Given a rooted tree T, with ‘n’ nodes, each node has a color denoted by the array color[](color[i] denotes the color of ith node in form of an integer). Respond to ‘Q’ queries of the following type:

**distinct**u – Print the number of distinct colored nodes under the subtree rooted under ‘u’

**Examples:**

1 / \ 2 3 /|\ | \ 4 5 6 7 8 /| \ 9 10 11 color[] = {0, 2, 3, 3, 4, 1, 3, 4, 3, 2, 1, 1} Indexes NA 1 2 3 4 5 6 7 8 9 10 11 (Node Values and colors start from index 1) distinct 3 -> output should be '4'. There are six different nodes in the subtree rooted with 3, nodes are 3, 7, 8, 9, 10 and 11. These nodes have four distinct colors (3, 4, 2 and 1) distinct 2 -> output should be '3'. distinct 7 -> output should be '3'.

Building a solution in steps:

- Flatten the tree using DFS; store the visiting time and ending time for every node in two arrays,
**vis_time[i]**stores the visiting time of the ith node while**end_time[i]**stores the ending time. - In the same DFS call, store the value of color of every node in an array
**flat_tree[]**, at indices:**vis_time[i]**and**end_time[i]**for ith node.

Note: size of the array**flat_tree[]**will be 2n.

Now the problem is reduced to finding the number of distinct elements in the range **[vis_time[u], end_time[u] ]** in the array **flat_tree[]** for each query of the specified type. To do so, we will process the queries off-line(processing the queries in an order different than the one provided in the question and storing the results, and finally printing the result for each in the order specified in the question).

**Steps:**

- First, we pre-process the array
**flat_tree[]**; we maintain a**table[]**(an array of vectors),**table[i]**stores the vector containing all the indices in**flat_tree[]**that have value i. That is, if**flat_tree[j] = i**, then**table[i]**will have one of its element**j**. - In BIT, we update ‘1’ at ith index if we want the ith element of
**flat_tree[]**to be counted in**query()**method. We now maintain another array**traverser[]**;**traverser[i]**contains the pointer to the next element of table[i] that is not marked in BIT yet. - We now update our BIT and set ‘1’ at first occurrence of every element in
**flat_tree[]**and increment corresponding**traverser[]**by ‘1’(if**flat_tree[i]**is occurring for the first time then**traverser[flat_tree[i]]**is incremented by ‘1’) to point to the next occurrence of that element. - Now our
**query(R)**function for BIT would return the number of distinct elements in**flat_tree[]**in the range**[1, R]**. - We sort all the queries in order of increasing
**vis_time[]**, let**l**denote_{i}**vis_time[i]**and**r**denote the_{i}**end_time[i]**. Sorting the queries in increasing order of**l**gives us an edge, as when processing the ith query we won’t see any query in future with its ‘_{i}**l**‘ smaller than**l**. So we can remove all the elements’ occurrences up to_{i}**l**from BIT and add their next occurrences using the_{i}– 1**traverser[]**array. And then**query(R)**would return the number of distinct elements in the range**[l**._{i}, r_{i}]

