Open in App
Not now

Count of distinct colors in a subtree of a Colored Tree with given min frequency for Q queries

• Difficulty Level : Hard
• Last Updated : 17 Mar, 2023

Given a N-ary tree with some colour associated with every node and Q queries. Each query contains two integers A and X. The task is to count all distinct colors in a subtree rooted at A, having frequency of colors greater than or equal to X in that subtree.
Examples:

Input: Tree:


1(1)
/       \
/          \
2(2)         5(3)
/   \       /  |   \
/     \     /   |    \
3(2)    4(3) 6(2) 7(3)  8(3)

query[] = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {5, 3}}
Output: {2, 2, 1, 0, 1}
Explanation:
In the subtree rooted at 1, the frequency of colour 2 is 3 and colour 3 is 4. So the answer is 2 for the 1 query, 2 for 2nd query and 1 for 3rd query.
For subtree rooted at 2, frequency of color 2 is 2 and color 3 is 1. So no color have frequency more than or equal to 4.
For subtree rooted at 5, frequency of color 2 is 1 and color 3 is 3. So color 3 have frequency equal to 3.

Naive Approach

• For each query, we’ll traverse the whole subtree of the given node.
• We’ll maintain a map which stores the frequency of every colour in the subtree of given node.
• Then, traverse the map and count number of colours such that it’s frequency is greater than given x.

Time Complexity: O(Q * N)
Space Complexity: O(Q * N)

Approach: (Using Mo’s Algorithm)

• We’ll first flat the tree using Euler Tour.
• We’ll give the number to every node, when it will go in the DFS and when it came out. Let’s denote this with tin[node] and tout[node] for every node.
• After we have flatten the tree into an array, every sub tree of can be denoted as some  array with start and end index as tin[node] and tout[node] respectively.
• Now the question changed into number of elements with frequency greater than equal to X in some subarray.
• We’ll use Mo’s algorithm to solve this problem.
• First, we’ll store the queries and sort them according to the tin[node] / SQ where SQ is the Square root of N.
• As we move the pointers, we’ll store the frequency at ith colour in an array and answer to the query is stored at Xth position of array as it stores the count of colours with frequency greater equal to X.

C++

 // C++ program to count distinct colors // in a subtree of a Colored Tree // with given min frequency for Q queries   #include  using namespace std;   const int N = 1e5 + 5;   // graph as adjacency list vector > v(N);   // colour written on node. vector<int> colour(N);   // order of node entering in DFS vector<int> in(N);   // order of node exiting in DFS vector<int> out(N);   // for counting the frequency of // nodes with colour i vector<int> cnt(N);   // for storing frequency of colours vector<int> freq(N);   // tree in a flatten // form (using Euler tour) vector<int> flatTree(N);   // number of nodes int n,       // square root of n     sq;   // indexes for in and // out of node in DFS int start = 0;   // DFS function to find // order of euler tour void DFSEuler(int a, int par) {       // storing the start index     in[a] = ++start;       // storing colour of node     // in flatten array     flatTree[start] = colour[a];       for (int i : v[a]) {           // doing DFS on its child         // skipping parent         if (i == par)             continue;           DFSEuler(i, a);     }       // out index of the node.     out[a] = start; }   // comparator for queries bool comparator(pair<int, int>& a,                 pair<int, int>& b) {     // comparator for queries to be     // sorted according to in[x] / sq     if (in[a.first] / sq != in[b.first] / sq)         return in[a.first] < in[b.first];       return out[a.first] < out[b.first]; }   // Function to answer the queries void solve(vector > arr,            int q) {     sq = sqrt(n) + 1;       // for storing answers     vector<int> answer(q);       // for storing indexes of queries     // in the order of input.     map, int> idx;       for (int i = 0; i < q; i++) {           // storing indexes of queries         idx[arr[i]] = i;     }       // doing depth first search to     // find indexes to flat the     // tree using euler tour.     DFSEuler(1, 0);       // After doing Euler tour,     // subtree of x can be     // represented as a subarray     // from in[x] to out[x];       // we'll sort the queries     // according to the in[i];     sort(arr.begin(),          arr.end(),          comparator);       // two pointers for     // sliding the window     int l = 1, r = 0;       for (int i = 0; i < q; i++) {           // finding answer to the query         int node = arr[i].first,             x = arr[i].second;         int id = idx[arr[i]];           while (l > in[node]) {               // decrementing the pointer as             // it is greater than start             // and adding answer             // to our freq array.             l--;             cnt[flatTree[l]]++;             freq[cnt[flatTree[l]]]++;         }           while (r < out[node]) {               // incrementing pointer as it is             // less than the end value and             // adding answer to our freq array.             r++;             cnt[flatTree[r]]++;             freq[cnt[flatTree[r]]]++;         }           while (l < in[node]) {               // removing the lth node from             // freq array and incrementing             // the pointer             freq[cnt[flatTree[l]]]--;             cnt[flatTree[l]]--;             l++;         }           while (r > out[node]) {               // removing the rth node from             // freq array and decrementing             // the pointer             freq[cnt[flatTree[r]]]--;             cnt[flatTree[r]]--;             r--;         }           // answer to this query         // is stored at freq[x]         // freq[x] stores the frequency         // of nodes greater equal to x         answer[id] = freq[x];     }       // printing the queries     for (int i = 0; i < q; i++)         cout << answer[i] << " "; }   int main() {     // Driver Code     /*                1(1)              /       \             /          \          2(2)         5(3)         /   \       /  |   \        /     \     /   |    \     3(2)    4(3) 6(2) 7(3)  8(3)     */     n = 8;     v[1].push_back(2);     v[2].push_back(1);     v[2].push_back(3);     v[3].push_back(2);     v[2].push_back(4);     v[4].push_back(2);     v[1].push_back(5);     v[5].push_back(1);     v[5].push_back(6);     v[6].push_back(5);     v[5].push_back(7);     v[7].push_back(5);     v[5].push_back(8);     v[8].push_back(5);       colour[1] = 1;     colour[2] = 2;     colour[3] = 2;     colour[4] = 3;     colour[5] = 3;     colour[6] = 2;     colour[7] = 3;     colour[8] = 3;       vector > queries         = { { 1, 2 },             { 1, 3 },             { 1, 4 },             { 2, 3 },             { 5, 3 } };     int q = queries.size();       solve(queries, q);     return 0; }

