Queries for number of distinct elements in a subarray | Set 2
Given an array arr[] of N integers and Q queries. Each query can be represented by two integers L and R. The task is to find the count of distinct integers in the subarray arr[L] to arr[R].
Examples:
Input: arr[] = {1, 1, 3, 3, 5, 5, 7, 7, 9, 9 }, L = 0, R = 4
Output: 3
Input: arr[] = { 1, 1, 2, 1, 3 }, L = 1, R = 3
Output : 2
Naive approach: In this approach, we will traverse through the range l, r, and use a set to find all the distinct elements in the range and print them.
Time Complexity: O(q*n)
Efficient approach: Idea is to form a segment tree in which the nodes will store all the distinct elements in the range. For this purpose, we can use a self-balancing BST or “set” data structure in C++.
Each query will return the size of the set.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Each segment of the segment tree would be a set// to maintain distinct elementsset<int>* segment;// Build the segment tree// i denotes current node, s denotes start and// e denotes the end of range for current nodevoid build(int i, int s, int e, int arr[]){ // If start is equal to end then // insert the array element if (s == e) { segment[i].insert(arr[s]); return; } // Else divide the range into two halves // (start to mid) and (mid+1 to end) // first half will be the left node // and the second half will be the right node build(2 * i, s, (s + e) / 2, arr); build(1 + 2 * i, 1 + (s + e) / 2, e, arr); // Insert the sets of right and left // node of the segment tree segment[i].insert(segment[2 * i].begin(), segment[2 * i].end()); segment[i].insert(segment[2 * i + 1].begin(), segment[2 * i + 1].end());}// Query in an range a to bset<int> query(int node, int l, int r, int a, int b){ set<int> left, right, result; // If the range is out of the bounds // of this segment if (b < l || a > r) return result; // If the range lies in this segment if (a <= l && r <= b) return segment[node]; // Else query for the right and left // leaf node of this subtree // and insert them into the set left = query(2 * node, l, (l + r) / 2, a, b); result.insert(left.begin(), left.end()); right = query(1 + 2 * node, 1 + (l + r) / 2, r, a, b); result.insert(right.begin(), right.end()); // Return the result return result;}// Initialize the segment treevoid init(int n){ // Get the height of the segment tree int h = (int)ceil(log2(n)); h = (2 * (pow(2, h))) - 1; // Initialize the segment tree segment = new set<int>[h];}// Function to get the result for the// subarray from arr[l] to arr[r]void getDistinct(int l, int r, int n){ // Query for the range set set<int> ans = query(1, 0, n - 1, l, r); cout << ans.size() << endl;}// Driver codeint main(){ int arr[] = { 1, 1, 2, 1, 3 }; int n = sizeof(arr) / sizeof(arr[0]); init(n); // Bulid the segment tree build(1, 0, n - 1, arr); // Query in range 0 to 4 getDistinct(0, 4, n); return 0;} |
Java
// Java implementation of above approachimport java.io.*;import java.util.*;class GFG{ // Each segment of the segment tree would be a set // to maintain distinct elements static HashSet<Integer>[] segment; // Build the segment tree // i denotes current node, s denotes start and // e denotes the end of range for current node static void build(int i, int s, int e, int[] arr) { // If start is equal to end then // insert the array element if (s == e) { segment[i].add(arr[s]); return; } // Else divide the range into two halves // (start to mid) and (mid+1 to end) // first half will be the left node // and the second half will be the right node build(2 * i, s, (s + e) / 2, arr); build(1 + 2 * i, 1 + (s + e) / 2, e, arr); // Insert the sets of right and left // node of the segment tree segment[i].addAll(segment[2 * i]); segment[i].addAll(segment[2 * i + 1]); } // Query in an range a to b static HashSet<Integer> query(int node, int l, int r, int a, int b) { HashSet<Integer> left = new HashSet<>(); HashSet<Integer> right = new HashSet<>(); HashSet<Integer> result = new HashSet<>(); // If the range is out of the bounds // of this segment if (b < l || a > r) return result; // If the range lies in this segment if (a <= l && r <= b) return segment[node]; // Else query for the right and left // leaf node of this subtree // and insert them into the set left = query(2 * node, l, (l + r) / 2, a, b); result.addAll(left); right = query(1 + 2 * node, 1 + (l + r) / 2, r, a, b); result.addAll(right); // Return the result return result; } // Initialize the segment tree @SuppressWarnings("unchecked") static void init(int n) { // Get the height of the segment tree int h = (int) Math.ceil(Math.log(n) / Math.log(2)); h = (int) (2 * Math.pow(2, h)) - 1; // Initialize the segment tree segment = new HashSet[h]; for (int i = 0; i < h; i++) segment[i] = new HashSet<>(); } // Function to get the result for the // subarray from arr[l] to arr[r] static void getDistinct(int l, int r, int n) { // Query for the range set HashSet<Integer> ans = query(1, 0, n - 1, l, r); System.out.println(ans.size()); } // Driver Code public static void main(String[] args) { int[] arr = { 1, 1, 2, 1, 3 }; int n = arr.length; init(n); // Bulid the segment tree build(1, 0, n - 1, arr); // Query in range 0 to 4 getDistinct(0, 4, n); }}// This code is contributed by// sanjeev2552 |
