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 :



title-img



                        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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