Java

 // Java program to count distinct colors // in a subtree of a Colored Tree // with given min frequency for Q queries import java.util.*;   public class Main {       static int N = 100005;     // graph as adjacency list     static List[] v = new List[N];             // colour written on node.     static int[] colour = new int[N];           // order of node entering in DFS     static int[] in_time = new int[N];           // order of node exiting in DFS     static int[] out_time = new int[N];           // for counting the frequency of     // nodes with colour i     static int[] cnt = new int[N];           // for storing frequency of colours     static int[] freq = new int[N];           // tree in a flatten     // form (using Euler tour)     static int[] flatTree = new int[N];         // number of nodes     static int n = 0;           // square root of n     static int sq = 0;           // indexes for in and     // out of node in DFS     static int start = 0;         // DFS function to find     // order of euler tour     static void DFSEuler(int a, int par)     {           // storing the start index         start++;         in_time[a] = start;                       // storing colour of node         // in flatten array         flatTree[start] = colour[a];         for (int i : v[a]) {               // doing DFS on its child             // skipping parent             if (i == par) {                 continue;             }             DFSEuler(i, a);         }           // out index of the node.         out_time[a] = start;     }             // comparator for queries     static int comparator(Pair a,                           Pair b)     {           // comparator for queries to be         // sorted according to in[x] / sq         if (in_time[a.getKey()] / sq != in_time[b.getKey()] / sq) {             return in_time[a.getKey()] - in_time[b.getKey()];         }         return out_time[a.getKey()] - out_time[b.getKey()];     }         // Function to answer the queries     static void solve(List > arr,                       int q)     {           // for storing answers         sq = (int)Math.sqrt(n) + 1;         int[] answer = new int[q];                   // for storing indexes of queries         // in the order of input.         Map, Integer> idx = new HashMap<>();         for (int i = 0; i < q; i++) {               // storing indexes of queries             idx.put(arr.get(i), i);         }                       // doing depth first search to         // find indexes to flat the         // tree using euler tour.             DFSEuler(1, 0);                       // After doing Euler tour,         // subtree of x can be         // represented as a subarray         // from in[x] to out[x];            // we'll sort the queries         // according to the in[i];         arr.sort(new Comparator >() {             @Override             public int compare(Pair a,                                Pair b)             {                 if (in[a.getFirst()] / sq                     != in[b.getFirst()] / sq)                     return in[a.getFirst()]                         - in[b.getFirst()];                 return out[a.getFirst()]                     - out[b.getFirst()];             }         });                   // two pointers for         // sliding the window         int l = 1, r = 0;                 for (int i = 0; i < q; i++) {               // finding answer to the query             int node = arr.get(i).getFirst(),                 x = arr.get(i).getSecond(),                 id = idx.get(arr.get(i));             while (l > in[node]) {                                   // decrementing the pointer as                 // it is greater than start                 // and adding answer                 // to our freq array.                 l--;                 cnt[flatTree[l]]++;                 freq[cnt[flatTree[l]]]++;             }                             while (r < out[node]) {                                   // incrementing pointer as it is                 // less than the end value and                 // adding answer to our freq array.                 