Formatted question description: https://leetcode.ca/all/2392.html

2392. Build a Matrix With Conditions

Difficulty: Hard.
Related Topics: Array, Graph, Topological Sort, Matrix.
Similar Questions: Course Schedule, Course Schedule II, Find Eventual Safe States, Loud and Rich.

Problem

You are given a positive integer k. You are also given:

a 2D integer array rowConditions of size n where rowConditions[i] = [abovei, belowi], and
a 2D integer array colConditions of size m where colConditions[i] = [lefti, righti].

The two arrays contain integers from 1 to k.

You have to build a k x k matrix that contains each of the numbers from 1 to k exactly once. The remaining cells should have the value 0.

The matrix should also satisfy the following conditions:

The number abovei should appear in a row that is strictly above the row at which the number belowi appears for all i from 0 to n - 1.
The number lefti should appear in a column that is strictly left of the column at which the number righti appears for all i from 0 to m - 1.

Return **any matrix that satisfies the conditions**. If no answer exists, return an empty matrix.

Example 1:

Input: k = 3, rowConditions = [[1,2],[3,2]], colConditions = [[2,1],[3,2]]
Output: [[3,0,0],[0,0,1],[0,2,0]]
Explanation: The diagram above shows a valid example of a matrix that satisfies all the conditions.
The row conditions are the following:
- Number 1 is in row 1, and number 2 is in row 2, so 1 is above 2 in the matrix.
- Number 3 is in row 0, and number 2 is in row 2, so 3 is above 2 in the matrix.
The column conditions are the following:
- Number 2 is in column 1, and number 1 is in column 2, so 2 is left of 1 in the matrix.
- Number 3 is in column 0, and number 2 is in column 1, so 3 is left of 2 in the matrix.
Note that there may be multiple correct answers.

Example 2:

Input: k = 3, rowConditions = [[1,2],[2,3],[3,1],[2,3]], colConditions = [[2,1]]
Output: []
Explanation: From the first two conditions, 3 has to be below 1 but the third conditions needs 3 to be above 1 to be satisfied.
No matrix can satisfy all the conditions, so we return the empty matrix.

Constraints:

2 <= k <= 400
1 <= rowConditions.length, colConditions.length <= 104
rowConditions[i].length == colConditions[i].length == 2
1 <= abovei, belowi, lefti, righti <= k
abovei != belowi
lefti != righti

Solution

class Solution {
    List<Integer>[] al;
    
    public int[][] buildMatrix(int k, int[][] rc, int[][] cc) {
        
        int ans[][] = new int[k][k];
        this.al = new ArrayList[k+1];
        
        for(int i=0;i<=k;i++) al[i] = new ArrayList<>();
        for(int r[] : rc) al[r[0]].add(r[1]);
        
        boolean[] vis = new boolean[k+1];
        boolean dfsvis[] = new boolean[k+1];
        List<Integer> rowStack = new ArrayList<>();
        
        for(int node=1;node<=k;node++){
            if(!vis[node]) {
                if(!dfs(node,vis,rowStack,dfsvis)) return new int[0][0];
            }
        }

        for(int i=0;i<=k;i++) al[i] = new ArrayList<>();
        for(int c[] : cc) al[c[0]].add(c[1]);
        List<Integer> colStack = new ArrayList<>();
        
        vis = new boolean[k+1];
        dfsvis = new boolean[k+1];
        
        for(int node=1;node<=k;node++){
            if(!vis[node]) {
                if(!dfs(node,vis,colStack,dfsvis)) return new int[0][0];
            }
        }
        
        HashMap<Integer,Integer> map = new HashMap<>();
        
        for(int i=0;i<colStack.size();i++){
            map.put(colStack.get(i),i);
        }
        
        int row = 0;
        
        for(int i=rowStack.size()-1;i>=0;i--){
			// should place this value leaving required cells ahead for other values to come to make column condition true. 
            ans[row++][k-map.get(rowStack.get(i))-1] = rowStack.get(i);
        }
        
        return ans;
    
