Absolute distinct count in a sorted array
Last Updated :
26 Mar, 2023
Given a sorted array of integers, return the number of distinct absolute values among the elements of the array. The input can contain duplicates values.
Examples:
Input: [-3, -2, 0, 3, 4, 5]
Output: 5
There are 5 distinct absolute values
among the elements of this array, i.e.
0, 2, 3, 4 and 5)
Input: [-1, -1, -1, -1, 0, 1, 1, 1, 1]
Output: 2
Input: [-1, -1, -1, -1, 0]
Output: 2
Input: [0, 0, 0]
Output: 1
The solution should do only one scan of the input array and should not use any extra space. i.e. expected time complexity is O(n) and auxiliary space is O(1).
One simple solution is to use set. For each element of the input array, we insert its absolute value in the set. As set doesn’t support duplicate elements, the element’s absolute value will be inserted only once. Therefore, the required count is size of the set.
Algorithm:
Step 1: Define a function named distinctCount which takes two parameters, an array of integers arr[] and an integer n.
Step 2: Declare an empty HashSet named s.
Step 3: Traverse the array arr[] using a for loop from i=0 to i<n.
Step 4: Insert the absolute value of arr[i] in the set s using the add() function.
Step 5: After the for loop, return the size of set s using the size() function.
Below is the implementation of the idea.
C++
#include <bits/stdc++.h>
using namespace std;
int distinctCount(int arr[], int n)
{
unordered_set<int> s;
for (int i = 0 ; i < n; i++)
s.insert(abs(arr[i]));
return s.size();
}
int main()
{
int arr[] = {-2, -1, 0, 1, 1};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Count of absolute distinct values : "
<< distinctCount(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int distinctCount(int arr[], int n)
{
Set<Integer> s = new HashSet<Integer> ();
for (int i = 0 ; i < n; i++)
s.add(Math.abs(arr[i]));
return s.size();
}
public static void main(String[] args)
{
int arr[] = {-2, -1, 0, 1, 1};
int n = arr.length;
System.out.println("Count of absolute distinct values : "
+ distinctCount(arr, n));
}
}
|
Python3
def distinctCount(arr, n):
s = set()
for i in range(n):
s.add(abs(arr[i]))
return len(s)
arr = [-2, -1, 0, 1, 1]
n = len(arr)
print("Count of absolute distinct values:",
distinctCount(arr, n))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int distinctCount(int []arr, int n)
{
HashSet<int> s = new HashSet<int>();
for (int i = 0 ; i < n; i++)
s.Add(Math.Abs(arr[i]));
return s.Count;
}
public static void Main()
{
int []arr = {-2, -1, 0, 1, 1};
int n = arr.Length;
Console.Write("Count of absolute distinct values : "
+ distinctCount(arr, n));
}
}
|
Javascript
<script>
function distinctCount(arr, n)
{
let s = new Set();
for (let i = 0 ; i < n; i++)
s.add(Math.abs(arr[i]));
return s.size;
}
let arr = [-2, -1, 0, 1, 1];
let n = arr.length;
document.write("Count of absolute distinct values : "
+ distinctCount(arr, n));
</script>
|
Output
Count of absolute distinct values : 3
Time Complexity : O(n)
Auxiliary Space : O(n)
The above implementation takes O(n) extra space, how to do in O(1) extra space?
The idea is to take advantage of the fact that the array is already Sorted. We initialize the count of distinct elements to number of elements in the array. We start with two index variables from two corners of the array and check for pair in the input array with sum as 0. If pair with 0 sum is found or duplicates are encountered, we decrement the count of distinct elements.Finally we return the updated count.
Below is the implementation of above approach.