r++;                 cnt[flatTree[r]]++;                 freq[cnt[flatTree[r]]]++;             }             while (l < in[node]) {                                   // removing the lth node from                 // freq array and incrementing                 // the pointer                 freq[cnt[flatTree[l]]]--;                 cnt[flatTree[l]]--;                 l++;             }             while (r > out[node]) {                                   // removing the rth node from                 // freq array and decrementing                 // the pointer                 freq[cnt[flatTree[r]]]--;                 cnt[flatTree[r]]--;                 r--;             }                           // answer to this query             // is stored at freq[x]             // freq[x] stores the frequency             // of nodes greater equal to x             answer[id] = freq[x];         }                       // printing the queries         for (int i = 0; i < q; i++)             System.out.print(answer[i] + " ");     }     public static void main(String[] args)     {         // Driver Code         /*                    1(1)                  /       \                /          \              2(2)         5(3)             /   \       /  |   \            /     \     /   |    \         3(2)    4(3) 6(2) 7(3)  8(3)         */         Scanner sc = new Scanner(System.in);         n = 8;         for (int i = 1; i <= n; i++) {             v[i] = new ArrayList();         }         v[1].add(2);         v[2].add(1);         v[2].add(3);         v[3].add(2);         v[2].add(4);         v[4].add(2);         v[1].add(5);         v[5].add(1);         v[5].add(6);         v[6].add(5);         v[5].add(7);         v[7].add(5);         v[5].add(8);         v[8].add(5);         colour[1] = 1;         colour[2] = 2;         colour[3] = 2;         colour[4] = 3;         colour[5] = 3;         colour[6] = 2;         colour[7] = 3;         colour[8] = 3;         List > queries             = Arrays.asList(                 new Pair(1, 2),                 new Pair(1, 3),                 new Pair(1, 4),                 new Pair(2, 3),                 new Pair(5, 3));         int q = queries.size();         solve(queries, q);     } }   // run this code in offline compiler because pair is not working in online compiler.

Python3

 import math   N = 100005   v = [[] for i in range(N)] colour = [0]*N in_time = [0]*N out_time = [0]*N cnt = [0]*N freq = [0]*N flatTree = [0]*N   n = 0 sq = 0 start = 0   def DFSEuler(a, par):     global start     start += 1     in_time[a] = start     flatTree[start] = colour[a]           for i in v[a]:         if i == par:             continue         DFSEuler(i, a)               out_time[a] = start   def comparator(a, b):     if in_time[a[0]] // sq != in_time[b[0]] // sq:         return in_time[a[0]] < in_time[b[0]]     return out_time[a[0]] < out_time[b[0]]   def solve(arr, q):     global n, sq     sq = int(math.sqrt(n)) + 1     answer = [0]*q     idx = {}     for i in range(q):         idx[arr[i]] = i     DFSEuler(1, 0)     arr.sort(key=lambda x: comparator(x, x))     l, r = 1, 0     for i in range(q):         node, x = arr[i]         id = idx[arr[i]]         while l > in_time[node]:             l -= 1             cnt[flatTree[l]] += 1             freq[cnt[flatTree[l]]] += 1         while r < out_time[node]:             r += 1             cnt[flatTree[r]] += 1             freq[cnt[flatTree[r]]] += 1         while l < in_time[node]:             freq[cnt[flatTree[l]]] -= 1             cnt[flatTree[l]] -= 1             l += 1         while r > out_time[node]:             freq[cnt[flatTree[r]]] -= 1             cnt[flatTree[r]] -= 1             r -= 1         answer[id] = freq[x]     for i in range(q):         print(answer[i], end=" ")   if __name__ == '__main__':     n = 8     v[1].append(2)     v[2].append(1)     v[2].append(3)     v[3].append(2)     v[2].append(4)     v[4].append(2)     v[1].append(5)     v[5].append(1)     v[5].append(6)     v[6].append(5)     v[5].append(7)     v[7].append(5)     v[5].append(8)     v[8].append(5)     colour[1] = 1     colour[2] = 2     colour[3] = 2     colour[4] = 3     colour[5] = 3     colour[6] = 2     colour[7] = 3     colour[8] = 3     queries = [(1, 2), (1, 3), (1, 4), (2, 3), (5, 3)]     q = len(queries)     solve(queries, q)

