Minimize Maximum Stadium Size - Google Top Interview Questions
Problem Statement :
You are given an integer n and a two-dimensional list of integers views.
There are n teams in a tournament and you want to build two stadiums to host the games.
Each element in views contains [a, b, people] meaning that when team a and team b compete, there are people number of people that will view that game.
The viewership numbers among every pair of teams will be in views.
You must permanently assign each team to a stadium. Every team in a stadium plays every other team in that stadium, and only within that stadium.
Return the smallest possible maximum capacity of the two stadiums, where the capacity of a stadium is the number of seats required to fit the largest audience of a game in that stadium.
Constraints
2 ≤ n ≤ 500
1 ≤ m ≤ 200,000 where m is the length of views
Example 1
Input
n = 3
views = [
[0, 1, 100],
[0, 2, 900],
[1, 2, 500]
]
Output
100
Explanation
We can put teams 0 and 1 in one stadium and team 2 in the other stadium. This means team 2 doesn't
have to compete with any other team. Then, for the first stadium we need capacity of 100 to host the
game between teams 0 and 1.
Example 2
Input
n = 2
views = [
[0, 1, 100]
]
Output
0
Explanation
If we put each team in separate stadiums, then they don't have to compete.
Example 3
Input
n = 4
views = [
[0, 1, 4],
[0, 2, 1],
[0, 3, 5],
[1, 2, 4],
[1, 3, 3],
[2, 3, 4]
]
Output
3
Explanation
We can place teams 0 and 2 in one stadium, and 1 and 3 in the other stadium. In the first stadium, we
need capacity of 1 and in the second stadium we need capacity of 3.
Solution :
Solution in C++ :
int n, graph[501][501], which_stadium[501];
bool possible(int team, int stadium, const int& max_people) {
which_stadium[team] = stadium;
for (int i = 0; i < n; i++) {
if (which_stadium[i] != -1 && graph[team][i] > max_people &&
which_stadium[i] != (stadium ^ 1))
return false;
else if (which_stadium[i] == -1 && graph[team][i] > max_people)
if (!possible(i, stadium ^ 1, max_people)) return false;
}
return true;
}
int solve(int n, vector<vector<int>>& views) {
::n = n;
int left = 0, right = 0;
for (auto& v : views) right = max(right, graph[v[0]][v[1]] = graph[v[1]][v[0]] = v[2]);
auto can = [&](int mid) {
fill(which_stadium, which_stadium + n, -1);
for (int i = 0; i < n; i++)
if (which_stadium[i] == -1 && !possible(i, 0, mid)) return false;
return true;
};
while (left < right) {
int mid = left + (right - left) / 2;
if (can(mid))
right = mid;
else
left = mid + 1;
}
return right;
}
Solution in Java :
import java.util.*;
class Solution {
static ArrayList<Integer>[] graph;
public int solve(int N, int[][] views) {
sort(views);
graph = new ArrayList[N];
for (int i = 0; i < N; i++) {
graph[i] = new ArrayList<Integer>();
}
for (int i = views.length - 1; i >= 0; i--) {
// update the graph
graph[views[i][0]].add(views[i][1]);
graph[views[i][1]].add(views[i][0]);
// if there's an odd cycle, its impossible to assign stadiums.
if (!bipartite()) {
return views[i][2];
}
}
return 0;
}
public static boolean bipartite() {
int N = graph.length;
// assign 1-2 colors
int[] color = new int[N];
for (int i = 0; i < N; i++) {
if (color[i] == 0) {
color[i] = 1;
ArrayDeque<Integer> bfs = new ArrayDeque<Integer>();
bfs.add(i);
while (!bfs.isEmpty()) {
int n = bfs.pollFirst();
for (int next : graph[n]) {
if (color[next] == 0) {
color[next] = 3 - color[n];
bfs.add(next);
} else if (color[next] == color[n]) {
return false;
}
}
}
}
}
return true;
}
public static void sort(int[][] arr) {
Arrays.sort(arr, new Comparator<int[]>() {
@Override
public int compare(int[] a, int[] b) {
return a[2] - b[2];
// Ascending order.
}
});
}
}
Solution in Python :
class Solution:
def solve(self, n, views):
if n <= 2:
return 0
dist = [[-1] * n for _ in range(n)]
e = collections.defaultdict(list)
for x, y, v in views:
dist[x][y] = v
dist[y][x] = v
def two_colorable(target):
color = [-1] * n
found = False
def go(x, c, target):
for y in range(n):
v = dist[x][y]
if v > target:
if color[x] == color[y]:
nonlocal found
found = True
return
elif color[y] == -1:
color[y] = 1 - c
go(y, 1 - c, target)
for x in range(n):
if color[x] == -1:
color[x] = 1
go(x, 1, target)
if found:
return False
return True
left = 0
right = 10 ** 10
while left < right:
mid = (left + right) // 2
if two_colorable(mid):
right = mid
else:
left = mid + 1
return left
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