explanation

We can go 0 to 2 for a cost of 7. Then, we can go 2 to 1 for a cost of 2. We already have the ability to go

Solution :

```                        ```Solution in C++ :

const int M = 1000000;
const int N = 1000;

struct edg {
int u, v;
int cost;
} E[M], E_copy[M];

int In[N], ID[N], vis[N], pre[N];

int Directed_MST(int root, int NV, int NE) {
for (int i = 0; i < NE; i++) {
E_copy[i] = E[i];
}
int ret = 0;
int u, v;
while (true) {
for (int i = 0; i < NV; i++) {
In[i] = 2000000000;
}
for (int i = 0; i < NE; i++) {
u = E_copy[i].u;
v = E_copy[i].v;
if (E_copy[i].cost < In[v] && u != v) {
In[v] = E_copy[i].cost;
pre[v] = u;
}
}

int cnt = 0;
for (int i = 0; i < NV; i++) {
ID[i] = -1;
vis[i] = -1;
}
In[root] = 0;

// check for cycles
for (int i = 0; i < NV; i++) {
ret += In[i];
int v = i;
while (vis[v] != i && ID[v] == -1 && v != root) {
vis[v] = i;
v = pre[v];
}
if (v != root && ID[v] == -1) {
for (u = pre[v]; u != v; u = pre[u]) {
ID[u] = cnt;
}
ID[v] = cnt++;
}
}
// check if there are no cycles
if (cnt == 0) {
break;
}

// condense cycles
for (int i = 0; i < NV; i++) {
if (ID[i] == -1) {
ID[i] = cnt++;
}
}
for (int i = 0; i < NE; i++) {
v = E_copy[i].v;
E_copy[i].u = ID[E_copy[i].u];
E_copy[i].v = ID[E_copy[i].v];
if (E_copy[i].u != E_copy[i].v) {
E_copy[i].cost -= In[v];
}
}
NV = cnt;
root = ID[root];
}
return ret;
}

int solve(vector<int>& costs_from, vector<int>& costs_to, vector<vector<int>>& edges) {
int n = costs_from.size();
vector<pair<int, int>> v;
for (auto& e : edges) {
v.emplace_back(e[0], e[1]);
}
sort(v.begin(), v.end());
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
pair<int, int> key = make_pair(i, j);
auto it = lower_bound(v.begin(), v.end(), key);
if (it != v.end() && *it == key) {
E[i * n + j].cost = 0;
} else {
E[i * n + j].cost = costs_from[i] + costs_to[j];
}
E[i * n + j].u = i;
E[i * n + j].v = j;
}
}
return Directed_MST(0, n, n * n);
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[] costs_from, int[] costs_to, int[][] edges) {
int n = costs_to.length;
ArrayList<Integer>[] es = new ArrayList[n];
int[] rd = new int[n], low = new int[n], st = new int[n], en = new int[n];
for (int i = 0; i < n; ++i) {
rd[i] = -1;
st[i] = costs_from[i];
en[i] = costs_to[i];
}
Arrays.fill(low, n + 1);
boolean[] path = new boolean[n], seen = new boolean[n];
for (int[] e : edges) {
if (es[e[0]] == null)
es[e[0]] = new ArrayList();
}
int from = mark(es, 0, seen, costs_from), res = 0;
for (int i = 1; i < n; ++i) {
if (!seen[i]) {
ArrayList<Integer> al = new ArrayList();
int[] ms = new int[1];
ms[0] = costs_from[i];
int t = costs_to[i];
traverse(es, st, rd, 0, low, new Stack<Integer>(), al, i, seen, path, ms, i);
for (int a : al) {
rd[a] = i;
st[a] = ms[0];
t = Math.min(t, costs_to[a]);
}
for (int a : al) {
en[a] = t;
}
}
}
ArrayList<int[]> al = new ArrayList();
for (int i = 1; i < n; ++i) {
int t = find(rd, i);
if (t >= 0) {
rd[t] = -1;
}
}
if (al.size() == 0)
return 0;
Collections.sort(al, (a, b) -> (a[0] - b[0]));
for (int[] d : al) {
res += from + d[1];
from = Math.min(from, d[0]);
}
return res;
}
private int mark(ArrayList<Integer>[] es, int idx, boolean[] seen, int[] fr) {
seen[idx] = true;
int res = fr[idx];
if (es[idx] != null) {
for (int a : es[idx]) {
if (!seen[a]) {
int t = mark(es, a, seen, fr);
res = Math.min(res, t);
}
}
}
return res;
}
private int find(int[] rd, int idx) {
while (idx >= 0 && idx != rd[idx]) {
int t = rd[idx];
if (t >= 0)
rd[idx] = rd[t];
idx = rd[idx];
}
return idx;
}
private void traverse(ArrayList<Integer>[] es, int[] st, int[] rd, int h, int[] low,
Stack<Integer> sta, ArrayList<Integer> al, int idx, boolean[] seen, boolean[] path,
int[] ms, int root) {
seen[idx] = true;
low[idx] = h;
path[idx] = true;
sta.push(idx);
if (es[idx] != null) {
for (int a : es[idx]) {
ms[0] = Math.min(ms[0], st[a]);
int t = find(rd, a);
if (t >= 0)
rd[t] = -1;
if (!seen[a]) {
traverse(es, st, rd, h + 1, low, sta, al, a, seen, path, ms, root);
low[idx] = Math.min(low[idx], low[a]);
} else if (path[a]) {
low[idx] = Math.min(low[idx], low[a]);
}
}
}
if (low[idx] == h) {
while (!sta.isEmpty()) {
int t = sta.pop();
if (idx == root)
path[t] = false;
if (t == idx)
break;
}
}
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, costs_from, costs_to, edges):
for u, v in edges:
if v != 0:

idx = -1
start = {}
s = []
in_stack = set()
scc = {}
root = {}

def tarjan(v):
nonlocal idx
idx += 1
start[v] = idx
s.append(v)

if n not in start:
tarjan(n)
elif n in in_stack:

new_scc = []
while s[-1] != v:
in_stack.remove(s[-1])
root[s[-1]] = v
new_scc.append(s.pop())

in_stack.remove(s[-1])
root[s[-1]] = v
new_scc.append(s.pop())

scc[v] = new_scc

for i in range(len(costs_from)):
if i not in start:
tarjan(i)

new_from = {}
new_to = {}

for r in scc:
to = fr = float("inf")
for n in scc[r]:
to = min(to, costs_to[n])
fr = min(fr, costs_from[n])

new_from[r] = fr
new_to[r] = to

visited = set()
sources = {}

def dfs(cur):
m = new_from[cur]
if n in sources:
m = min(m, sources[n])
del sources[n]
elif n not in visited:
m = min(m, dfs(n))

return m

for v in new_to.keys():
if v not in visited:
sources[v] = dfs(v)

src = min(sources.keys(), key=sources.get)
res = sources[0] - sources[src]
for v in sources:
if v != 0:
res += sources[src] + new_to[v]

return res```
```

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