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Count distinct elements from a range of a sorted sequence from a given frequency array

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  • Difficulty Level : Expert
  • Last Updated : 28 Jul, 2022

Given two integers L and R and an array arr[] consisting of N positive integers( 1-based indexing ) such that the frequency of ith element of a sorted sequence, say A[], is arr[i]. The task is to find the number of distinct elements from the range [L, R] in the sequence A[].

Examples:

Input: arr[] = {3, 6, 7, 1, 8}, L = 3, R = 7
Output: 2
Explanation: From the given frequency array, the sorted array will be {1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, ….}. Now, the number of distinct elements from the range [3, 7] is 2( = {1, 2}).

Input: arr[] = {1, 2, 3, 4}, L = 3, R = 4
Output: 1

 

Naive Approach: The simplest approach to solve the given problem is to construct the sorted sequence from the given array arr[] using the given frequencies and then traverse the constructed array over the range [L, R] to count the number of distinct elements.

Time Complexity: O(N + R – L)
Auxiliary Space: O(S), where S is the sum of the array elements.

Efficient Approach: The above approach can be optimized by using the Binary Search and the prefix sum technique to find the number of distinct elements over the range [L, R]. Follow the steps below to solve the given problem:

  • Initialize an auxiliary array, say prefix[] that stores the prefix sum of the given array elements.
  • Find the prefix sum of the given array and stored it in the array prefix[].
  • By using binary search, find the first index at which the value in prefix[] is at least L, say left.
  • By using binary search, find the first index at which the value in prefix[] is at least R, say right.
  • After completing the above steps, print the value of (right – left + 1) as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to find the first index
// with value is at least element
int binarysearch(int array[], int right,
                 int element)
{
     
    // Update the value of left
    int left = 1;
 
    // Update the value of right
      
    // Binary search for the element
    while (left <= right)
    {
         
        // Find the middle element
        int mid = (left + right / 2);
 
        if (array[mid] == element)
        {
            return mid;
        }
 
        // Check if the value lies
        // between the elements at
        // index mid - 1 and mid
        if (mid - 1 > 0 && array[mid] > element &&
           array[mid - 1] < element)
        {
            return mid;
        }
 
        // Check in the right subarray
        else if (array[mid] < element)
        {
             
            // Update the value
            // of left
            left = mid + 1;
        }
 
        // Check in left subarray
        else
        {
             
            // Update the value of
            // right
            right = mid - 1;
        }
    }
    return 1;
}
 
// Function to count the number of
// distinct elements over the range
// [L, R] in the sorted sequence
void countDistinct(vector<int> arr,
                          int L, int R)
{
     
    // Stores the count of distinct
    // elements
    int count = 0;
 
    // Create the prefix sum array
    int pref[arr.size() + 1];
 
    for(int i = 1; i <= arr.size(); ++i)
    {
         
        // Update the value of count
        count += arr[i - 1];
 
        // Update the value of pref[i]
        pref[i] = count;
    }
 
    // Calculating the first index
    // of L and R using binary search
    int left = binarysearch(pref, arr.size() + 1, L);
    int right = binarysearch(pref, arr.size() + 1, R);
 
    // Print the resultant count
    cout << right - left + 1;
}
 
// Driver Code
int main()
{
    vector<int> arr{ 3, 6, 7, 1, 8 };
    int L = 3;
    int R = 7;
     
    countDistinct(arr, L, R);
}
 
// This code is contributed by ipg2016107


Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to find the first index
    // with value is at least element
    static int binarysearch(int array[],
                            int element)
    {
        // Update the value of left
        int left = 1;
 
        // Update the value of right
        int right = array.length - 1;
 
        // Binary search for the element
        while (left <= right) {
 
            // Find the middle element
            int mid = (int)(left + right / 2);
 
            if (array[mid] == element) {
                return mid;
            }
 
            // Check if the value lies
            // between the elements at
            // index mid - 1 and mid
            if (mid - 1 > 0
                && array[mid] > element
                && array[mid - 1] < element) {
 
                return mid;
            }
 
            // Check in the right subarray
            else if (array[mid] < element) {
 
                // Update the value
                // of left
                left = mid + 1;
            }
 
            // Check in left subarray
            else {
 
                // Update the value of
                // right
                right = mid - 1;
            }
        }
 
        return 1;
    }
 
    // Function to count the number of
    // distinct elements over the range
    // [L, R] in the sorted sequence
    static void countDistinct(int arr[],
                              int L, int R)
    {
        // Stores the count of distinct
        // elements
        int count = 0;
 
        // Create the prefix sum array
        int pref[] = new int[arr.length + 1];
 
        for (int i = 1; i <= arr.length; ++i) {
 
            // Update the value of count
            count += arr[i - 1];
 
            // Update the value of pref[i]
            pref[i] = count;
        }
 
        // Calculating the first index
        // of L and R using binary search
        int left = binarysearch(pref, L);
        int right = binarysearch(pref, R);
 
        // Print the resultant count
        System.out.println(
            (right - left) + 1);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 3, 6, 7, 1, 8 };
        int L = 3;
        int R = 7;
        countDistinct(arr, L, R);
    }
}


Python3




# Python3 program for the above approach
 
# Function to find the first index
# with value is at least element
def binarysearch(array,  right,
                 element):
 
