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Count of distinct numbers in an Array in a range for Online Queries using Merge Sort Tree

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  • Last Updated : 26 May, 2022
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Given an array arr[] of size N and Q queries of the form [L, R], the task is to find the number of distinct values in this array in the given range. Examples:

Input: arr[] = {4, 1, 9, 1, 3, 3}, Q = {{1, 3}, {1, 5}} Output: 3 4 Explanation: For query {1, 3}, elements are {4, 1, 9}. Therefore, count of distinct elements = 3 For query {1, 5}, elements are {4, 1, 9, 1, 3}. Therefore, count of distinct elements = 4 Input: arr[] = {4, 2, 1, 1, 4}, Q = {{2, 4}, {3, 5}} Output: 3 2

Naive Approach: A simple solution is that for every Query, iterate array from L to R and insert elements in a set. Finally, the Size of the set gives the number of distinct elements from L to R. Time Complexity: O(Q * N) Efficient Approach: The idea is to use Merge Sort Tree to solve this problem.

  1. We will store the next occurrence of the element in a temporary array.
  2. Then for every query from L to R, we will find the number of elements in the temporary array whose values are greater than R in range L to R.

Step 1: Take an array next_right, where next_right[i] holds the next right index of the number i in the array a. Initialize this array as N(length of the array). Step 2: Make a Merge Sort Tree from next_right array and make queries. Queries to calculate the number of distinct elements from L to R is equivalent to find the number of elements from L to R which are greater than R.

Construction of Merge Sort Tree from given array

  • We start with a segment arr[0 . . . n-1].
  • Every time we divide the current segment into two halves if it has not yet become a segment of length 1. Then call the same procedure on both halves, and for each such segment, we store the sorted array in each segment as in merge sort.
  • Also, the tree will be a Full Binary Tree because we always divide segments into two halves at every level.
  • Since the constructed tree is always a full binary tree with n leaves, there will be N-1 internal nodes. So the total number of nodes will be 2*N – 1.

Here is an example. Say 1 5 2 6 9 4 7 1 be an array.

|1 1 2 4 5 6 7 9|
|1 2 5 6|1 4 7 9|
|1 5|2 6|4 9|1 7|
|1|5|2|6|9|4|7|1|

Construction of next_right array

  • We store the next right occurrence of every element.
  • If the element has the last occurrence then we store ‘N'(Length of the array) Example:
arr = [2, 3, 2, 3, 5, 6];
next_right = [2, 3, 6, 6, 6, 6]

Below is the implementation of the above approach: 

C++




// C++ implementation to find
// count of distinct elements
// in a range L to R for Q queries
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to merge the right
// and the left tree
void merge(vector<int> tree[],
                 int treeNode)
{
    int len1 =
      tree[2 * treeNode].size();
    int len2 =
      tree[2 * treeNode + 1].size();
    int index1 = 0, index2 = 0;
 
    // Fill this array in such a
    // way such that values
    // remain sorted similar to mergesort
    while (index1 < len1 && index2 < len2) {
 
        // If the element on the left part
        // is greater than the right part
        if (tree[2 * treeNode][index1] >
              tree[2 * treeNode + 1][index2]) {
 
            tree[treeNode].push_back(
                tree[2 * treeNode + 1][index2]
                );
            index2++;
        }
        else {
            tree[treeNode].push_back(
                tree[2 * treeNode][index1]
                );
            index1++;
        }
    }
 
    // Insert the leftover elements
    // from the left part
    while (index1 < len1) {
        tree[treeNode].push_back(
            tree[2 * treeNode][index1]
            );
        index1++;
    }
 
    // Insert the leftover elements
    // from the right part
    while (index2 < len2) {
        tree[treeNode].push_back(
            tree[2 * treeNode + 1][index2]
            );
        index2++;
    }
    return;
}
 
// Recursive function to build
// segment tree by merging the
// sorted segments in sorted way
void build(vector<int> tree[],
    int* arr, int start, int end,
                  int treeNode)
{
    // Base case
    if (start == end) {
        tree[treeNode].push_back(
            arr[start]);
        return;
    }
    int mid = (start + end) / 2;
 
    // Building the left tree
    build(tree, arr, start,
          mid, 2 * treeNode);
 
    // Building the right tree
    build(tree, arr, mid + 1, end,
                 2 * treeNode + 1);
 
    // Merges the right tree
    // and left tree
    merge(tree, treeNode);
    return;
}
 
// Function similar to query() method
// as in segment tree
int query(vector<int> tree[],
     int treeNode, int start, int end,
            int left, int right)
{
 
    // Current segment is out of the range
    if (start > right || end < left) {
        return 0;
    }
    // Current segment completely
    // lies inside the range
    if (start >= left && end <= right) {
 
        // as the elements are in sorted order
        // so number of elements greater than R
        // can be find using binary
        // search or upper_bound
        return tree[treeNode].end() -
          upper_bound(tree[treeNode].begin(),
            tree[treeNode].end(), right);
    }
 
    int mid = (start + end) / 2;
 
    // Query on the left tree
    int op1 = query(tree, 2 * treeNode,
              start, mid, left, right);
    // Query on the Right tree
    int op2 = query(tree, 2 * treeNode + 1,
            mid + 1, end, left, right);
    return op1 + op2;
}
 
// Driver Code
int main()
{
 
    int n = 5;
    int arr[] = { 1, 2, 1, 4, 2 };
 
    int next_right[n];
    // Initialising the tree
    vector<int> tree[4 * n];
 
    unordered_map<int, int> ump;
 
    // Construction of next_right
    // array to store the
    // next index of occurrence
    // of elements
    for (int i = n - 1; i >= 0; i--) {
        if (ump[arr[i]] == 0) {
            next_right[i] = n;
            ump[arr[i]] = i;
        }
        else {
            next_right[i] = ump[arr[i]];
            ump[arr[i]] = i;
        }
    }
    // building the mergesort tree
    // by using next_right array
    build(tree, next_right, 0, n - 1, 1);
 
    int ans;
    // Queries one based indexing
    // Time complexity of each
    // query is log(N)
 
    // first query
    int left1 = 0;
    int right1 = 2;
    ans = query(tree, 1, 0, n - 1,
                  left1, right1);
    cout << ans << endl;
 
    // Second Query
    int left2 = 1;
    int right2 = 4;
    ans = query(tree, 1, 0, n - 1,
                  left2, right2);
    cout << ans << endl;
}

Output:

2
3

Time Complexity: O(Q*log N)


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