C#

 // C# program to count distinct colors // in a subtree of a Colored Tree // with given min frequency for Q queries using System; using System.Collections.Generic; using System.Linq;   // Array Equality Comparer class ArrayEqualityComparer : IEqualityComparer {     // Implementing the Equals method to compare two arrays     public bool Equals(T[] x, T[] y)     {         if (x == null && y == null)             return true;         if (x == null || y == null)             return false;         if (x.Length != y.Length)             return false;         for (int i = 0; i < x.Length; i++) {             if (!EqualityComparer.Default.Equals(x[i],                                                     y[i]))                 return false;         }         return true;     }       // Implementing the GetHashCode method to obtain a hash     // code value for the array     public int GetHashCode(T[] obj)     {         int hashCode = 0;         foreach(T item in obj)         {             hashCode                 ^= EqualityComparer.Default.GetHashCode(                     item);         }         return hashCode;     } }   class GFG {     private static readonly int N = 100005;       private static readonly List<int>[] v         = Enumerable.Range(0, N)               .Select(_ => new List<int>())               .ToArray();     private static readonly int[] colour = new int[N];     private static readonly int[] in_time = new int[N];     private static readonly int[] out_time = new int[N];     private static readonly int[] cnt = new int[N];     private static readonly int[] freq = new int[N];     private static readonly int[] flatTree = new int[N];       private static int n = 0;     private static int sq = 0;     private static int start = 0;       // DFS function to find     // order of euler tour     private static void DFSEuler(int a, int par)     {         start += 1;         in_time[a] = start;         flatTree[start] = colour[a];           foreach(int i in v[a])         {             if (i == par) {                 continue;             }             DFSEuler(i, a);         }           out_time[a] = start;     }       // comparator for queries     private static int Comparator(int[] a, int[] b)     {         // comparator for queries to be         // sorted according to in[x] / sq         if (in_time[a[0]] / sq != in_time[b[0]] / sq) {             return in_time[a[0]] < in_time[b[0]] ? -1 : 1;         }         return out_time[a[0]] < out_time[b[0]] ? -1 : 1;     }       // Function to answer the queries     private static void Solve(int[][] arr, int q)     {         sq = (int)Math.Floor(Math.Sqrt(n)) + 1;         int[] answer = new int[q];         Dictionary<int[], int> idx             = new Dictionary<int[], int>(                 new ArrayEqualityComparer<int>());         for (int i = 0; i < q; i++) {             idx.Add(arr[i], i);         }           // doing depth first search to         // find indexes to flat the         // tree using euler tour.         DFSEuler(1, 0);           // After doing Euler tour,         // subtree of x can be         // represented as a subarray         // from in[x] to out[x];           // we'll sort the queries         // according to the in[i];         Array.Sort(arr, Comparator);         int l = 1;         int r = 0;         for (int i = 0; i < q; i++) {             int[] query = arr[i];             int node = query[0];             int x = query[1];             int id = idx[query];             while (l > in_time[node]) {                 l -= 1;                 cnt[flatTree[l]] += 1;                 freq[cnt[flatTree[l]]] += 1;             }             while (r < out_time[node]) {                 r += 1;                 cnt[flatTree[r]] += 1;                 freq[cnt[flatTree[r]]] += 1;             }             while (l < in_time[node]) {                 freq[cnt[flatTree[l]]] -= 1;                 cnt[flatTree[l]] -= 1;                 l += 1;             }             while (r > out_time[node]) {                 freq[cnt[flatTree[r]]] -= 1;                 cnt[flatTree[r]] -= 1;                 r -= 1;             }             answer[id] = freq[x];         }           foreach(var val in answer) Console.Write(val + " ");     }       // Driver code     public static void Main(string[] args)     {         int n = 8;         v[1].Add(2);         v[2].Add(1);         v[2].Add(3);         v[3].Add(2);         v[2].Add(4);         v[4].Add(2);         v[1].Add(5);         v[5].Add(1);         v[5].Add(6);         v[6].Add(5);         v[5].Add(7);         v[7].Add(5);         v[5].Add(8);         v[8].Add(5);         colour[1] = 1;         colour[2] = 2;         colour[3] = 2;         colour[4] = 3;         colour[5] = 3;         colour[6] = 2;         colour[7] = 3;         colour[8] = 3;           int[][] queries             = { new int[] { 1, 2 }, new int[] { 1, 3 },                 new int[] { 1, 4 }, new int[] { 2, 3 },                 new int[] { 5, 3 } };         int q = queries.Length;           // Function call         Solve(queries, q);     } }