`// A C++ program implementing the above design ` `#include<bits/stdc++.h> ` `#define max_color 1000005 ` `#define maxn 100005 ` `using` `namespace` `std; ` ` ` `// Note: All elements of global arrays are ` `// initially zero ` `// All the arrays have been described above ` `int` `bit[maxn], vis_time[maxn], end_time[maxn]; ` `int` `flat_tree[2 * maxn]; ` `vector<` `int` `> tree[maxn]; ` `vector<` `int` `> table[max_color]; ` `int` `traverser[max_color]; ` ` ` `bool` `vis[maxn]; ` `int` `tim = 0; ` ` ` `//li, ri and index are stored in queries vector ` `//in that order, as the sort function will use ` `//the value li for comparison ` `vector< pair< pair<` `int` `, ` `int` `>, ` `int` `> > queries; ` ` ` `//ans[i] stores answer to ith query ` `int` `ans[maxn]; ` ` ` `//update function to add val to idx in BIT ` `void` `update(` `int` `idx, ` `int` `val) ` `{ ` ` ` `while` `( idx < maxn ) ` ` ` `{ ` ` ` `bit[idx] += val; ` ` ` `idx += idx & -idx; ` ` ` `} ` `} ` ` ` `//query function to find sum(1, idx) in BIT ` `int` `query(` `int` `idx) ` `{ ` ` ` `int` `res = 0; ` ` ` `while` `( idx > 0 ) ` ` ` `{ ` ` ` `res += bit[idx]; ` ` ` `idx -= idx & -idx; ` ` ` `} ` ` ` `return` `res; ` `} ` ` ` `void` `dfs(` `int` `v, ` `int` `color[]) ` `{ ` ` ` `//mark the node visited ` ` ` `vis[v] = 1; ` ` ` ` ` `//set visiting time of the node v ` ` ` `vis_time[v] = ++tim; ` ` ` ` ` `//use the color of node v to fill flat_tree[] ` ` ` `flat_tree[tim] = color[v]; ` ` ` ` ` `vector<` `int` `>::iterator it; ` ` ` `for` `(it=tree[v].begin(); it!=tree[v].end(); it++) ` ` ` `if` `(!vis[*it]) ` ` ` `dfs(*it, color); ` ` ` ` ` ` ` `// set ending time for node v ` ` ` `end_time[v] = ++tim; ` ` ` ` ` `// setting its color in flat_tree[] again ` ` ` `flat_tree[tim] = color[v]; ` `} ` ` ` `//function to add an edge(u, v) to the tree ` `void` `addEdge(` `int` `u, ` `int` `v) ` `{ ` ` ` `tree[u].push_back(v); ` ` ` `tree[v].push_back(u); ` `} ` ` ` `//function to build the table[] and also add ` `//first occurrences of elements to the BIT ` `void` `hashMarkFirstOccurrences(` `int` `n) ` `{ ` ` ` `for` `(` `int` `i = 1 ; i <= 2 * n ; i++) ` ` ` `{ ` ` ` `table[flat_tree[i]].push_back(i); ` ` ` ` ` `//if it is the first occurrence of the element ` ` ` `//then add it to the BIT and increment traverser ` ` ` `if` `(table[flat_tree[i]].size() == 1) ` ` ` `{ ` ` ` `//add the occurrence to bit ` ` ` `update(i, 1); ` ` ` ` ` `//make traverser point to next occurrence ` ` ` `traverser[flat_tree[i]]++; ` ` ` `} ` ` ` `} ` `} ` ` ` `//function to process all the queries and store their answers ` `void` `processQueries() ` `{ ` ` ` `int` `j = 1; ` ` ` `for` `(` `int` `i=0; i<queries.size(); i++) ` ` ` `{ ` ` ` `//for each query remove all the ocurences before its li ` ` ` `//li is the visiting time of the node ` ` ` `//which is stored in first element of first pair ` ` ` `for` `( ; j < queries[i].first.first ; j++ ) ` ` ` `{ ` ` ` `int` `elem = flat_tree[j]; ` ` ` ` ` `//update(i, -1) removes an element at ith index ` ` ` `//in the BIT ` ` ` `update( table[elem][traverser[elem] - 1], -1); ` ` ` ` ` `//if there is another occurrence of the same element ` ` ` `if` `( traverser[elem] < table[elem].size() ) ` ` ` `{ ` ` ` `//add the occurrence to the BIT and ` ` ` `//increment traverser ` ` ` `update(table[elem][ traverser[elem] ], 1); ` ` ` `traverser[elem]++; ` ` ` `} ` ` ` `} ` ` ` ` ` `//store the answer for the query, the index of the query ` ` ` `//is the second element of the pair ` ` ` `//And ri is stored in second element of the first pair ` ` ` `ans[queries[i].second] = query(queries[i].first.second); ` ` ` `} ` `} ` ` ` `// Count distinct colors in subtrees rooted with qVer[0], ` `// qVer[1], ...qVer[qn-1] ` `void` `countDistinctColors(` `int` `color[], ` `int` `n, ` `int` `qVer[], ` `int` `qn) ` `{ ` ` ` `// build the flat_tree[], vis_time[] and end_time[] ` ` ` `dfs(1, color); ` ` ` ` ` `// add query for u = 3, 2 and 7 ` ` ` `for` `(` `int` `i=0; i<qn; i++) ` ` ` `queries.push_back(make_pair(make_pair(vis_time[qVer[i]], ` ` ` `end_time[qVer[i]]), i) ); ` ` ` ` ` `// sort the queries in order of increasing vis_time ` ` ` `sort(queries.begin(), queries.end()); ` ` ` ` ` `// make table[] and set '1' at first occurrences of elements ` ` ` `hashMarkFirstOccurrences(n); ` ` ` ` ` `// process queries ` ` ` `processQueries(); ` ` ` ` ` `// print all the answers, in order asked ` ` ` `// in the question ` ` ` `for` `(` `int` `i=0; i<queries.size() ; i++) ` ` ` `{ ` ` ` `cout << ` `"Distinct colors in the corresponding subtree"` ` ` `"is: "` `<< ans[i] << endl; ` ` ` `} ` `} ` ` ` `//driver code ` `int` `main() ` `{ ` ` ` `/* ` ` ` `1 ` ` ` `/ \ ` ` ` `2 3 ` ` ` `/|\ | \ ` ` ` `4 5 6 7 8 ` ` ` `/| \ ` ` ` `9 10 11 */` ` ` `int` `n = 11; ` ` ` `int` `color[] = {0, 2, 3, 3, 4, 1, 3, 4, 3, 2, 1, 1}; ` ` ` ` ` `// add all the edges to the tree ` ` ` `addEdge(1, 2); ` ` ` `addEdge(1, 3); ` ` ` `addEdge(2, 4); ` ` ` `addEdge(2, 5); ` ` ` `addEdge(2, 6); ` ` ` `addEdge(3, 7); ` ` ` `addEdge(3, 8); ` ` ` `addEdge(7, 9); ` ` ` `addEdge(7, 10); ` ` ` `addEdge(7, 11); ` ` ` ` ` ` ` `int` `qVer[] = {3, 2, 7}; ` ` ` `int` `qn = ` `sizeof` `(qVer)/` `sizeof` `(qVer[0]); ` ` ` ` ` `countDistinctColors(color, n, qVer, qn); ` ` ` ` ` `return` `0; ` `} ` |

Output:

Distinct colors in the corresponding subtree is:4 Distinct colors in the corresponding subtree is:3 Distinct colors in the corresponding subtree is:3

Time Complexity:O( Q * log(n) )

This article is contributed by **Saumye Malhotra**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.