    }
    
	// toposort dfs along with cycle check (returns false if cycle exists)
    private boolean dfs(int node,boolean[] vis,List<Integer> stack,boolean[] dfsvis){
        vis[node] = true;
        dfsvis[node] = true;
        for(int next : al[node]){
            if(!vis[next]) {
                if(!dfs(next,vis,stack,dfsvis)) return false;
            }else if(dfsvis[next]) return false;
        }
        stack.add(node);
        dfsvis[node] = false;
        return true;
    }
}

############

class Solution {
    private int k;

    public int[][] buildMatrix(int k, int[][] rowConditions, int[][] colConditions) {
        this.k = k;
        List<Integer> row = f(rowConditions);
        List<Integer> col = f(colConditions);
        if (row == null || col == null) {
            return new int[0][0];
        }
        int[][] ans = new int[k][k];
        int[] m = new int[k + 1];
        for (int i = 0; i < k; ++i) {
            m[col.get(i)] = i;
        }
        for (int i = 0; i < k; ++i) {
            ans[i][m[row.get(i)]] = row.get(i);
        }
        return ans;
    }

    private List<Integer> f(int[][] cond) {
        List<Integer>[] g = new List[k + 1];
        Arrays.setAll(g, key -> new ArrayList<>());
        int[] indeg = new int[k + 1];
        for (var e : cond) {
            int a = e[0], b = e[1];
            g[a].add(b);
            ++indeg[b];
        }
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 1; i < indeg.length; ++i) {
            if (indeg[i] == 0) {
                q.offer(i);
            }
        }
        List<Integer> res = new ArrayList<>();
        while (!q.isEmpty()) {
            for (int n = q.size(); n > 0; --n) {
                int i = q.pollFirst();
                res.add(i);
                for (int j : g[i]) {
                    if (--indeg[j] == 0) {
                        q.offer(j);
                    }
                }
            }
        }
        return res.size() == k ? res : null;
    }
}

class Solution:
    def buildMatrix(
        self, k: int, rowConditions: List[List[int]], colConditions: List[List[int]]
    ) -> List[List[int]]:
        def f(cond):
            g = defaultdict(list)
            indeg = [0] * (k + 1)
            for a, b in cond:
                g[a].append(b)
                indeg[b] += 1
            q = deque([i for i, v in enumerate(indeg[1:], 1) if v == 0])
            res = []
            while q:
                for _ in range(len(q)):
                    i = q.popleft()
                    res.append(i)
                    for j in g[i]:
                        indeg[j] -= 1
                        if indeg[j] == 0:
                            q.append(j)
            return None if len(res) != k else res

        row = f(rowConditions)
        col = f(colConditions)
        if row is None or col is None:
            return []
        ans = [[0] * k for _ in range(k)]
        m = [0] * (k + 1)
        for i, v in enumerate(col):
            m[v] = i
        for i, v in enumerate(row):
            ans[i][m[v]] = v
        return ans

############

# 2392. Build a Matrix With Conditions
# https://leetcode.com/problems/build-a-matrix-with-conditions

class Solution:
    def buildMatrix(self, k: int, rowConditions: List[List[int]], colConditions: List[List[int]]) -> List[List[int]]:
        res = [[0] * k for _ in range(k)]
        
        def topoSort(A):
            adj = [[] for _ in range(k + 1)] 
            
            for a, b in A:
                adj[a].append(b)
            
            seen = [False] * (k + 1)
            order = []
            
            def dfs(x):
                if seen[x]: return
                
                seen[x] = True
                
                for nei in adj[x]:
                    if not seen[nei]:
                        dfs(nei)
                
                order.append(x)
            
            for x in range(1, k + 1):
                dfs(x)
            
            order.reverse()
            M = {j: i for i, j in enumerate(order)}
            
            if all(M[i] < M[j] for i, j in A):
                return order
            
            return None
        
        R = topoSort(rowConditions)
        C = topoSort(colConditions)
        
        if R is None or C is None:
            return []
        
