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Querying the number of distinct colors in a subtree of a colored tree using BIT

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  • Difficulty Level : Hard
  • Last Updated : 06 Aug, 2021
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Prerequisites: BIT, DFS

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’


           / \
          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:  

  1. 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.
  2. 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).


  1. 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.
  2. 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.
  3. 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.
  4. Now our query(R) function for BIT would return the number of distinct elements in flat_tree[] in the range [1, R].
  5. We sort all the queries in order of increasing vis_time[], let li denote vis_time[i] and ri denote the end_time[i]. Sorting the queries in increasing order of li gives us an edge, as when processing the ith query we won’t see any query in future with its ‘l‘ smaller than li. So we can remove all the elements’ occurrences up to li – 1 from BIT and add their next occurrences using the traverser[] array. And then query(R) would return the number of distinct elements in the range [li, ri ]


// A C++ program implementing the above design
#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)
//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++)
        //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
//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 occurrences 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);
        //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++)
                                    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
    // process queries
    // 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()
           / \
          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;


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) )

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