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 
to 3 from 0 for free.



Solution :



title-img




                        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();
            es[e[0]].add(e[1]);
        }
        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;
                al.add(new int[] {st[t], en[t]});
            }
        }
        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)
                    al.add(t);
                path[t] = false;
                if (t == idx)
                    break;
            }
        }
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, costs_from, costs_to, edges):
        adj = defaultdict(list)
        for u, v in edges:
            if v != 0:
                adj[u].append(v)

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

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

            for n in adj[v]:
                if n not in start:
                    tarjan(n)
                    lowlink[v] = min(lowlink[v], lowlink[n])
                elif n in in_stack:
                    lowlink[v] = min(lowlink[v], start[n])

            if lowlink[v] == start[v]:
                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_adj = defaultdict(set)
        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_adj[r].update(root[k] for k in adj[n])

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

        visited = set()
        sources = {}

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

            return m

        for v in new_to.keys():
            if v not in visited:
                visited.add(v)
                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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