Javascript

 // JavaScript program to count distinct colors // in a subtree of a Colored Tree // with given min frequency for Q queries const N = 100005;   const v = Array.from({ length: N }, () => []); const colour = new Array(N).fill(0); const in_time = new Array(N).fill(0); const out_time = new Array(N).fill(0); const cnt = new Array(N).fill(0); const freq = new Array(N).fill(0); const flatTree = new Array(N).fill(0);   let n = 0; let sq = 0; let start = 0;   // DFS function to find // order of euler tour function DFSEuler(a, par) {   start += 1;   in_time[a] = start;   flatTree[start] = colour[a];     for (const i of v[a]) {     if (i === par) {       continue;     }     DFSEuler(i, a);   }     out_time[a] = start; }   // comparator for queries function comparator(a, b) { // comparator for queries to be // sorted according to in[x] / sq   if (Math.floor(in_time[a[0]] / sq) !== Math.floor(in_time[b[0]] / sq)) {     return in_time[a[0]] < in_time[b[0]];   }   return out_time[a[0]] < out_time[b[0]]; }   // Function to answer the queries function solve(arr, q) {   sq = Math.floor(Math.sqrt(n)) + 1;   const answer = new Array(q).fill(0);   const idx = {};   for (let i = 0; i < q; i++) {     idx[arr[i]] = i;   }       // doing depth first search to   // find indexes to flat the   // tree using euler tour.   DFSEuler(1, 0);       // After doing Euler tour,   // subtree of x can be   // represented as a subarray   // from in[x] to out[x];     // we'll sort the queries   // according to the in[i];   arr.sort((a, b) => comparator(a, b));   let l = 1;   let r = 0;   for (let i = 0; i < q; i++) {     const [node, x] = arr[i];     const id = idx[arr[i]];     while (l > in_time[node]) {       l -= 1;       cnt[flatTree[l]] += 1;       freq[cnt[flatTree[l]]] += 1;     }     while (r < out_time[node]) {       r += 1;       cnt[flatTree[r]] += 1;       freq[cnt[flatTree[r]]] += 1;     }     while (l < in_time[node]) {       freq[cnt[flatTree[l]]] -= 1;       cnt[flatTree[l]] -= 1;       l += 1;     }     while (r > out_time[node]) {       freq[cnt[flatTree[r]]] -= 1;       cnt[flatTree[r]] -= 1;       r -= 1;     }     answer[id] = freq[x];   }       console.log(answer.join(' ')); }   n = 8; v[1].push(2); v[2].push(1); v[2].push(3); v[3].push(2); v[2].push(4); v[4].push(2); v[1].push(5); v[5].push(1); v[5].push(6); v[6].push(5); v[5].push(7); v[7].push(5); v[5].push(8); v[8].push(5); colour[1] = 1; colour[2] = 2; colour[3] = 2; colour[4] = 3; colour[5] = 3; colour[6] = 2; colour[7] = 3; colour[8] = 3;   const queries = [[1, 2], [1, 3], [1, 4], [2, 3], [5, 3]]; const q = queries.length; solve(queries, q);   //contributed by adityasharmadev01

Output:

2 2 1 0 1

Time Complexity:
Auxiliary Space:

My Personal Notes arrow_drop_up
Related Articles