Matrix Relations - Google Top Interview Questions
Problem Statement :
You are given an integer n and a two-dimensional list of integers relations.
You want to fill an n by n matrix using relations, which defines the spatial ordering of some cell values in the matrix. Each element relations[i] contains (x, y, type) which means that
x is left of y if type = 0
x is right of y if type = 1
x is above y if type = 2
x is below y if type = 3
Return the n by n matrix following the constraints in relations. Since some cells may not have a value,
set it to -1. You can assume each defined cell in the matrix will have a unique value. Also, there is one
unique solution.
Constraints
n ≤ 500
1 ≤ m ≤ 100,000 where m is the length of relations
Example 1
Input
n = 3
relations = [
[1, 2, 0],
[2, 3, 0],
[1, 2, 2],
[2, 3, 2]
]
Output
[
[1, -1, -1],
[-1, 2, -1],
[-1, -1, 3]
]
Explanation
The relations says:
1 is left of 2
2 is left of 3
1 is above 2
2 is above 3
Other unfilled squares are set to -1.
Solution :
Solution in C++ :
vector<vector<int>> solve(int n, vector<vector<int>> &relations) {
unordered_map<int, vector<int>> left, above;
unordered_map<int, int> indegLeft, indegAbove;
unordered_map<int, pair<int, int>> mp;
for (int i = 0; i < relations.size(); i++) {
int x = relations[i][0], y = relations[i][1], type = relations[i][2];
mp[x] = {-1, -1};
mp[y] = {-1, -1};
if (type == 0 || type == 1) {
if (type == 1) {
swap(x, y);
}
left[y].push_back(x);
indegLeft[x]++;
indegLeft[y] = indegLeft[y];
} else {
if (type == 3) {
swap(x, y);
}
above[y].push_back(x);
indegAbove[x]++;
indegAbove[y] = indegAbove[y];
}
}
queue<int> q;
for (auto it = indegLeft.begin(); it != indegLeft.end(); it++) {
if (it->second == 0) {
q.push(it->first);
}
}
int j = n - 1;
while (!q.empty()) {
int s = q.size();
while (s--) {
int p = q.front();
q.pop();
mp[p].second = j;
vector<int> &l = left[p];
for (int i = 0; i < l.size(); i++) {
indegLeft[l[i]]--;
if (indegLeft[l[i]] == 0) {
q.push(l[i]);
}
}
}
j--;
}
for (auto it = indegAbove.begin(); it != indegAbove.end(); it++) {
if (it->second == 0) {
q.push(it->first);
}
}
int i = n - 1;
while (!q.empty()) {
int s = q.size();
while (s--) {
int p = q.front();
q.pop();
mp[p].first = i;
vector<int> &l = above[p];
for (int j = 0; j < l.size(); j++) {
indegAbove[l[j]]--;
if (indegAbove[l[j]] == 0) {
q.push(l[j]);
}
}
}
i--;
}
vector<vector<int>> res(n, vector<int>(n, -1));
for (auto it = mp.begin(); it != mp.end(); it++) {
int val = it->first;
int x = it->second.first, y = it->second.second;
res[x][y] = val;
}
return res;
}
Solution in Java :
import java.util.*;
class Solution {
public int[][] solve(int n, int[][] r) {
int[][] ans = new int[n][n];
// Initialize the matrix
for (int[] row : ans) {
Arrays.fill(row, -1);
}
Map<Integer, Set<Integer>> downTo = new HashMap<>();
Map<Integer, Set<Integer>> rightTo = new HashMap<>();
Map<Integer, Set<Integer>> upTo = new HashMap<>();
Map<Integer, Set<Integer>> leftTo = new HashMap<>();
Map<Integer, Integer> xaxis = new HashMap<>();
Map<Integer, Integer> yaxis = new HashMap<>();
for (int i = 0; i < r.length; i++) {
// I change all the right or below direction to left or above
if (r[i][2] == 1 || r[i][2] == 3) {
r[i][2]--;
int tmp = r[i][0];
r[i][0] = r[i][1];
r[i][1] = tmp;
}
int x = r[i][0];
int y = r[i][1];
// Build set if a map doesn't have the key.
rightTo.computeIfAbsent(x, k -> new HashSet<>());
rightTo.computeIfAbsent(y, k -> new HashSet<>());
leftTo.computeIfAbsent(x, k -> new HashSet<>());
leftTo.computeIfAbsent(y, k -> new HashSet<>());
downTo.computeIfAbsent(x, k -> new HashSet<>());
downTo.computeIfAbsent(y, k -> new HashSet<>());
upTo.computeIfAbsent(x, k -> new HashSet<>());
upTo.computeIfAbsent(y, k -> new HashSet<>());
if (r[i][2] == 0) {
// x left to y
rightTo.get(x).add(y);
leftTo.get(y).add(x);
} else {
// x up to y
downTo.get(x).add(y);
upTo.get(y).add(x);
}
}
Queue<Integer> q = new LinkedList<>();
// If degree is 0 then we add it to the queue
for (int k : rightTo.keySet()) {
if (rightTo.get(k).size() == 0) {
q.offer(k);
}
}
int idx = n - 1;
while (!q.isEmpty()) {
int size = q.size();
while (size > 0) {
size--;
int cur = q.poll();
// System.out.println(cur);
yaxis.put(cur, idx);
for (int p : leftTo.get(cur)) {
rightTo.get(p).remove(cur);
if (rightTo.get(p).size() == 0) {
q.offer(p);
}
}
}
idx--;
}
q = new LinkedList<>();
for (int k : downTo.keySet()) {
if (downTo.get(k).size() == 0) {
q.offer(k);
}
}
idx = n - 1;
while (!q.isEmpty()) {
int size = q.size();
while (size > 0) {
size--;
int cur = q.poll();
// System.out.println(cur);
xaxis.put(cur, idx);
for (int p : upTo.get(cur)) {
downTo.get(p).remove(cur);
if (downTo.get(p).size() == 0) {
q.offer(p);
}
}
}
idx--;
}
for (int key : xaxis.keySet()) {
ans[xaxis.get(key)][yaxis.get(key)] = key;
}
return ans;
}
}
Solution in Python :
class Solution:
def solve(self, N, relations):
rindegree = collections.Counter()
cindegree = collections.Counter()
rlist = collections.defaultdict(list)
clist = collections.defaultdict(list)
keys = set()
# Parsing queries
for x, y, t in relations:
keys.add(x)
keys.add(y)
if t == 0:
rindegree[y] += 1
rlist[x].append(y)
elif t == 1:
rindegree[x] += 1
rlist[y].append(x)
if t == 2:
cindegree[y] += 1
clist[x].append(y)
elif t == 3:
cindegree[x] += 1
clist[y].append(x)
# Get the rows (or cols) mapping of the indegree and list
def topsort(indegree, elist):
ans = {}
q = collections.deque()
for x in keys:
if indegree[x] == 0:
q.append((x, 0))
while len(q) > 0:
index, d = q.popleft()
ans[index] = d
for v in elist[index]:
if indegree[v] > 0:
indegree[v] -= 1
if indegree[v] == 0:
q.append((v, d + 1))
return ans
rans = topsort(rindegree, rlist)
cans = topsort(cindegree, clist)
ans = [[-1] * N for _ in range(N)]
for x in keys:
ans[cans[x]][rans[x]] = x
return ans
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