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
View More Similar Problems
Self Balancing Tree
An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ
View Solution →Array and simple queries
Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty
View Solution →Median Updates
The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o
View Solution →Maximum Element
You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each
View Solution →Balanced Brackets
A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra
View Solution →Equal Stacks
ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of
View Solution →