Count distinct elements from a range of a sorted sequence from a given frequency array
Last Updated :
24 Aug, 2023
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: 2
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.
Code-
C++
#include<bits/stdc++.h>
using namespace std;
void countDistinct(vector< int > arr,
int L, int R)
{
vector< int > vec;
for ( int i=0;i<arr.size();i++){
int temp=arr[i];
for ( int j=0;j<temp;j++){
vec.push_back(i+1);
}
}
int curr=INT_MIN;
int count=0;
for ( int i=L-1;i<R;i++){
if (curr!=vec[i]){
count++;
curr=vec[i];
}
}
cout<<count<<endl;
}
int main()
{
vector< int > arr{ 3, 6, 7, 1, 8 };
int L = 3;
int R = 7;
countDistinct(arr, L, R);
}
|
Java
import java.util.*;
class GFG {
static void countDistinct(ArrayList<Integer> arr, int L,
int R)
{
ArrayList<Integer> vec = new ArrayList<Integer>();
for ( int i = 0 ; i < arr.size(); i++) {
int temp = arr.get(i);
for ( int j = 0 ; j < temp; j++) {
vec.add(i + 1 );
}
}
int curr = Integer.MIN_VALUE;
int count = 0 ;
for ( int i = L - 1 ; i < R; i++) {
if (curr != vec.get(i)) {
count++;
curr = vec.get(i);
}
}
System.out.println(count);
}
public static void main(String[] args)
{
ArrayList<Integer> arr = new ArrayList<Integer>(
Arrays.asList( 3 , 6 , 7 , 1 , 8 ));
int L = 3 ;
int R = 7 ;
countDistinct(arr, L, R);
}
}
|
Python3
def countDistinct(arr, L, R):
vec = []
for i in range ( len (arr)):
temp = arr[i]
for j in range (temp):
vec.append(i + 1 )
curr = float ( '-inf' )
count = 0
for i in range (L - 1 , R):
if curr ! = vec[i]:
count + = 1
curr = vec[i]
print (count)
if __name__ = = '__main__' :
arr = [ 3 , 6 , 7 , 1 , 8 ]
L = 3
R = 7
countDistinct(arr, L, R)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void CountDistinct(List< int > arr, int L, int R)
{
List< int > vec = new List< int >();
foreach ( int temp in arr)
{
for ( int j = 0; j < temp; j++)
{
vec.Add(arr.IndexOf(temp) + 1);
}
}
int curr = int .MinValue;
int count = 0;
for ( int i = L - 1; i < R; i++)
{
if (curr != vec[i])
{
count++;
curr = vec[i];
}
}
Console.WriteLine(count);
}
static void Main()
{
List< int > arr = new List< int > { 3, 6, 7, 1, 8 };
int L = 3;
int R = 7;
CountDistinct(arr, L, R);
}
}
|
Javascript
function countDistinct(arr, L, R) {
let vec = [];
for (let i = 0; i < arr.length; i++) {
let temp = arr[i];
for (let j = 0; j < temp; j++) {
vec.push(i + 1);
}
}
let curr = Number.MIN_SAFE_INTEGER;
let count = 0;
for (let i = L - 1; i < R; i++) {
if (curr !== vec[i]) {
count++;
curr = vec[i];
}
}
console.log(count);
}
let arr = [3, 6, 7, 1, 8];
let L = 3;
let R = 7;
countDistinct(arr, L, R);
|
Time Complexity: O(S + 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++
#include<bits/stdc++.h>
using namespace std;
int binarysearch( int array[], int right,
int element)
{
int left = 1;
while (left <= right)
{
int mid = (left + right / 2);
if (array[mid] == element)
{
return mid;
}
if (mid - 1 > 0 && array[mid] > element &&
array[mid - 1] < element)
{
return mid;
}
else if (array[mid] < element)
{
left = mid + 1;
}
else
{
right = mid - 1;
}
}
return 1;
}
void countDistinct(vector< int > arr,
int L, int R)
{
int count = 0;
int pref[arr.size() + 1];
for ( int i = 1; i <= arr.size(); ++i)
{
count += arr[i - 1];
pref[i] = count;
}
int left = binarysearch(pref, arr.size() + 1, L);
int right = binarysearch(pref, arr.size() + 1, R);
cout << right - left + 1;