C++
#include <bits/stdc++.h>
using namespace std;
int distinctCount(int arr[], int n)
{
int count = n;
int i = 0, j = n - 1, sum = 0;
while (i < j)
{
while (i != j && arr[i] == arr[i + 1])
count--, i++;
while (i != j && arr[j] == arr[j - 1])
count--, j--;
if (i == j)
break;
sum = arr[i] + arr[j];
if (sum == 0)
{
count--;
i++, j--;
}
else if(sum < 0)
i++;
else
j--;
}
return count;
}
int main()
{
int arr[] = {-2, -1, 0, 1, 1};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Count of absolute distinct values : "
<< distinctCount(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int distinctCount(int arr[], int n)
{
int count = n;
int i = 0, j = n - 1, sum = 0;
while (i < j)
{
while (i != j && arr[i] == arr[i + 1])
{
count--;
i++;
}
while (i != j && arr[j] == arr[j - 1])
{
count--;
j--;
}
if (i == j)
break;
sum = arr[i] + arr[j];
if (sum == 0)
{
count--;
i++;
j--;
}
else if(sum < 0)
i++;
else
j--;
}
return count;
}
public static void main (String[] args) {
int arr[] = {-2, -1, 0, 1, 1};
int n = arr.length;
System.out.println ("Count of absolute distinct values : "+
distinctCount(arr, n));
}
}
|
Python3
def distinctCount(arr, n):
count = n;
i = 0; j = n - 1; sum = 0;
while (i < j):
while (i != j and arr[i] == arr[i + 1]):
count = count - 1;
i = i + 1;
while (i != j and arr[j] == arr[j - 1]):
count = count - 1;
j = j - 1;
if (i == j):
break;
sum = arr[i] + arr[j];
if (sum == 0):
count = count - 1;
i = i + 1;
j = j - 1;
elif(sum < 0):
i = i + 1;
else:
j = j - 1;
return count;
arr = [-2, -1, 0, 1, 1];
n = len(arr);
print("Count of absolute distinct values : ",
distinctCount(arr, n));
|
C#
using System;
class GFG {
static int distinctCount(int []arr, int n)
{
int count = n;
int i = 0, j = n - 1, sum = 0;
while (i < j)
{
while (i != j && arr[i] == arr[i + 1])
{
count--;
i++;
}
while (i != j && arr[j] == arr[j - 1])
{
count--;
j--;
}
if (i == j)
break;
sum = arr[i] + arr[j];
if (sum == 0)
{
count--;
i++;
j--;
}
else if(sum < 0)
i++;
else
j--;
}
return count;
}
public static void Main () {
int []arr = {-2, -1, 0, 1, 1};
int n = arr.Length;
Console.WriteLine("Count of absolute distinct values : "+
distinctCount(arr, n));
}
}
|
PHP
<?php
function distinctCount($arr, $n)
{
$count = $n;
$i = 0; $j = $n - 1; $sum = 0;
while ($i < $j)
{
while ($i != $j && $arr[$i] == $arr[$i + 1])
{$count--; $i++;}
while ($i != $j && $arr[$j] == $arr[$j - 1])
{$count--; $j--;}
if ($i == $j)
break;
$sum = $arr[$i] + $arr[$j];
if ($sum == 0)
{
$count--;
$i++; $j--;
}
else if($sum < 0)
$i++;
else
$j--;
}
return $count;
}
$arr = array(-2, -1, 0, 1, 1);
$n = sizeof($arr);
echo "Count of absolute distinct values : " .
distinctCount($arr, $n);
?>
|
Javascript
<script>
function distinctCount(arr, n)
{
let count = n;
let i = 0, j = n - 1, sum = 0;
while (i < j)
{
while (i != j && arr[i] == arr[i + 1])
count--, i++;
while (i != j && arr[j] == arr[j - 1])
count--, j--;
if (i == j)
break;
sum = arr[i] + arr[j];
if (sum == 0)
{
count--;
i++, j--;
}
else if(sum < 0)
i++;
else
j--;
}
return count;
}
let arr = [-2, -1, 0, 1, 1];
let n = arr.length;
document.write(
"Count of absolute distinct values : " + distinctCount(arr, n));
</script>
|
Output
Count of absolute distinct values : 3
Time Complexity: O(n)
Auxiliary Space: O(1)
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