Python3
# python3 implementation of above approachfrom math import ceil,log,floor# Each segment of the segment tree would be a set# to maintain distinct elementssegment=[[] for i in range(1000)]# Build the segment tree# i denotes current node, s denotes start and# e denotes the end of range for current nodedef build(i, s, e, arr): # If start is equal to end then # append the array element if (s == e): segment[i].append(arr[s]) return # Else divide the range into two halves # (start to mid) and (mid+1 to end) # first half will be the left node # and the second half will be the right node build(2 * i, s, (s + e) // 2, arr) build(1 + 2 * i, 1 + (s + e) // 2, e, arr) # Insert the sets of right and left # node of the segment tree segment[i].append(segment[2 * i]) segment[i].append(segment[2 * i + 1])# Query in an range a to bdef query(node, l, r, a, b): left, right, result=[],[],[] # If the range is out of the bounds # of this segment if (b < l or a > r): return result # If the range lies in this segment if (a <= l and r <= b): return segment[node] # Else query for the right and left # leaf node of this subtree # and append them into the set left = query(2 * node, l, (l + r) // 2, a, b) result.append(left) right = query(1 + 2 * node, 1 + (l + r) // 2, r, a, b) result.append(right) # Return the result return resultdef answer(ans): d = {} for i in str(ans): if i not in ['[',',',']',' ']: d[i]=1 return len(d)# Initialize the segment treedef init(n): # Get the height of the segment tree h = ceil(log(n, 2)) h = (2 * (pow(2, h))) - 1# Function to get the result for the# subarray from arr[l] to arr[r]def getDistinct(l, r, n): # Query for the range set ans = query(1, 0, n - 1, l, r) print(answer(str(ans)))# Driver codearr=[1, 1, 2, 1, 3]n = len(arr)init(n)# Bulid the segment treebuild(1, 0, n - 1, arr)# Query in range 0 to 4getDistinct(0, 4, n)# This code is contributed by mohit kumar 29 |
C#
// C# implementation of// the above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{ // Each segment of the segment// tree would be a set to maintain// distinct elementsstatic HashSet<int>[] segment; // Build the segment tree// i denotes current node,// s denotes start and// e denotes the end of// range for current nodestatic void build(int i, int s, int e, int[] arr){ // If start is equal to end then // insert the array element if (s == e) { segment[i].Add(arr[s]); return; } // Else divide the range // into two halves (start // to mid) and (mid+1 to end) // first half will be the left // node and the second half // will be the right node build(2 * i, s, (s + e) / 2, arr); build(1 + 2 * i, 1 + (s + e) / 2, e, arr); // Insert the sets of // right and left node // of the segment tree foreach(int x in segment[2 * i]) { segment[i].Add(x); } foreach(int x in segment[2 * i + 1]) { segment[i].Add(x); }} // Query in an range a to bstatic HashSet<int> query(int node, int l, int r, int a, int b){ HashSet<int> left = new HashSet<int>(); HashSet<int> right = new HashSet<int>(); HashSet<int> result = new HashSet<int>(); // If the range is out // of the bounds // of this segment if (b < l || a > r) return result; // If the range lies // in this segment if (a <= l && r <= b) return segment[node]; // Else query for the right and left // leaf node of this subtree // and insert them into the set left = query(2 * node, l, (l + r) / 2, a, b); foreach(int x in left) { result.Add(x); } right = query(1 + 2 * node, 1 + (l + r) / 2, r, a, b); foreach(int x in right) { result.Add(x); } // Return the result return result;}// Initialize the segment treestatic void init(int n){ // Get the height of the segment tree int h = (int) Math.Ceiling(Math.Log(n) / Math.Log(2)); h = (int) (2 * Math.Pow(2, h)) - 1; // Initialize the segment tree segment = new HashSet<int>[h]; for (int i = 0; i < h; i++) segment[i] = new HashSet<int>();}// Function to get the result for the// subarray from arr[l] to arr[r]static void getDistinct(int l, int r, int n){ // Query for the range set HashSet<int> ans = query(1, 0, n - 1, l, r); Console.Write(ans.Count);}// Driver Codepublic static void Main(string[] args){ int[] arr = {1, 1, 2, 1, 3}; int n = arr.Length; init(n); // Bulid the segment tree build(1, 0, n - 1, arr); // Query in range 0 to 4 getDistinct(0, 4, n);}}// This code is contributed by rutvik_56 |
Output:
3
Time Complexity: O(q*n*log(n))
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