1766. Tree of Coprimes

Description

There is a tree (i.e., a connected, undirected graph that has no cycles) consisting of n nodes numbered from 0 to n - 1 and exactly n - 1 edges. Each node has a value associated with it, and the root of the tree is node 0.

To represent this tree, you are given an integer array nums and a 2D array edges. Each nums[i] represents the i^th node's value, and each edges[j] = [u_j, v_j] represents an edge between nodes u_j and v_j in the tree.

Two values x and y are coprime if gcd(x, y) == 1 where gcd(x, y) is the greatest common divisor of x and y.

An ancestor of a node i is any other node on the shortest path from node i to the root. A node is not considered an ancestor of itself.

Return an array ans of size n, where ans[i] is the closest ancestor to node i such that nums[i] and nums[ans[i]] are coprime, or -1 if there is no such ancestor.

Example 1:

Input: nums = [2,3,3,2], edges = [[0,1],[1,2],[1,3]]
Output: [-1,0,0,1]
Explanation: In the above figure, each node's value is in parentheses.
- Node 0 has no coprime ancestors.
- Node 1 has only one ancestor, node 0. Their values are coprime (gcd(2,3) == 1).
- Node 2 has two ancestors, nodes 1 and 0. Node 1's value is not coprime (gcd(3,3) == 3), but node 0's
  value is (gcd(2,3) == 1), so node 0 is the closest valid ancestor.
- Node 3 has two ancestors, nodes 1 and 0. It is coprime with node 1 (gcd(3,2) == 1), so node 1 is its
  closest valid ancestor.

Example 2:

Input: nums = [5,6,10,2,3,6,15], edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]]
Output: [-1,0,-1,0,0,0,-1]

Constraints:

nums.length == n
1 <= nums[i] <= 50
1 <= n <= 10⁵
edges.length == n - 1
edges[j].length == 2
0 <= u_j, v_j < n
u_j != v_j

Solutions

Solution 2: Preprocessing + Enumeration + Explicit Stack

Thinking

For each node we want the closest ancestor whose value is coprime with it. With $n\le 10^5$ and values in $[1,50]$, one ancestor stack per value is enough. Recursing into each child from the root is too deep: a chain makes the call depth $n$.

Precompute coprime pairs in $1..50$. On a visit, look at the stack top of every value coprime to $nums[i]$ and take the deepest.

The explicit stack stores $(node, parent, depth, next child)$. Before walking to a child we push $(i,\textit{depth})$ onto the stack of $nums[i]$ and pop it after that child is finished, so the top stays the nearest ancestor.

Since the range of $nums[i]$ in the problem is $[1, 50]$, we can preprocess all the coprime numbers for each number and record them in the array $f$, where $f[i]$ represents all the coprime numbers of $i$.

Next, an explicit stack walks the tree in preorder from the root. For each node $i$, enumerate the values coprime to $nums[i]$ and take the ancestor $t$ of maximum depth among those stack tops; that is the nearest coprime ancestor. The stack array $stks$ has length $51$, and $stks[v]$ records nodes of value $v$ on the current root path together with their depths. Before walking to a child, push $(i,\textit{depth})$ onto $stks[nums[i]]$, and pop it after that child is finished.

The time complexity is $O(n \times M)$, and the space complexity is $O(M^2 + n)$. Where $n$ is the number of nodes, and $M$ is the maximum value of $nums[i]$, in this problem $M = 50$.

class Solution {
    private List<Integer>[] g;
    private List<Integer>[] f;
    private Deque<int[]>[] stks;
    private int[] nums;
    private int[] ans;

    public int[] getCoprimes(int[] nums, int[][] edges) {
        int n = nums.length;
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int u = e[0], v = e[1];
            g[u].add(v);
            g[v].add(u);
        }
        f = new List[51];
        stks = new Deque[51];
        Arrays.setAll(f, k -> new ArrayList<>());
        Arrays.setAll(stks, k -> new ArrayDeque<>());
        for (int i = 1; i < 51; ++i) {
            for (int j = 1; j < 51; ++j) {
                if (gcd(i, j) == 1) {
                    f[i].add(j);
                }
            }
        }
        this.nums = nums;
        ans = new int[n];
        dfs(0, -1, 0);
        return ans;
    }

    private void dfs(int i, int fa, int depth) {
        int t = -1, k = -1;
        for (int v : f[nums[i]]) {
            var stk = stks[v];
            if (!stk.isEmpty() && stk.peek()[1] > k) {
                t = stk.peek()[0];
                k = stk.peek()[1];
            }
        }
        ans[i] = t;
        for (int j : g[i]) {
            if (j != fa) {
                stks[nums[i]].push(new int[] {i, depth});
                dfs(j, i, depth + 1);
                stks[nums[i]].pop();
            }
        }
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}