        RM = {j : i for i, j in enumerate(R)}
        CM = {j : i for i, j in enumerate(C)}
        
        for x in range(1, k + 1):
            res[RM[x]][CM[x]] = x
            
        return res

class Solution {
public:
    int k;

    vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
        this->k = k;
        auto row = f(rowConditions);
        auto col = f(colConditions);
        if (row.empty() || col.empty()) return {};
        vector<vector<int>> ans(k, vector<int>(k));
        vector<int> m(k + 1);
        for (int i = 0; i < k; ++i) {
            m[col[i]] = i;
        }
        for (int i = 0; i < k; ++i) {
            ans[i][m[row[i]]] = row[i];
        }
        return ans;
    }

    vector<int> f(vector<vector<int>>& cond) {
        vector<vector<int>> g(k + 1);
        vector<int> indeg(k + 1);
        for (auto& e : cond) {
            int a = e[0], b = e[1];
            g[a].push_back(b);
            ++indeg[b];
        }
        queue<int> q;
        for (int i = 1; i < k + 1; ++i) {
            if (!indeg[i]) {
                q.push(i);
            }
        }
        vector<int> res;
        while (!q.empty()) {
            for (int n = q.size(); n; --n) {
                int i = q.front();
                res.push_back(i);
                q.pop();
                for (int j : g[i]) {
                    if (--indeg[j] == 0) {
                        q.push(j);
                    }
                }
            }
        }
        return res.size() == k ? res : vector<int>();
    }
};

func buildMatrix(k int, rowConditions [][]int, colConditions [][]int) [][]int {
	f := func(cond [][]int) []int {
		g := make([][]int, k+1)
		indeg := make([]int, k+1)
		for _, e := range cond {
			a, b := e[0], e[1]
			g[a] = append(g[a], b)
			indeg[b]++
		}
		q := []int{}
		for i, v := range indeg[1:] {
			if v == 0 {
				q = append(q, i+1)
			}
		}
		res := []int{}
		for len(q) > 0 {
			for n := len(q); n > 0; n-- {
				i := q[0]
				q = q[1:]
				res = append(res, i)
				for _, j := range g[i] {
					indeg[j]--
					if indeg[j] == 0 {
						q = append(q, j)
					}
				}
			}
		}
		if len(res) == k {
			return res
		}
		return []int{}
	}

	row := f(rowConditions)
	col := f(colConditions)
	if len(row) == 0 || len(col) == 0 {
		return [][]int{}
	}
	m := make([]int, k+1)
	for i, v := range col {
		m[v] = i
	}
	ans := make([][]int, k)
	for i := range ans {
		ans[i] = make([]int, k)
	}
	for i, v := range row {
		ans[i][m[v]] = v
	}
	return ans
}

function buildMatrix(
    k: number,
    rowConditions: number[][],
    colConditions: number[][],
): number[][] {
    function f(cond) {
        const g = Array.from({ length: k + 1 }, () => []);
        const indeg = new Array(k + 1).fill(0);
        for (const [a, b] of cond) {
            g[a].push(b);
            ++indeg[b];
        }
        const q = [];
        for (let i = 1; i < indeg.length; ++i) {
            if (indeg[i] == 0) {
                q.push(i);
            }
        }
        const res = [];
        while (q.length) {
            for (let n = q.length; n; --n) {
                const i = q.shift();
                res.push(i);
                for (const j of g[i]) {
                    if (--indeg[j] == 0) {
                        q.push(j);
                    }
                }
            }
        }
        return res.length == k ? res : [];
    }

    const row = f(rowConditions);
    const col = f(colConditions);
    if (!row.length || !col.length) return [];
    const ans = Array.from({ length: k }, () => new Array(k).fill(0));
    const m = new Array(k + 1).fill(0);
    for (let i = 0; i < k; ++i) {
        m[col[i]] = i;
    }
    for (let i = 0; i < k; ++i) {
        ans[i][m[row[i]]] = row[i];
    }
    return ans;
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).

2392 - Build a Matrix With Conditions

2392. Build a Matrix With Conditions

Problem

Solution

All Problems

All Solutions

Welcome to Subscribe On Youtube

2392. Build a Matrix With Conditions

Problem

Solution

All Problems

All Solutions