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

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;
}

Python3

from bisect import *
#function to merge the right and the left tree
def merge(tree, treeNode):
    len1 = len(tree[2 * treeNode])
    len2 = len(tree[2 * treeNode + 1])
    index1 = 0
    index2 = 0

    # Fill this array in such a
    # way such that values
    # remain sorted similar to mergesort
    while index1 < len1 and 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].append(tree[2 * treeNode + 1][index2])
            index2 += 1
        else:
            tree[treeNode].append(tree[2 * treeNode][index1])
            index1 += 1

    # Insert the leftover elements
    # from the left part
    while index1 < len1:
        tree[treeNode].append(tree[2 * treeNode][index1])
        index1 += 1

    # Insert the leftover elements
    # from the right part
    while index2 < len2:
        tree[treeNode].append(tree[2 * treeNode + 1][index2])
        index2 += 1

    return

# Recursive function to build
# segment tree by merging the
# sorted segments in sorted way
def build(tree, arr, start, end, treeNode):

    # Base case
    if start == end:
        tree[treeNode].append(arr[start])
        return

    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
def query(tree, treeNode, start, end, left, right):

    # Current segment is out of the range
    if start > right or end < left:
        return 0

    # Current segment completely lies inside the range
    if start >= left and 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 len(tree[treeNode]) - bisect_right(tree[treeNode], right)

    mid = (start + end) // 2

    # Query on the left tree
    op1 = query(tree, 2 * treeNode, start, mid, left, right)

    # Query on the Right tree
    op2 = query(tree, 2 * treeNode + 1, mid + 1,
                            end, left, right)
    return op1 + op2

# Driver code
if __name__ == '__main__':

    n = 5
    arr = [1, 2, 1, 4, 2]
    next_right = [0] * n

    # Initialising the tree
    tree = [[] for i in range(4 * n)]

    ump = dict()

    # Construction of next_right
    # array to store the
    # next index of occurrence
    # of elements
    for i in range(n - 1, -1, -1):
        if arr[i] not in ump:
            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)

    ans = 0
    # Queries one based indexing
    # Time complexity of each
    # query is log(N)

    # first query
    left1 = 0
    right1 = 2
    ans = query(tree, 1, 0, n - 1,
                          left1, right1)
    print(ans)

    # Second Query
    left2 = 1
    right2 = 4
    ans = query(tree, 1, 0, n - 1,
                          left2, right2)
    print(ans)

C#

using System;
using System.Collections.Generic;
using System.Linq;

public class GFG {

  // Function to merge the right
  // and the left tree
  static void merge(List<int>[] tree, int treeNode) {
    int len1 = tree[2 * treeNode].Count();
    int len2 = tree[2 * treeNode + 1].Count();
    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].Add(tree[2 * treeNode + 1][index2]);
        index2++;
      } else {
        tree[treeNode].Add(tree[2 * treeNode][index1]);
        index1++;
      }
    }

    // Insert the leftover elements
    // from the left part
    while (index1 < len1) {
      tree[treeNode].Add(tree[2 * treeNode][index1]);
      index1++;
    }

    // Insert the leftover elements
    // from the right part
    while (index2 < len2) {
      tree[treeNode].Add(tree[2 * treeNode + 1][index2]);
      index2++;
    }
  }

  // Recursive function to build
  // segment tree by merging the
  // sorted segments in sorted way
  static void build(List<int>[] tree, int[] arr, int start, int end, int treeNode) {
    // Base case
    if (start == end) {
      tree[treeNode].Add(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);
  }

  // Function similar to query() method
  // as in segment tree
  static int query(List<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].Count() - tree[treeNode].BinarySearch(right, Comparer<int>.Create((x, y) => x.CompareTo(y)));
    }

    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)/2+1);
  }

  // Driver Code
  static void Main(string[] args) {
    int n = 5;
    int[] arr = new int[] {1, 2, 1, 4, 2};


    int[] next_right = new int[n];
    // Initialising the tree
    List<int>[] tree = new List<int>[4 * n];
    for (int i = 0; i < tree.Length; i++)
    {
      tree[i] = new List<int>();
    }
    Dictionary<int, int> ump = new Dictionary<int, int>();

    // Construction of next_right
    // array to store the
    // next index of occurrence
    // of elements
    for (int i = n - 1; i >= 0; i--) {
      if (!ump.ContainsKey(arr[i])) {
        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);
    Console.WriteLine(ans);

    // Second Query
    int left2 = 1;
    int right2 = 4;
    ans = query(tree, 1, 0, n - 1, left2, right2);
    Console.WriteLine(ans);
  }
}

Output:

2
3

Time Complexity: O(Q*log N)

Space complexity: The space complexity of the above algorithm is O(N), which is used to store the segment tree.