// Solution 2
class Solution {
    public int[] getCoprimes(int[] nums, int[][] edges) {
        int n = nums.length;
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int u = e[0], v = e[1];
            g[u].add(v);
            g[v].add(u);
        }
        List<Integer>[] f = new List[51];
        Deque<int[]>[] stks = new Deque[51];
        Arrays.setAll(f, k -> new ArrayList<>());
        Arrays.setAll(stks, k -> new ArrayDeque<>());
        for (int i = 1; i < 51; ++i) {
            for (int j = 1; j < 51; ++j) {
                if (gcd(i, j) == 1) {
                    f[i].add(j);
                }
            }
        }
        int[] ans = new int[n];
        Deque<int[]> stk = new ArrayDeque<>();
        stk.push(new int[] {0, -1, 0, 0});
        while (!stk.isEmpty()) {
            int[] cur = stk.peek();
            int i = cur[0], fa = cur[1], depth = cur[2], k = cur[3];
            if (k == 0) {
                int t = -1, mx = -1;
                for (int v : f[nums[i]]) {
                    Deque<int[]> s = stks[v];
                    if (!s.isEmpty() && s.peek()[1] > mx) {
                        t = s.peek()[0];
                        mx = s.peek()[1];
                    }
                }
                ans[i] = t;
            } else {
                int jprev = g[i].get(k - 1);
                if (jprev != fa) {
                    stks[nums[i]].pop();
                }
            }
            while (k < g[i].size() && g[i].get(k) == fa) {
                ++k;
            }
            if (k == g[i].size()) {
                stk.pop();
                continue;
            }
            int j = g[i].get(k);
            cur[3] = k + 1;
            stks[nums[i]].push(new int[] {i, depth});
            stk.push(new int[] {j, i, depth + 1, 0});
        }
        return ans;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

class Solution {
public:
    vector<int> getCoprimes(vector<int>& nums, vector<vector<int>>& edges) {
        int n = nums.size();
        vector<vector<int>> g(n);
        vector<vector<int>> f(51);
        vector<stack<pair<int, int>>> stks(51);
        for (auto& e : edges) {
            int u = e[0], v = e[1];
            g[u].emplace_back(v);
            g[v].emplace_back(u);
        }
        for (int i = 1; i < 51; ++i) {
            for (int j = 1; j < 51; ++j) {
                if (__gcd(i, j) == 1) {
                    f[i].emplace_back(j);
                }
            }
        }
        vector<int> ans(n);
        function<void(int, int, int)> dfs = [&](int i, int fa, int depth) {
            int t = -1, k = -1;
            for (int v : f[nums[i]]) {
                auto& stk = stks[v];
                if (!stk.empty() && stk.top().second > k) {
                    t = stk.top().first;
                    k = stk.top().second;
                }
            }
            ans[i] = t;
            for (int j : g[i]) {
                if (j != fa) {
                    stks[nums[i]].push({i, depth});
                    dfs(j, i, depth + 1);
                    stks[nums[i]].pop();
                }
            }
        };
        dfs(0, -1, 0);
        return ans;
    }
};


// Solution 2
class Solution {
public:
    vector<int> getCoprimes(vector<int>& nums, vector<vector<int>>& edges) {
        int n = nums.size();
        vector<vector<int>> g(n);
        vector<vector<int>> f(51);
        vector<vector<pair<int, int>>> stks(51);
        for (auto& e : edges) {
            int u = e[0], v = e[1];
            g[u].emplace_back(v);
            g[v].emplace_back(u);
        }
        for (int i = 1; i < 51; ++i) {
            for (int j = 1; j < 51; ++j) {
                if (__gcd(i, j) == 1) {
                    f[i].emplace_back(j);
                }
            }
        }
        vector<int> ans(n);
        vector<array<int, 4>> stk{{0, -1, 0, 0}};
        while (!stk.empty()) {
            auto& cur = stk.back();
            int i = cur[0], fa = cur[1], depth = cur[2], k = cur[3];
            if (k == 0) {
                int t = -1, mx = -1;
                for (int v : f[nums[i]]) {
                    auto& s = stks[v];
                    if (!s.empty() && s.back().second > mx) {
                        t = s.back().first;
                        mx = s.back().second;
                    }
                }
                ans[i] = t;
            } else {
                int jprev = g[i][k - 1];
                if (jprev != fa) {
                    stks[nums[i]].pop_back();
                }
            }
            while (k < (int) g[i].size() && g[i][k] == fa) {
                ++k;
            }
            if (k == (int) g[i].size()) {
                stk.pop_back();
                continue;
            }
            int j = g[i][k];
            cur[3] = k + 1;
            stks[nums[i]].emplace_back(i, depth);
            stk.push_back({j, i, depth + 1, 0});
        }
        return ans;
    }
};

class Solution:
    def getCoprimes(self, nums: List[int], edges: List[List[int]]) -> List[int]:
        def dfs(i, fa, depth):
            t = k = -1
            for v in f[nums[i]]:
                stk = stks[v]
                if stk and stk[-1][1] > k:
                    t, k = stk[-1]
            ans[i] = t
            for j in g[i]:
                if j != fa:
                    stks[nums[i]].append((i, depth))
                    dfs(j, i, depth + 1)
                    stks[nums[i]].pop()

        g = defaultdict(list)
        for u, v in edges:
            g[u].append(v)
            g[v].append(u)
        f = defaultdict(list)
        for i in range(1, 51):
            for j in range(1, 51):
                if gcd(i, j) == 1:
                    f[i].append(j)
        stks = defaultdict(list)
        ans = [-1] * len(nums)
        dfs(0, -1, 0)
        return ans