}
int main()
{
vector< int > arr{ 3, 6, 7, 1, 8 };
int L = 3;
int R = 7;
countDistinct(arr, L, R);
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static int binarysearch( int array[],
int element)
{
int left = 1 ;
int right = array.length - 1 ;
while (left <= right) {
int mid = ( int )(left + right / 2 );
if (array[mid] == element) {
return mid;
}
if (mid - 1 > 0
&& array[mid] > element
&& array[mid - 1 ] < element) {
return mid;
}
else if (array[mid] < element) {
left = mid + 1 ;
}
else {
right = mid - 1 ;
}
}
return 1 ;
}
static void countDistinct( int arr[],
int L, int R)
{
int count = 0 ;
int pref[] = new int [arr.length + 1 ];
for ( int i = 1 ; i <= arr.length; ++i) {
count += arr[i - 1 ];
pref[i] = count;
}
int left = binarysearch(pref, L);
int right = binarysearch(pref, R);
System.out.println(
(right - left) + 1 );
}
public static void main(String[] args)
{
int arr[] = { 3 , 6 , 7 , 1 , 8 };
int L = 3 ;
int R = 7 ;
countDistinct(arr, L, R);
}
}
|
Python3
def binarysearch(array, right,
element):
left = 1
while (left < = right):
mid = (left + right / / 2 )
if (array[mid] = = element):
return mid
if (mid - 1 > 0 and array[mid] > element and
array[mid - 1 ] < element):
return mid
elif (array[mid] < element):
left = mid + 1
else :
right = mid - 1
return 1
def countDistinct(arr, L, R):
count = 0
pref = [ 0 ] * ( len (arr) + 1 )
for i in range ( 1 , len (arr) + 1 ):
count + = arr[i - 1 ]
pref[i] = count
left = binarysearch(pref, len (arr) + 1 , L)
right = binarysearch(pref, len (arr) + 1 , R)
print (right - left + 1 )
if __name__ = = "__main__" :
arr = [ 3 , 6 , 7 , 1 , 8 ]
L = 3
R = 7
countDistinct(arr, L, R)
|
C#
using System;
using System.Collections.Generic;
class GFG{
static int binarysearch( int []array, int right,
int element)
{
int left = 1;
while (left <= right)
{
int mid = (left + right / 2);
if (array[mid] == element)
{
return mid;
}
if (mid - 1 > 0 && array[mid] > element &&
array[mid - 1] < element)
{
return mid;
}
else if (array[mid] < element)
{
left = mid + 1;
}
else
{
right = mid - 1;
}
}
return 1;
}
static void countDistinct(List< int > arr,
int L, int R)
{
int count = 0;
int []pref = new int [arr.Count + 1];
for ( int i = 1; i <= arr.Count; ++i)
{
count += arr[i - 1];
pref[i] = count;
}
int left = binarysearch(pref, arr.Count + 1, L);
int right = binarysearch(pref, arr.Count + 1, R);
Console.Write(right - left + 1);
}
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);
}
}
|
Javascript
<script>
function binarysearch(array, right, element)
{
let left = 1;
while (left <= right)
{
let mid = Math.floor((left + right / 2));
if (array[mid] == element)
{
return mid;
}
if (mid - 1 > 0 && array[mid] > element &&
array[mid - 1] < element)
{
return mid;
}
else if (array[mid] < element)
{
left = mid + 1;
}
else
{
right = mid - 1;
}
}
return 1;
}
function countDistinct(arr, L, R)
{
let count = 0;
let pref = Array.from(
{length: arr.length + 1}, (_, i) => 0);
for (let i = 1; i <= arr.length; ++i)
{
count += arr[i - 1];
pref[i] = count;
}
let left = binarysearch(pref, arr.length + 1, L);
let right = binarysearch(pref, arr.length + 1, R);
document.write((right - left) + 1);
}
let arr = [ 3, 6, 7, 1, 8 ];
let L = 3;
let R = 7;
countDistinct(arr, L, R);
</script>
|
Time Complexity: O(log(N))
Auxiliary Space: O(N)
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