    # Update the value of left
    left = 1
 
    # Update the value of right
 
    # Binary search for the element
    while (left <= right):
 
        # Find the middle element
        mid = (left + right // 2)
 
        if (array[mid] == element):
            return mid
 
        # Check if the value lies
        # between the elements at
        # index mid - 1 and mid
        if (mid - 1 > 0 and array[mid] > element and
                        array[mid - 1] < element):
            return mid
 
        # Check in the right subarray
        elif (array[mid] < element):
 
            # Update the value
            # of left
            left = mid + 1
 
        # Check in left subarray
        else:
 
            # Update the value of
            # right
            right = mid - 1
 
    return 1
 
# Function to count the number of
# distinct elements over the range
# [L, R] in the sorted sequence
def countDistinct(arr, L, R):
 
    # Stores the count of distinct
    # elements
    count = 0
 
    # Create the prefix sum array
    pref = [0] * (len(arr) + 1)
 
    for i in range(1, len(arr) + 1):
 
        # Update the value of count
        count += arr[i - 1]
 
        # Update the value of pref[i]
        pref[i] = count
 
    # Calculating the first index
    # of L and R using binary search
    left = binarysearch(pref, len(arr) + 1, L)
    right = binarysearch(pref, len(arr) + 1, R)
 
    # Print the resultant count
    print(right - left + 1)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ 3, 6, 7, 1, 8 ]
    L = 3
    R = 7
 
    countDistinct(arr, L, R)
 
# This code is contributed by ukasp


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the first index
// with value is at least element
static int binarysearch(int []array, int right,
                        int element)
{
     
    // Update the value of left
    int left = 1;
 
    // Update the value of right
      
    // Binary search for the element
    while (left <= right)
    {
         
        // Find the middle element
        int mid = (left + right / 2);
 
        if (array[mid] == element)
        {
            return mid;
        }
 
        // Check if the value lies
        // between the elements at
        // index mid - 1 and mid
        if (mid - 1 > 0 && array[mid] > element &&
            array[mid - 1] < element)
        {
            return mid;
        }
 
        // Check in the right subarray
        else if (array[mid] < element)
        {
             
            // Update the value
            // of left
            left = mid + 1;
        }
 
        // Check in left subarray
        else
        {
             
            // Update the value of
            // right
            right = mid - 1;
        }
    }
    return 1;
}
 
// Function to count the number of
// distinct elements over the range
// [L, R] in the sorted sequence
static void countDistinct(List<int> arr,
                          int L, int R)
{
     
    // Stores the count of distinct
    // elements
    int count = 0;
 
    // Create the prefix sum array
    int []pref = new int[arr.Count + 1];
 
    for(int i = 1; i <= arr.Count; ++i)
    {
         
        // Update the value of count
        count += arr[i - 1];
 
        // Update the value of pref[i]
        pref[i] = count;
    }
 
    // Calculating the first index
    // of L and R using binary search
    int left = binarysearch(pref, arr.Count + 1, L);
    int right = binarysearch(pref, arr.Count + 1, R);
 
    // Print the resultant count
    Console.Write(right - left + 1);
}
 
// Driver Code
public static void Main()
{
    List<int> arr = new List<int>(){ 3, 6, 7, 1, 8 };
    int L = 3;
    int R = 7;
     
    countDistinct(arr, L, R);
}
}
 
// This code is contributed by SURENDRA_GANGWAR


Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find the first index
// with value is at least element
function binarysearch(array, right, element)
{
     
    // Update the value of left
    let left = 1;
  
    // Update the value of right
       
    // Binary search for the element
    while (left <= right)
    {
         
        // Find the middle element
        let mid = Math.floor((left + right / 2));
  
        if (array[mid] == element)
        {
            return mid;
        }
  
        // Check if the value lies
        // between the elements at
        // index mid - 1 and mid
        if (mid - 1 > 0 && array[mid] > element &&
            array[mid - 1] < element)
        {
            return mid;
        }
  
        // Check in the right subarray
        else if (array[mid] < element)
        {
              
            // Update the value
            // of left
            left = mid + 1;
        }
  
        // Check in left subarray
        else
        {
              
            // Update the value of
            // right
            right = mid - 1;
        }
    }
    return 1;
}
  
// Function to count the number of
// distinct elements over the range
// [L, R] in the sorted sequence
function countDistinct(arr, L, R)
{
     
    // Stores the count of distinct
    // elements
    let count = 0;
  
    // Create the prefix sum array
    let pref = Array.from(
        {length: arr.length + 1}, (_, i) => 0);
  
    for(let i = 1; i <= arr.length; ++i)
    {
         
        // Update the value of count
        count += arr[i - 1];
  
        // Update the value of pref[i]
        pref[i] = count;
    }
  
    // Calculating the first index
    // of L and R using binary search
    let left = binarysearch(pref, arr.length + 1, L);
    let right = binarysearch(pref, arr.length + 1, R);
  
        // Print the resultant count
        document.write((right - left) + 1);
}
 
// Driver Code
let arr = [ 3, 6, 7, 1, 8 ];
let L = 3;
let R = 7;
 
countDistinct(arr, L, R);
 
// This code is contributed by susmitakundugoaldanga
 
</script>


Output: 

2

 

Time Complexity: O(log(N))
Auxiliary Space: O(N)


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