# Solution 2
class Solution:
    def getCoprimes(self, nums: List[int], edges: List[List[int]]) -> List[int]:
        n = len(nums)
        g = [[] for _ in range(n)]
        for u, v in edges:
            g[u].append(v)
            g[v].append(u)
        f = [[] for _ in range(51)]
        for i in range(1, 51):
            for j in range(1, 51):
                if gcd(i, j) == 1:
                    f[i].append(j)
        stks = [[] for _ in range(51)]
        ans = [-1] * n
        stk = [(0, -1, 0, 0)]
        while stk:
            i, fa, depth, k = stk[-1]
            if k == 0:
                t = mx = -1
                for v in f[nums[i]]:
                    cur = stks[v]
                    if cur and cur[-1][1] > mx:
                        t, mx = cur[-1]
                ans[i] = t
            else:
                jprev = g[i][k - 1]
                if jprev != fa:
                    stks[nums[i]].pop()
            while k < len(g[i]) and g[i][k] == fa:
                k += 1
            if k == len(g[i]):
                stk.pop()
                continue
            j = g[i][k]
            stk[-1] = (i, fa, depth, k + 1)
            stks[nums[i]].append((i, depth))
            stk.append((j, i, depth + 1, 0))
        return ans

func getCoprimes(nums []int, edges [][]int) []int {
	n := len(nums)
	g := make([][]int, n)
	f := [51][]int{}
	type pair struct{ first, second int }
	stks := [51][]pair{}
	for _, e := range edges {
		u, v := e[0], e[1]
		g[u] = append(g[u], v)
		g[v] = append(g[v], u)
	}
	for i := 1; i < 51; i++ {
		for j := 1; j < 51; j++ {
			if gcd(i, j) == 1 {
				f[i] = append(f[i], j)
			}
		}
	}
	ans := make([]int, n)
	var dfs func(i, fa, depth int)
	dfs = func(i, fa, depth int) {
		t, k := -1, -1
		for _, v := range f[nums[i]] {
			stk := stks[v]
			if len(stk) > 0 && stk[len(stk)-1].second > k {
				t, k = stk[len(stk)-1].first, stk[len(stk)-1].second
			}
		}
		ans[i] = t
		for _, j := range g[i] {
			if j != fa {
				stks[nums[i]] = append(stks[nums[i]], pair{i, depth})
				dfs(j, i, depth+1)
				stks[nums[i]] = stks[nums[i]][:len(stks[nums[i]])-1]
			}
		}
	}
	dfs(0, -1, 0)
	return ans
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}


// Solution 2
func getCoprimes(nums []int, edges [][]int) []int {
	n := len(nums)
	g := make([][]int, n)
	f := [51][]int{}
	type pair struct{ first, second int }
	stks := [51][]pair{}
	for _, e := range edges {
		u, v := e[0], e[1]
		g[u] = append(g[u], v)
		g[v] = append(g[v], u)
	}
	for i := 1; i < 51; i++ {
		for j := 1; j < 51; j++ {
			if gcd(i, j) == 1 {
				f[i] = append(f[i], j)
			}
		}
	}
	ans := make([]int, n)
	type frame struct{ i, fa, depth, k int }
	stk := []frame{{0, -1, 0, 0}}
	for len(stk) > 0 {
		cur := &stk[len(stk)-1]
		i, fa, depth, k := cur.i, cur.fa, cur.depth, cur.k
		if k == 0 {
			t, mx := -1, -1
			for _, v := range f[nums[i]] {
				s := stks[v]
				if len(s) > 0 && s[len(s)-1].second > mx {
					t, mx = s[len(s)-1].first, s[len(s)-1].second
				}
			}
			ans[i] = t
		} else {
			jprev := g[i][k-1]
			if jprev != fa {
				stks[nums[i]] = stks[nums[i]][:len(stks[nums[i]])-1]
			}
		}
		for k < len(g[i]) && g[i][k] == fa {
			k++
		}
		if k == len(g[i]) {
			stk = stk[:len(stk)-1]
			continue
		}
		j := g[i][k]
		cur.k = k + 1
		stks[nums[i]] = append(stks[nums[i]], pair{i, depth})
		stk = append(stk, frame{j, i, depth + 1, 0})
	}
	return ans
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

1766 - Tree of Coprimes

1766. Tree of Coprimes

Description

Solutions

Solution 2: Preprocessing + Enumeration + Explicit Stack

All Problems

All Solutions

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1766. Tree of Coprimes

Description

Solutions

Solution 2: Preprocessing + Enumeration + Explicit Stack

All Problems

All Solutions